Correction to: The number of different true permutation polynomial based interleavers under Zhao and Fan sufficient conditions

Lucian Trifina; Daniela Tarniceriu

doi:10.1007/s11235-018-0528-z

Telecommunication Systems

January 2019, Volume 70, Issue 1, pp 141–158 | Cite as

Correction to: The number of different true permutation polynomial based interleavers under Zhao and Fan sufficient conditions

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Lucian Trifina
Daniela Tarniceriu

Correction

First Online: 11 December 2018

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1 Correction to: Telecommun Syst (2016) 63:593–623 https://doi.org/10.1007/s11235-016-0144-8

In the original version of this article [1] unfortunately, there are mistakes in some formulas for determining the number of true different cubic, fourth degree, and fifth degree permutation polynomial based interleavers under Zhao and Fan sufficient conditions. The reason for these mistakes was considering all null polynomials. However under these sufficient conditions, the null polynomials have to be restricted in some cases. These particular cases are indicated, and the corresponding formulas are corrected.

In the original version, we have determined the number of true different cubic, fourth degree, and fifth degree permutation polynomial (PP) based interleavers under Zhao and Fan (ZF) sufficient conditions given in [2]. To determine the number of true different PPs of a certain degree n, we took into account the equivalence conditions for PPs considering null polynomials (NPs) of degree up to n. In [1], we have not considered the fact that by adding a NP to a PP fulfilling ZF sufficient conditions does not always result a PP that also fulfills ZF sufficient conditions. This omission generated a mistake in some formulas for the number of true different PPs of degree 3 to 5 and for the lengths leading this number to be equal to zero. In this paper, we will correct these formulas.

2 Null polynomials under Zhao and Fan sufficient conditions

Let there be a PP of degree n modulo a positive integer N defined as:

$$\begin{aligned}&\pi (x) = q_0 +q_1 \cdot x +q_2 \cdot x^2 + \dots + q_n \cdot x^n \ (\mathrm {mod}\ N), \nonumber \\&x = \overline{0, N-1}, \end{aligned}$$

(1)

and let the prime factorization of N be

$$\begin{aligned} N = \prod _{\begin{array}{c} p\in {\mathcal {P}},\\ p \mid N \end{array}} p^{n_{N,p}}, \end{aligned}$$

(2)

where ${\mathcal {P}}$ is the set of prime numbers, the notation $p \mid N$ means that p divides N and $n_{N,p} \ge 1$ for a finite number of prime numbers.

ZF sufficient conditions on the coefficients of the polynomial in (1) so that it is PP modulo N are given in Table 1, which is similar to Theorem 1 in [2].

Table 1

Zhao and Fan sufficient conditions for coefficients $q_1, q_2, \dots , q_n$ so that $\pi (x)$ in (1) is a PP

1.a	$p = 2$	$n_{N,2}=1$	$(q_1 + q_2 + \dots + q_n) \ne 0 \ (\mathrm {mod}\ 2)$
1.b		$n_{N,2}>1$	$q_1 \ne 0$, $q_2+q_4+q_6+ \dots = 0 \ (\mathrm {mod}\ 2)$ and $q_3+q_5+q_7+ \dots = 0 \ (\mathrm {mod}\ 2)$
2	$p > 2$	$n_{N,p} \ge 1$	$q_1 \ne 0, q_2 = q_3 = \dots = q_n = 0 \ (\mathrm {mod}\ p)$

Let there be

$$\begin{aligned} z(x) = \sum _{k=1}^{n} z_k \cdot x^k \ (\mathrm {mod}\ N) \end{aligned}$$

(3)

a NP of degree n modulo N, i.e., $z(x) = 0, \forall x = \overline{0, N-1}$.

We define a NP valid under ZF sufficient conditions (denoted by ZF NP) a NP z(x) fulfilling the condition that if $\pi (x)$ is PP under ZF sufficient conditions then $\pi (x) +z(x)$ is also a PP under ZF sufficient conditions. From Table 1, we can determine the conditions on the coefficients of a NP valid under ZF sufficient conditions. Since ZF sufficient conditions are also necessary for the prime $p = 2$, we have to take into account only the conditions for primes $p > 2$ (i.e., the last row in Table 1). This means:

$$\begin{aligned}&z_1 \ne (-q_1) \ (\mathrm {mod}\ p), z_2 = z_3 = \cdots = z_n = 0 \ (\mathrm {mod}\ p), \nonumber \\&\quad \forall p \mid N, p> 2. \end{aligned}$$

(4)

We know that the general form of a NP of degree n is [3]:

$$\begin{aligned} \begin{aligned} z(x)&= \sum _{k=1}^{n} \left\{ \frac{N}{\gcd (k!, N)} \cdot \tau _k \cdot \prod _{m=0}^{k-1} (x - m) \right\} \ (\mathrm {mod}\ N), \\&\quad \text {where } 0 \le \tau _k \le \gcd (k!, N) - 1 \end{aligned}\nonumber \\ \end{aligned}$$

(5)

In (5), gcd stands for the greatest common divisor. The expanded form of (5) for $n=5$ was given in [1] as

$$\begin{aligned} \begin{aligned} z_{n=5}(x)&= \frac{N}{\gcd (5!,N)} \cdot \tau _5 \cdot x^5 \\&\quad + \left( \frac{N}{\gcd (4!,N)} \cdot \tau _4 - \frac{N}{\gcd (5!,N)} \cdot 10 \cdot \tau _5 \right) \cdot x^4 \\&\quad + \left( \frac{N}{\gcd (3!,N)} \cdot \tau _3 - \frac{N}{\gcd (4!,N)} \cdot 6 \cdot \tau _4 \right. \\&\quad \qquad \left. + \frac{N}{\gcd (5!,N)} \cdot 35 \cdot \tau _5 \right) \cdot x^3 \\&\quad + \left( \frac{N}{\gcd (2!,N)} \cdot \tau _2 - \frac{N}{\gcd (3!,N)} \cdot 3 \cdot \tau _3 \right. \\&\quad \qquad + \frac{N}{\gcd (4!,N)} \cdot 11 \cdot \tau _4 \\&\quad \qquad \left. - \frac{N}{\gcd (5!,N)} \cdot 50 \cdot \tau _5 \right) \\&\quad \cdot x^2 + \left( \frac{N}{\gcd (3!,N)} \cdot 2 \cdot \tau _3 - \frac{N}{\gcd (2!,N)} \cdot \tau _2 \right. \\&\qquad \qquad - \frac{N}{\gcd (4!,N)} \cdot 6 \cdot \tau _4 \\&\qquad \qquad \left. + \frac{N}{\gcd (5!,N)} \cdot 24 \cdot \tau _5 \right) \cdot x, \\&\quad \text {with}~ 0 \le \tau _2 \le \gcd (2!,N)-1, \\&\quad 0 \le \tau _3 \le \gcd (3!,N)-1, \\&\quad 0 \le \tau _4 \le \gcd (4!,N)-1 ~\text {and} \\&\quad 0 \le \tau _5 \le \gcd (5!,N)-1. \end{aligned} \end{aligned}$$

(6)

For $\tau _5 = 0$ in (6), we obtain the expanded form of a fourth degree NP $z_{n=4}(x) $, and for $\tau _5 = \tau _4 = 0$ in (6), we obtain the expanded form of a third degree NP $z_{n=3}(x)$ (a cubic NP).

We have to impose additional constraints for $z_{n=5}(x)$ to be a valid ZF NP only if $3 \mid N$ and/or $5 \mid N$ because for every prime $p > 5$, with $p \mid N$, the five coefficients of $z_{n=5}(x)$ will be divisible by p, and thus, they fulfill conditions (4).

We denote by $g_k$ the value $\gcd (k!,N)$. Let the prime factorization of $g_k$, for $k = 3, 4, 5$, be of the form

$$\begin{aligned}&g_k = \gcd (k!,N) = 2^{n_{g_k,2}} \cdot 3^{n_{g_k,3}} \cdot 5^{n_{g_k,5}}, ~\text {where} \nonumber \\&\quad 0 \le n_{g_k,2} \le 3, 0 \le n_{g_k,3} \le 1, ~\text {and}~ 0 \le n_{g_k,5} \le 1. \end{aligned}$$

(7)

Obviously, for $k = 3$ or $k = 4$, we have $n_{g_k,5} = 0$ in (7).

We write the prime factorization of N as

$$\begin{aligned}&N = 2^{n_{N,2}} \cdot 3^{n_{N,3}} \cdot 5^{n_{N,5}} \cdot \prod _{j=4}^{s} p_j^{n_{N,p_j}}, ~\text {with} \nonumber \\&\quad n_{N,p} \ge 0 ~\text {for}~ p = 2, 3, 5, ~\text {and}~ n_{N,p_j} \ge 1\quad \forall j = \overline{4,s}.\nonumber \\ \end{aligned}$$

(8)

We note that if $n_{N,3} = 0$, then $3 \not \mid N$ and we have not to impose additional constraints for prime $p = 3$ for $z_{n=k}(x)$, $k = 3, 4, 5$, to be a valid ZF NP. Also, if $n_{N,5} = 0$, then $5 \not \mid N$ and we have not to impose additional constraints for prime $p = 5$ for $z_{n=5}(x)$.

If $n_{N,3} > 1$, we have $3 \mid \frac{N}{\gcd (k!,N)}, \forall k = 3, 4, 5$ and from (6) it results that 3 divides all coefficients of $z_k(x)$, thus fulfilling the conditions for prime $p = 3$ for a valid ZF NP. If $n_{N,5} > 1$, we have $5 \mid \frac{N}{\gcd (5!,N)}$ and from (6) it results that 5 divides all coefficients of $z_k(x)$, thus fulfilling the conditions for prime $p = 5$ for a valid ZF NP.

From those above it results that we have to impose additional constraints for $z_{n=5}(x)$ to be a valid ZF NP only if $n_{N,3} = 1$ and/or $n_{N,5} = 1$. We analyze these cases below.

As it results from (5), the coefficient of the maximum degree term of a NP, i.e., $z_n$, is any multiple of $\frac{N}{\gcd (n!, N)}$. Thus, as it is shown in Sect. 3 from [1], any PP modulo N is equivalent to a PP having the coefficient of the k degree term $q_k < \frac{N}{\gcd (k!, N)}$, $\forall k = \overline{2, n}$. However, for a PP under ZF sufficient conditions this condition is not always valid. We detail below these conditions for a NP of degree three, four, and five.

If $n_{N,3} = 1$, then $3 \not \mid \frac{N}{\gcd (k!,N)}$, $\forall k = 3, 4, 5$. In this case for $z_{n=k}(x)$, $\forall k = 3, 4, 5$, to be a valid ZF NP, we have to impose the condition that $3 \mid \tau _k$. Similarly, if $n_{N,5} = 1$, then $5 \not \mid \frac{N}{\gcd (5!,N)}$. In this case for $z_{n=5}(x)$ to be a valid ZF NP we have to impose the condition that $5 \mid \tau _5$.

Thus, when $n_{N,3} = 1$ any PP modulo N under ZF sufficient conditions is equivalent to a PP under ZF sufficient conditions having the coefficient of the third degree term $q_3 < 3 \cdot \frac{N}{\gcd (3!, N)}$.

For similar reasons when $n_{N,3} = 1$, any PP modulo N under ZF sufficient conditions is equivalent to a PP under ZF sufficient conditions having the coefficient of the fourth degree term $q_4 < 3 \cdot \frac{N}{\gcd (4!, N)}$.

When $n_{N,3} = 1$ and $n_{N,5} = 0$, any PP modulo N under ZF sufficient conditions is equivalent to a PP under ZF sufficient conditions having the coefficient of the fifth degree term $q_5 < 3 \cdot \frac{N}{\gcd (5!, N)}$.

When $n_{N,3} = 0$ and $n_{N,5} = 1$, any PP modulo N under ZF sufficient conditions is equivalent to a PP under ZF sufficient conditions having the coefficient of the fifth degree term $q_5 < 5 \cdot \frac{N}{\gcd (5!, N)}$.

When $n_{N,3} = 1$ and $n_{N,5} = 1$, any PP modulo N under ZF sufficient conditions is equivalent to a PP under ZF sufficient conditions having the coefficient of the fifth degree term $q_5 < 15 \cdot \frac{N}{\gcd (5!, N)}$.

We have not to impose additional constraints for coefficients $z_1$ and $z_2$ of the NP $z_{n=5}(x)$ because when $3 \mid \tau _k \cdot \frac{N}{\gcd (k!,N)}$, $\forall k = 3, 4, 5$ and $5 \mid \tau _5 \cdot \frac{N}{\gcd (5!,N)}$, then $p \mid z_1$ and $p \mid z_2$, for $p = 3 \text { or } 5$. Thus, $z_2 = z_1 = 0 \ (\mathrm {mod}\ p)$, for $p = 3 \text { or } 5$, and the conditions from (4) for primes 3 and 5 are fulfilled.

The analysis from Sect. 4 in [1] regarding the three types of prime factors from Table 1 remains valid. For primes 3 and 5, we can summarize the formula for the number of coefficients as follows.

For $n=3$ and $n_{N,3} \ge 1$, there are $3^{n_{N,3}-1}$ coefficients $q_{2,j}$ and $3^{\max \{0; n_{N,3}-2\} }$ coefficients $q_{3,j}$, out of which one is zero.

For $n=4$ and $n_{N,3} \ge 1$, there are $3^{n_{N,3}-1}$ coefficients $q_{2,j}$ and $3^{\max \{0; n_{N,3}-2\} }$ coefficients $q_{3,j}$ and $q_{4,j}$, respectively, out of which one is zero.

For $n=5$, if $p_j=3$ (i.e., $n_{N,3} \ge 1$) there are $3^{n_{N,3}-1}$ coefficients $q_{2,j}$ and $3^{\max \{0; n_{N,3}-2\} }$ coefficients $q_{3,j}$, $q_{4,j}$, and $q_{5,j}$, respectively, out of which one is zero.

If $p_j=5$ (i.e., $n_{N,5} \ge 1$), there are $5^{n_{N,5}-1}$ coefficients $q_{2,j}$, $q_{3,j}$, and $q_{4,j}$, respectively, and $5^{\max \{0; n_{N,5}-2\} }$ coefficients $q_{5,j}$, out of which one is zero.

3 Corrections for formulas in Sects. 4.1, 4.2, and 4.3 from [1] considering null polynomials under Zhao and Fan sufficient conditions

In the following, we correct the formulas in Sects. 4.1, 4.2, and 4.3 from [1] in cases when the power of 3 and/or 5 in the factorization of N is equal to one. As in some cases the formulas become simpler, we restate and proof the complete theorems.

3.1 Corrections for formulas in Sect. 4.1 from [1]

In Sect. 4.1, the corrected equivalence conditions for ZF CPPs are:

$q_2 < N/2$ and $q_3 < N/2$, when $2 \mid N$ and $3 \not \mid N$, or when $6 \mid N$ and $9 \not \mid N$;
$q_3 < N/3$, when $2 \not \mid N$ and $9 \mid N$;
$q_2 < N/2$ and $q_3 < N/6$, when $18 \mid N$.

The formulas from Theorems 4.1, 4.2, and 4.3 remain valid. The formulas from Theorems 4.4, 4.5, and 4.6 remain valid only when $n_{N,3} > 1$. We correct these formulas for the case when $n_{N,3} = 1$, restating and proving the complete theorems.

Theorem 2.1

(Updated Theorem 4.4 from [1]) If $2 \not \mid N$ and $3 \mid N$, i.e., $\displaystyle N = 3^{n_{N,3}} \cdot \prod \nolimits _{j=2}^{s} p_j^{n_{N,p_j}}$, with $n_{N,3} \ge 1$, $p_j > 3$ and $n_{N,p_j} \ge 1$, $\forall j = \overline{2, s}$, the number of ZF CPPs will be equal to:

$$\begin{aligned} \begin{aligned} C_{N, CPPs, ZF}&= 2 \cdot 3^{2 \cdot (n_{N,3}-1)} \cdot \prod _{j=2}^{s} \left( p_j^{2 \cdot (n_{N,p_j}-1)} \cdot (p_j - 1) \right) \\&\quad \cdot \left( 3^{\max \{0; n_{N,3}-2 \} } \cdot \prod _{j=2}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \end{aligned} \end{aligned}$$

(9)

Proof

See Appendix A.1. $\square $

From (9), we see that the number of ZF CPPs is equal to 0, if N is three or nine times a product of prime numbers greater than three, each of them to the power of 1.

Theorem 2.2

(Updated Theorem 4.5 from [1]) If $2 \mid N$, $4 \not \mid N$ and $3 \mid N$, i.e., $\displaystyle N = 2 \cdot 3^{n_{N,3}} \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}}$, with $n_{N,3} \ge 1$, $p_j > 3$ and $n_{N,p_j} \ge 1$, $\forall j = \overline{3, s}$, the number of ZF CPPs will be equal to:

$$\begin{aligned} \begin{aligned} C_{N, CPPs, ZF}&= 2 \cdot 3^{2 \cdot (n_{N,3}-1)} \cdot \prod _{j=3}^{s} \left( p_j^{2 \cdot (n_{N,p_j}-1)} \cdot (p_j - 1) \right) \\&\quad \cdot \left( 3^{\max \{0; n_{N,3}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \end{aligned} \end{aligned}$$

(10)

Proof

See Appendix A.2. $\square $

From (10), we see that the number of ZF CPPs is equal to 0 if N is 6 or 18 times a product of prime numbers greater than three, each of them to the power of 1.

Theorem 2.3

(Updated Theorem 4.6 from [1]) If $4 \mid N$ and $3 \mid N$, i.e., $\displaystyle N = 2^{n_{N,2}} \cdot 3^{n_{N,3}} \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}}$, with $n_{N,2} > 1$, $n_{N,3} \ge 1$, $p_j > 3$ and $n_{N,p_j} \ge 1$, $\forall j = \overline{3, s}$, the number of ZF CPPs will be equal to:

$$\begin{aligned} \begin{aligned}&C_{N, CPPs, ZF} \\&\quad = 2^{2 \cdot (n_{N,2}-1)} \cdot 3^{2 \cdot (n_{N,3}-1)} \\&\quad \quad \cdot \prod _{j=3}^{s} \left( p_j^{2 \cdot (n_{N,p_j}-1)} \cdot (p_j - 1) \right) \\&\quad \quad \cdot \left( 2^{n_{N,2}-2} \cdot 3^{\max \{0; n_{N,3}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \end{aligned} \end{aligned}$$

(11)

Proof

See Appendix A.3. $\square $

From (11), we see that the number of ZF CPPs is equal to 0, if N is 12 or 36 times a product of prime numbers greater than three, each of them to the power of 1.

The conclusions from the end of Sect. 4.1 in [1] have to be corrected as follows.

Under ZF sufficient conditions, the number of true different CPPs is 0, when the interleaver length is

$$\begin{aligned} \begin{aligned}&N = 2^{n_{N,2}} \cdot 3^{n_{N,3}} \cdot \prod _{j=3}^{s} p_j, n_{N,2} = \overline{0, 2}, \\&\quad n_{N,3} = \overline{0, 2}, p_j > 3, \forall j = \overline{3, s} \end{aligned} \end{aligned}$$

(12)

Because all the lengths from the LTE standard [4] are multiples of 8, from (12) it results that for all these lengths there exist CPPs under ZF sufficient conditions.

The remark that the LTE standard contains 58 of all 188 lengths for which ZF sufficient conditions of CPPs become also necessary remains valid. However, for all lengths for which in their prime factorization there exists the prime 3 at power of 1, but no primes p of the form $(p = 2 \ (\mathrm {mod}\ 3)$ at power of 1, the number of all true different CPPs is equal to the number of true different CPPs under ZF sufficient conditions. There exist 25 such lengths among the LTE lengths (namely 48, 96, 168, 192, 312, 336, 384, 456, 624, 672, 768, 912, 1248, 1344, 1536, 1824, 2496, 2688, 3072, 3648, 4800, 4992, 5376, 5952, and 6144). Thus for 83 lengths from the LTE standard, the number of all true different CPPs is equal to the number of true different CPPs under ZF sufficient conditions. For another 105 lengths, the ratio (in percentage) between the number of true different CPPs under ZF sufficient conditions and the number of all true different CPPs is less than 50%. This is shown in Fig. 1 (the corrected Fig. 1 from [1]).

The example for Theorem 2.3 (i.e., updated Theorem 4.6 from [1]) has to be corrected as in Appendix D.1.

Fig. 1
The ratio (in percentage) between the number of true different CPPs under Zhao and Fan sufficient conditions and the number of all true different CPPs, for all lengths of LTE standard

3.2 Corrections for formulas in Sect. 4.2 from [1]

In Sect. 4.2, the corrected equivalence conditions for 4-PPs under ZF sufficient conditions are written as

$q_2 < N/2$, $q_3 < N/2$ and $q_4 < N/2$, when $2 \mid N$, $3 \not \mid N$ and $4 \not \mid N$, or when $4 \not \mid N$, $6 \mid N$, and $9 \not \mid N$;
$q_2 < N/2$, $q_3 < N/2$ and $q_4 < N/4$, when $3 \not \mid N$, $4 \mid N$ and $8 \not \mid N$, or when $8 \not \mid N$, $9 \not \mid N$, and $12 \mid N$;
$q_2 < N/2$, $q_3 < N/2$ and $q_4 < N/8$, when $3 \not \mid N$ and $8 \mid N$, or when $9 \not \mid N$ and $24 \mid N$;
$q_3 < N/3$ and $q_4 < N/3$, when $2 \not \mid N$ and $9 \mid N$;
$q_2 < N/2$, $q_3 < N/6$ and $q_4 < N/6$, when $4 \not \mid N$ and $18 \mid N$;
$q_2 < N/2$, $q_3 < N/6$ and $q_4 < N/12$, when $8 \not \mid N$ and $36 \mid N$;
$q_2 < N/2$, $q_3 < N/6$ and $q_4 < N/24$, when $72 \mid N$.

The formulas from Theorems 4.7, 4.8, 4.10, and 4.12 remain valid. The formulas from Theorems 4.9, 4.11, 4.13, and 4.14 remain valid only when $n_{N,3} > 1$. We correct these formulas for the case when $n_{N,3} = 1$, restating and proving the complete theorems.

Theorem 2.4

(Updated Theorem 4.9 from [1]) If $2 \not \mid N$ and $3 \mid N$, i.e., $\displaystyle N = 3^{n_{N,3}} \cdot \prod \nolimits _{j=2}^{s} p_j^{n_{N,p_j}}$, with $n_{N,3} \ge 1$, $p_j > 3$ and $n_{N,p_j} \ge 1$, $\forall j = \overline{2, s}$, the number of ZF 4-PPs will be equal to:

$$\begin{aligned} \begin{aligned} C_{N, 4-PPs, ZF}&= 2 \cdot 3^{2 \cdot (n_{N,3}-1) + \max \{ 0; n_{N,3}-2 \}} \\&\quad \cdot \prod _{j=2}^{s} \left( p_j^{3 \cdot (n_{N,p_j}-1)} \cdot (p_j - 1) \right) \\&\quad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod _{j=2}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \end{aligned} \end{aligned}$$

(13)

Proof

See Appendix B.1. $\square $

From (13), we see that the number of ZF 4-PPs is equal to 0 if N is three or nine times a product of prime numbers greater than three, each of them to the power of 1.

Theorem 2.5

(Updated Theorem 4.11 from [1]) If $6 \mid N$ and $4 \not \mid N$, i.e., $\displaystyle N = 2 \cdot 3^{n_{N,3}} \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}}$, with $n_{N,3} \ge 1$, $p_j > 3$ and $n_{N,p_j} \ge 1$, $\forall j = \overline{3, s}$, the number of ZF 4-PPs will be equal to:

$$\begin{aligned} \begin{aligned} C_{N, 4-PPs, ZF}&= 2 \cdot 3^{2 \cdot (n_{N,3}-1) + \max \{ 0; n_{N,3}-2 \}} \\&\quad \cdot \prod _{j=3}^{s} \left( p_j^{3 \cdot (n_{N,p_j}-1)} \cdot (p_j - 1) \right) \\&\quad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \end{aligned} \end{aligned}$$

(14)

Proof

See Appendix B.2. $\square $

From (14), we see that the number of ZF 4-PPs is equal to 0 if N is 6 or 18 times a product of prime numbers greater than three, each of them to the power of 1.

Theorem 2.6

(Updated Theorem 4.13 from [1]) If $12 \mid N$ and $8 \not \mid N$, i.e., $\displaystyle N = 4 \cdot 3^{n_{N,3}} \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}}$, with $n_{N,3} \ge 1$, $p_j > 3$ and $n_{N,p_j} \ge 1$, $\forall j = \overline{3, s}$, the number of ZF 4-PPs will be equal to:

$$\begin{aligned} \begin{aligned} C_{N, 4-PPs, ZF}&= 4 \cdot 3^{2 \cdot (n_{N,3}-1) + \max \{ 0; n_{N,3}-2 \}} \\&\quad \cdot \prod _{j=3}^{s} \left( p_j^{3 \cdot (n_{N,p_j}-1)} \cdot (p_j - 1) \right) \\&\quad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \end{aligned} \end{aligned}$$

(15)

Proof

See Appendix B.3. $\square $

From (15), we see that the number of ZF 4-PPs is equal to 0 if N is 12 or 36 times a product of prime numbers, greater than three, each of them to the power of 1.

Theorem 2.7

(Updated Theorem 4.14 from [1]) If $24 \mid N$, i.e., $\displaystyle N = 2^{n_{N,2}} \cdot 3^{n_{N,3}} \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}}$, with $n_{N,2} \ge 3$, $n_{N,3} \ge 1$, $p_j > 3$ and $n_{N,p_j} \ge 1$, $\forall j = \overline{3, s}$, the number of ZF 4-PPs will be equal to:

$$\begin{aligned} \begin{aligned}&C_{N, 4-PPs, ZF} \\&\quad = 2^{3 \cdot n_{N,2} - 4} \cdot 3^{2 \cdot (n_{N,3}-1) + \max \{ 0; n_{N,3}-2\} } \\&\quad \quad \cdot \prod _{j=3}^{s} \left( p_j^{3 \cdot (n_{N,p_j}-1)} \cdot (p_j - 1) \right) \\&\quad \quad \cdot \left( 2^{n_{N,2}-3} \cdot 3^{ \max \{ 0; n_{N,3}-2\} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \end{aligned} \end{aligned}$$

(16)

Proof

See Appendix B.4. $\square $

From (16), we see that the number of ZF 4-PPs is equal to 0 if N is 24 or 72 times a product of prime numbers greater than three, each of them to the power of 1.

The conclusions from the end of Sect. 4.2 have to be corrected as follows.

Under ZF sufficient conditions, the number of true different 4-PPs is 0, when the interleaver length is of the form

$$\begin{aligned} \begin{aligned}&N = 2^{n_{N,2}} \cdot 3^{n_{N,3}} \cdot \prod _{j=3}^{s} p_j, n_{N,2} = \overline{0, 3}, n_{N,3} = \overline{0, 2}, \\&\quad p_j > 3, \forall j = \overline{3, s}. \end{aligned} \end{aligned}$$

(17)

The LTE standard [4] contains 27 lengths (of form (17) with $n_{N,2} = 3$) for which there are no 4-PPs under ZF sufficient conditions, namely the lengths 40, 56, 72, 88, 104, 120, 136, 152, 168, 184, 232, 248, 264, 280, 296, 312, 328, 344, 360, 376, 408, 424, 440, 456, 472, 488, and 504.

Fig. 2
The ratio (in percentage) between the number of true different 4-PPs under Zhao and Fan sufficient conditions and the number of all true different 4-PPs, for all lengths used in the LTE standard

We compared the total number of true different 4-PPs, obtained with the algorithm from [5], for all lengths from the LTE standard with the number of true different 4-PPs under ZF sufficient conditions. Figure 2 shows the ratio (in percentage) between the number of true different 4-PPs under ZF sufficient conditions and the number of all true different 4-PPs for all the 188 lengths of the LTE standard. We note that for 23 from the 188 lengths, the number of all true different 4-PPs is zero (i.e., there are no true 4-PPs for these lengths), and thus, the percentage is 100%. For 84 from the 188 lengths (from which 23 lengths are those mentioned above), the ZF sufficient conditions are also necessary (i.e., this percentage is 100%). For other 4 lengths, this percentage is 0% (i.e., there are no true 4-PPs under ZF sufficient conditions, but there are true different 4-PPs, for these lengths, namely 56, 168, 280, and 504), for 12 lengths the percentage is equal to 50%, and for the other 88 lengths, the percentage is less than 20%.

In the example for Theorem 2.7 (i.e., updated Theorem 4.14 from [1]), the number of ZF 4-PPs computed according to updated formula (16) will be equal to:

$$\begin{aligned} \begin{aligned} C_{1080, 4-PPs, ZF}&= 2^{3 \cdot 3 - 4} \cdot 3^{2 \cdot (3-1)+ \max \{0; 3-2 \} } \cdot 5^{3 \cdot (1-1)} \cdot (5 - 1) \\&\quad \cdot \left( 2^{3-3} \cdot 3^{ \max \{0; 3-2 \} } \cdot 5^{1-1} - 1 \right) \\&= 32 \cdot 243 \cdot 1 \cdot 4 \cdot ( 3-1 ) = 62{,}208, \end{aligned} \end{aligned}$$

which is the same as that obtained by means of the old formula from [1], because in this case $n_{N,3} > 1$.

3.3 Corrections for formulas in Sect. 4.3 from [1]

In Sect. 4.3, the corrected equivalence conditions for 5-PPs under ZF sufficient conditions are written as

$q_2 < N/2$, $q_3 < N/2$, $q_4 < N/2$, and $q_5 < N/2$ when $2 \mid N$, $3 \not \mid N$, $4 \not \mid N$, and $5 \not \mid N$, or when $4 \not \mid N$, $5 \not \mid N$, $6 \mid N$, and $9 \not \mid N$, or when $3 \not \mid N$, $4 \not \mid N$, $10 \mid N$, and $25 \not \mid N$, or when $4 \not \mid N$, $9 \not \mid N$, $25 \not \mid N$, and $30 \mid N$;
$q_2 < N/2$, $q_3 < N/2$, $q_4 < N/4$, and $q_5 < N/4$ when $3 \not \mid N$, $4 \mid N$, $5 \not \mid N$, and $8 \not \mid N$, or when $5 \not \mid N$, $8 \not \mid N$, $9 \not \mid N$, and $12 \mid N$, or when $3 \not \mid N$, $8 \not \mid N$, $20 \mid N$, and $25 \not \mid N$, or when $8 \not \mid N$, $9 \not \mid N$, $25 \not \mid N$, and $60 \mid N$;
$q_2 < N/2$, $q_3 < N/2$, $q_4 < N/8$, and $q_5 < N/8$ when $3 \not \mid N$, $5 \not \mid N$, and $8 \mid N$, or when $5 \not \mid N$, $9 \not \mid N$, and $24 \mid N$, or when $3 \not \mid N$, $25 \not \mid N$, and $40 \mid N$, or when $9 \not \mid N$, $25 \not \mid N$, and $120 \mid N$;
$q_3 < N/3$, $q_4 < N/3$, and $q_5 < N/3$ when $2 \not \mid N$, $5 \not \mid N$, and $9 \mid N$, or when $2 \not \mid N$, $25 \not \mid N$, and $45 \mid N$;
$q_2 < N/2$, $q_3 < N/6$, $q_4 < N/6$, and $q_5 < N/6$ when $4 \not \mid N$, $5 \not \mid N$, and $18 \mid N$, or when $4 \not \mid N$, $25 \not \mid N$, and $90 \mid N$;
$q_5 < N/5$ when $2 \not \mid N$, $3 \not \mid N$, and $25 \mid N$, or when $2 \not \mid N$, $9 \not \mid N$, and $75 \mid N$;
$q_2 < N/2$, $q_3 < N/6$, $q_4 < N/12$, and $q_5 < N/12$ when $5 \not \mid N$, $8 \not \mid N$, and $36 \mid N$, or when $8 \not \mid N$, $25 \not \mid N$, and $180 \mid N$;
$q_2 < N/2$, $q_3 < N/2$, $q_4 < N/2$, and $q_5 < N/10$ when $3 \not \mid N$, $4 \not \mid N$, and $50 \mid N$, or when $4 \not \mid N$, $9 \not \mid N$, and $150 \mid N$;
$q_2 < N/2$, $q_3 < N/6$, $q_4 < N/24$, and $q_5 < N/24$ when $5 \not \mid N$ and $72 \mid N$, or when $25 \not \mid N$ and $360 \mid N$;
$q_2 < N/2$, $q_3 < N/2$, $q_4 < N/4$, and $q_5 < N/20$ when $3 \not \mid N$, $8 \not \mid N$, and $100 \mid N$, or when $8 \not \mid N$, $9 \not \mid N$, and $300 \mid N$;
$q_2 < N/2$, $q_3 < N/2$, $q_4 < N/8$, and $q_5 < N/40$ when $3 \not \mid N$ and $200 \mid N$, or when $9 \not \mid N$ and $600 \mid N$;
$q_3 < N/3$, $q_4 < N/3$, and $q_5 < N/15$ when $2 \not \mid N$ and $225 \mid N$;
$q_2 < N/2$, $q_3 < N/6$, $q_4 < N/6$, and $q_5 < N/30$ when $4 \not \mid N$ and $450 \mid N$;
$q_2 < N/2$, $q_3 < N/6$, $q_4 < N/12$, and $q_5 < N/60$ when $8 \not \mid N$ and $900 \mid N$;
$q_2 < N/2$, $q_3 < N/6$, $q_4 < N/24$, and $q_5 < N/120$ when $1800 \mid N$.

The formulas from Theorems 4.15, 4.16, 4.18, and 4.21 remain valid. The formulas from Theorems 4.17, 4.20, 4.23, and 4.26 remain valid only when $n_{N,3} > 1$. The formulas from Theorems 4.19, 4.22, 4.25, and 4.28 remain valid only when $n_{N,5} > 1$. The formulas from Theorems 4.24, 4.27, 4.29, and 4.30 remain valid only when $n_{N,3} > 1$ and $n_{N,5} > 1$. We correct these formulas for the cases when $n_{N,3} = 1$ and/or $n_{N,5} > 1$, restating and proving the complete theorems.

Theorem 2.8

(Updated Theorem 4.17 from [1]) If $2 \not \mid N$, $3 \mid N$ and $5 \not \mid N$, i.e., $\displaystyle N = 3^{n_{N,3}} \cdot \prod \nolimits _{j=2}^{s} p_j^{n_{N,p_j}}$, with $n_{N,3} \ge 1$, $p_j > 5$ and $n_{N,p_j} \ge 1$, $\forall j = \overline{2, s}$, the number of ZF 5-PPs will be equal to:

$$\begin{aligned} \begin{aligned} C_{N, 5-PPs, ZF}&= 2 \cdot 3^{2 \cdot (n_{N,3}-1) + 2 \cdot \max \{ 0; n_{N,3}-2 \} } \\&\quad \cdot \prod _{j=2}^{s} \left( p_j^{4 \cdot (n_{N,p_j}-1)} \cdot (p_j - 1) \right) \\&\quad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod _{j=2}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \end{aligned} \end{aligned}$$

(18)

Proof

See Appendix C.1. $\square $

From (18), we see that the number of ZF 5-PPs is equal to 0 if N is three or nine times a product of prime numbers greater than five, each of them to the power of 1.

Theorem 2.9

(Updated Theorem 4.19 from [1]) If $2 \not \mid N$, $3 \not \mid N$ and $5 \mid N$, i.e., $\displaystyle N = 5^{n_{N,5}} \cdot \prod \nolimits _{j=2}^{s} p_j^{n_{N,p_j}}$, with $n_{N,5} \ge 1$, $p_j > 5$ and $n_{N,p_j} \ge 1$, $\forall j = \overline{2, s}$, the number of ZF 5-PPs will be equal to:

$$\begin{aligned} \begin{aligned} C_{N, 5-PPs, ZF}&= 4 \cdot 5^{4 \cdot (n_{N,5}-1)} \\&\quad \cdot \prod _{j=2}^{s} \left( p_j^{4 \cdot (n_{N,p_j}-1)} \cdot (p_j - 1) \right) \\&\quad \cdot \left( 5^{\max \{ 0; n_{N,5}-2 \} } \cdot \prod _{j=2}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \end{aligned} \end{aligned}$$

(19)

Proof

See Appendix C.2. $\square $

From (19), we see that the number of ZF 5-PPs is equal to 0 if N is 5 or 25 times a product of prime numbers greater than five, each of them to the power of 1.

Theorem 2.10

(Updated Theorem 4.20 from [1]) If $4 \not \mid N$, $5 \not \mid N$ and $6 \mid N$, i.e., $\displaystyle N = 2 \cdot 3^{n_{N,3}} \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}}$, with $n_{N,3} \ge 1$, $p_j > 5$ and $n_{N,p_j} \ge 1$, $\forall j = \overline{3, s}$, the number of ZF 5-PPs will be equal to:

$$\begin{aligned} \begin{aligned} C_{N, 5-PPs, ZF}&= 2 \cdot 3^{2 \cdot (n_{N,3}-1) + 2 \cdot \max \{ 0; n_{N,3}-2 \} } \\&\quad \cdot \prod _{j=3}^{s} \left( p_j^{4 \cdot (n_{N,p_j}-1)} \cdot (p_j - 1) \right) \\&\quad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \end{aligned} \end{aligned}$$

(20)

Proof

See Appendix C.3. $\square $

From (20), we see that the number of ZF 5-PPs is equal to 0 if N is 6 or 18 times a product of prime numbers greater than five, each of them to the power of 1.

Theorem 2.11

(Updated Theorem 4.22 from [1]) If $2 \mid N$, $3 \not \mid N$, $4 \not \mid N$ and $5 \mid N$, i.e., $\displaystyle N = 2 \cdot 5^{n_{N,5}} \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}}$, with $n_{N,5} \ge 1$, $p_j > 5$ and $n_{N,p_j} \ge 1$, $\forall j = \overline{3, s}$, the number of ZF 5-PPs will be equal to:

$$\begin{aligned} \begin{aligned} C_{N, 5-PPs, ZF}&= 4 \cdot 5^{4 \cdot (n_{N,5}-1)} \\&\quad \cdot \prod _{j=3}^{s} \left( p_j^{4 \cdot (n_{N,p_j}-1)} \cdot (p_j - 1) \right) \\&\quad \cdot \left( 5^{\max \{ 0; n_{N,5}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \end{aligned} \end{aligned}$$

(21)

Proof

See Appendix C.4. $\square $

From (21), we see that the number of ZF 5-PPs is equal to 0 if N is 10 or 50 times a product of prime numbers greater than five, each of them to the power of 1.

Theorem 2.12

(Updated Theorem 4.23 from [1]) If $5 \not \mid N$, $8 \not \mid N$ and $12 \mid N$, i.e., $\displaystyle N = 4 \cdot 3^{n_{N,3}} \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}}$, with $n_{N,3} \ge 1$, $p_j > 5$ and $n_{N,p_j} \ge 1$, $\forall j = \overline{3, s}$, the number of ZF 5-PPs will be equal to:

$$\begin{aligned} \begin{aligned}&C_{N, 5-PPs, ZF} \\&\quad = 4 \cdot 3^{2 \cdot (n_{N,3}-1) + 2 \cdot \max \{ 0; n_{N,3}-2 \}} \\&\qquad \cdot \prod _{j=3}^{s} \left( p_j^{4 \cdot (n_{N,p_j}-1)} \cdot (p_j - 1) \right) \\&\qquad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \end{aligned} \end{aligned}$$

(22)

Proof

See Appendix C.5. $\square $

From (22), we see that the number of ZF 5-PPs is equal to 0 if N is 12 or 36 times a product of prime numbers greater than five, each of them to the power of 1.

Theorem 2.13

(Updated Theorem 4.24 from [1]) If $2 \not \mid N$, $3 \mid N$ and $5 \mid N$, i.e., $\displaystyle N = 3^{n_{N,3}} \cdot 5^{n_{N,5}} \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}}$, with $n_{N,3} \ge 1$, $n_{N,5} \ge 1$, $p_j > 5$ and $n_{N,p_j} \ge 1$, $\forall j = \overline{3, s}$, the number of ZF 5-PPs will be equal to:

$$\begin{aligned} \begin{aligned}&C_{N, 5-PPs, ZF} \\&\quad = 8 \cdot 3^{2 \cdot (n_{N,3}-1) + 2 \cdot \max \{ 0; n_{N,3}-2 \}} \cdot 5^{4 \cdot (n_{N,5}-1)} \\&\qquad \cdot \prod _{j=3}^{s} \left( p_j^{4 \cdot (n_{N,p_j}-1)} \cdot (p_j - 1) \right) \\&\qquad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot 5^{\max \{ 0; n_{N,5}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \end{aligned} \end{aligned}$$

(23)

Proof

See Appendix C.6. $\square $

From (23), we see that the number of ZF 5-PPs is 0 if N is 15, 45, 75, or 225 times a product of prime numbers greater than five, each of them to the power of 1.

Theorem 2.14

(Updated Theorem 4.25 from [1]) If $3 \not \mid N$, $4 \mid N$, $5 \mid N$ and $8 \not \mid N$, i.e., $\displaystyle N = 4 \cdot 5^{n_{N,5}} \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}}$, with $n_{N,5} \ge 1$, $p_j > 5$ and $n_{N,p_j} \ge 1$, $\forall j = \overline{3, s}$, the number of ZF 5-PPs will be equal to:

$$\begin{aligned} \begin{aligned}&C_{N, 5-PPs, ZF} \\&\quad = 8 \cdot 5^{4 \cdot (n_{N,5}-1)} \cdot \prod _{j=3}^{s} \left( p_j^{4 \cdot (n_{N,p_j}-1)} \cdot (p_j - 1) \right) \\&\qquad \cdot \left( 5^{\max \{ 0; n_{N,5}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \end{aligned} \end{aligned}$$

(24)

Proof

See Appendix C.7. $\square $

From (24), we see that the number of ZF 5-PPs is equal to 0 if N is 20 or 100 times a product of prime numbers greater than five, each of them to the power of 1.

Theorem 2.15

(Updated Theorem 4.26 from [1]) If $8 \mid N$, $3 \mid N$ and $5 \not \mid N$, i.e., $\displaystyle N = 2^{n_{N,2}} \cdot 3^{n_{N,3}} \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}}$, with $n_{N,2} \ge 3$, $n_{N,3} \ge 1$, $p_j > 5$ and $n_{N,p_j} \ge 1$, $\forall j = \overline{3, s}$, the number of ZF 5-PPs will be equal to:

$$\begin{aligned} \begin{aligned}&C_{N, 5-PPs, ZF} \\&\quad = 2^{4 \cdot n_{N,2} - 7} \cdot 3^{2 \cdot (n_{N,3}-1) + 2 \cdot \max \{ 0; n_{N,3}-2 \}} \\&\qquad \cdot \prod _{j=3}^{s} \left( p_j^{4 \cdot (n_{N,p_j}-1)} \cdot (p_j - 1) \right) \\&\qquad \cdot \left( 2^{n_{N,2}-3} \cdot 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \end{aligned} \end{aligned}$$

(25)

Proof

See Appendix C.8. $\square $

From (25), we see that the number of ZF 5-PPs is equal to 0 if N is 24 or 72 times a product of prime numbers greater than five, each of them to the power of 1.

Theorem 2.16

(Updated Theorem 4.27 from [1]) If $2 \mid N$, $3 \mid N$ and $4 \not \mid N$ and $5 \mid N$, i.e., $\displaystyle N = 2 \cdot 3^{n_{N,3}} \cdot 5^{n_{N,5}} \cdot \prod \nolimits _{j=4}^{s} p_j^{n_{N,p_j}}$, with $n_{N,3} \ge 1$, $n_{N,5} \ge 1$, $p_j > 5$ and $n_{N,p_j} \ge 1$, $\forall j = \overline{4, s}$, the number of ZF 5-PPs will be equal to:

$$\begin{aligned} \begin{aligned}&C_{N, 5-PPs, ZF} \\&\quad = 8 \cdot 3^{2 \cdot (n_{N,3}-1) + 2 \cdot \max \{ 0; n_{N,3}-2 \}} \cdot 5^{4 \cdot (n_{N,5}-1)} \\&\qquad \cdot \left( \prod _{j=4}^{s} p_j^{4 \cdot (n_{N,p_j}-1)} \cdot (p_j - 1) \right) \\&\qquad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot 5^{\max \{ 0; n_{N,5}-2 \} } \cdot \prod _{j=4}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \end{aligned} \end{aligned}$$

(26)

Proof

See Appendix C.9. $\square $

From (26), we see that the number of ZF 5-PPs is 0 if N is 30, 90, 150, or 450 times a product of prime numbers greater than five, each of them to the power of 1.

Theorem 2.17

(Updated Theorem 4.28 from [1]) If $3 \not \mid N$, $5 \mid N$ and $8 \mid N$, i.e., $\displaystyle N = 2^{n_{N,2}} \cdot 5^{n_{N,5}} \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}}$, with $n_{N,2} \ge 3$, $n_{N,5} \ge 1$, $p_j > 5$ and $n_{N,p_j} \ge 1$, $\forall j = \overline{3, s}$, the number of ZF 5-PPs will be equal to:

$$\begin{aligned} \begin{aligned}&C_{N, 5-PPs, ZF} \\&\quad = 2^{4 \cdot n_{N,2}-6} \cdot 5^{4 \cdot (n_{N,1}-1)} \cdot \prod _{j=3}^{s} \left( p_j^{4 \cdot (n_{N,p_j}-1)} \cdot (p_j - 1) \right) \\&\qquad \cdot \left( 2^{n_{N,2}-3} \cdot 5^{\max \{ 0; n_{N,5}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \end{aligned} \end{aligned}$$

(27)

Proof

See Appendix C.10. $\square $

From (27), we see that the number of ZF 5-PPs is equal to 0 if N is 40 or 200 times a product of prime numbers greater than five, each of them to the power of 1.

Theorem 2.18

(Updated Theorem 4.29 from [1]) If $3 \mid N$, $4 \mid N$, $5 \mid N$ and $8 \not \mid N$, i.e., $\displaystyle N = 4 \cdot 3^{n_{N,3}} \cdot 5^{n_{N,5}} \cdot \prod \nolimits _{j=4}^{s} p_j^{n_{N,p_j}}$, with $n_{N,3} \ge 1$, $n_{N,5} \ge 1$, $p_j > 5$ and $n_{N,p_j} \ge 1$, $\forall j = \overline{4, s}$, the number of ZF 5-PPs will be equal to:

$$\begin{aligned} \begin{aligned}&C_{N, 5-PPs, ZF} \\&\quad = 16 \cdot 3^{2 \cdot (n_{N,3}-1) + 2 \cdot \max \{ 0; n_{N,3}-2 \}} \cdot 5^{4 \cdot (n_{N,5}-1)} \\&\qquad \cdot \prod _{j=4}^{s} \left( p_j^{4 \cdot (n_{N,p_j}-1)} \cdot (p_j - 1) \right) \\&\qquad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot 5^{\max \{ 0; n_{N,5}-2 \} } \cdot \prod _{j=4}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \end{aligned} \end{aligned}$$

(28)

Proof

See Appendix C.11. $\square $

From (28), we see that the number of ZF 5-PPs is equal to 0 if N is 60, 180, 300, or 900 times a product of prime numbers greater than five, each of them to the power of 1.

Theorem 2.19

(Updated Theorem 4.30 from [1]) If $3 \mid N$, $5 \mid N$ and $8 \mid N$, i.e., $\displaystyle N = 2^{n_{N,2}} \cdot 3^{n_{N,3}} \cdot 5^{n_{N,5}} \cdot \prod \nolimits _{j=4}^{s} p_j^{n_{N,p_j}}$, with $n_{N,2} \ge 3$, $n_{N,3} \ge 1$, $n_{N,5} \ge 1$, $p_j > 5$ and $n_{N,p_j} \ge 1$, $\forall j = \overline{4, s}$, the number of ZF 5-PPs will be equal to:

$$\begin{aligned} \begin{aligned}&C_{N, 5-PPs, ZF} \\&\quad = 2^{4 \cdot n_{N,2} - 5} \cdot 3^{2 \cdot (n_{N,3}-1) + 2 \cdot \max \{ 0; n_{N,3}-2 \}} \cdot 5^{4 \cdot (n_{N,5}-1)} \\&\qquad \cdot \prod _{j=4}^{s} \left( p_j^{4 \cdot (n_{N,p_j}-1)} \cdot (p_j - 1) \right) \\&\qquad \cdot \left( 2^{ n_{N,2}- 3} \cdot 3^{\max \{ 0; n_{N,3}-2 \} } \cdot 5^{\max \{ 0; n_{N,5}-2 \} } \cdot \prod _{j=4}^{s} p_j^{n_{N,p_j}-1} - 1 \right) . \end{aligned} \end{aligned}$$

(29)

Proof

See Appendix C.12. $\square $

From (29), we see that the number of ZF 5-PPs is equal to 0 if N is 120, 360, 600, or 1800 times a product of prime numbers greater than five, each of them to the power of 1.

The conclusions from the end of Sect. 4.3 have to be corrected as follows.

Under ZF sufficient conditions, the number of true different 5-PPs is 0, when the interleaver length is of the form

$$\begin{aligned} \begin{aligned}&N = 2^{n_{N,2}} \cdot 3^{n_{N,3}} \cdot 5^{n_{N,5}} \cdot \prod _{j=4}^{s} p_j, n_{N,2} = \overline{0, 3}, n_{N,3} = \overline{0, 2}, \\&\quad n_{N,5} = \overline{0, 2}, p_j > 5, \forall j = \overline{4, s}. \end{aligned} \end{aligned}$$

(30)

The LTE standard [4] contains 28 lengths (of form (30) with $n_{N,2} = 3$) for which there are no 5-PPs under ZF sufficient conditions, namely the lengths 40, 56, 72, 88, 104, 120, 136, 152, 168, 184, 200, 232, 248, 264, 280, 296, 312, 328, 344, 360, 376, 408, 424, 440, 456, 472, 488, and 504.

Fig. 3
The ratio (in percentage) between the number of true different 5-PPs under Zhao and Fan sufficient conditions and the number of all true different 5-PPs, for all lengths used in the LTE standard

We compared the total number of true different 5-PPs, obtained with the algorithm from [5], for all lengths from the LTE standard with the number of true different 5-PPs under ZF sufficient conditions. Figure 3 shows the ratio (in percentage) between the number of true different 5-PPs under ZF sufficient conditions and the number of all true different 5-PPs for all the 188 lengths of the LTE standard. We note that for 8 from the 188 lengths, the number of all true different 5-PPs is zero (i.e., there are no true 5-PPs for these lengths, namely 40, 88, 120, 248, 264, 328, 440, and 488), and thus, the percentage is 100%. For 31 from the 188 lengths (from which 8 lengths are those mentioned above), the ZF sufficient conditions are also necessary (i.e., this percentage is 100%). For other 20 lengths, this percentage is 0% (i.e., there are no true 5-PPs under ZF sufficient conditions, but there are true different 5-PPs, for these lengths) and for the other 137 lengths, the percentage is less than 20%.

In the first example for Theorem 2.19 (i.e., updated Theorem 4.30 from [1]), the number of ZF 5-PPs computed according to updated formula (29) will be equal to:

$$\begin{aligned} \begin{aligned}&C_{25200, 5-PPs, ZF} \\&\quad = 2^{4 \cdot 4 - 5} \cdot 3^{2 \cdot (2-1) + 2 \cdot \max \{ 0; 2-2 \}} \cdot 5^{4 \cdot (2-1)} \cdot 7^{4 \cdot (1-1)} \cdot (7 - 1) \\&\qquad \cdot \left( 2^{ 4- 3} \cdot 3^{\max \{ 0; 2-2 \}} \cdot 5^{\max \{ 0; 2-2 \}} \cdot 7^{1-1} - 1 \right) \\&\quad = (2048 \cdot 9 \cdot 625 \cdot 1 \cdot 6) \cdot (2 \cdot 1 \cdot 1 \cdot 1 - 1) = 6912000, \end{aligned} \end{aligned}$$

which is the same as that obtained by means of the old formula from [1], because in this case $n_{N,3} > 1$ and $n_{N,5} > 1$.

The second example for Theorem 2.19 has to be corrected as in Appendix D.2.

Appendix A

A.1 Proof [Theorem 2.1 (Updated Theorem 4.4 from [1])]

The number of possible coefficients $q_1$ is equal to $\displaystyle 2 \cdot 3^{n_{N,3}-1} \cdot \prod \nolimits _{j=2}^{s} \left( p_j^{n_{N,p_j}-1} \cdot (p_j - 1) \right) $, the number of coefficients $q_2$ is equal to $\displaystyle 3^{n_{N,3}-1} \cdot \prod \nolimits _{j=2}^{s} p_j^{n_{N,p_j}-1}$, and the number of coefficients $q_3$ is equal to $\displaystyle 3^{\max \{0; n_{N,3}-2 \} } \cdot \prod \nolimits _{j=2}^{s} p_j^{n_{N,p_j}-1}$, out of which one value is zero. By removing the value $q_3 = 0$, the number of ZF CPPs will be equal to:

$$\begin{aligned} \begin{aligned} C_{N, CPPs, ZF}&= \left( 2 \cdot 3^{n_{N,3}-1} \cdot \prod _{j=2}^{s} p_j^{n_{N,p_j}-1} \cdot (p_j - 1) \right) \\&\quad \cdot \left( 3^{n_{N,3}-1} \cdot \prod _{j=2}^{s} p_j^{n_{N,p_j}-1} \right) \\&\quad \cdot \left( 3^{\max \{0; n_{N,3}-2 \} } \cdot \prod _{j=2}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \\&= 2 \cdot 3^{2 \cdot (n_{N,3}-1)} \cdot \prod _{j=2}^{s} \left( p_j^{2 \cdot (n_{N,p_j}-1)} \cdot (p_j - 1) \right) \\&\quad \cdot \left( 3^{\max \{0; n_{N,3}-2 \} } \cdot \prod _{j=2}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \end{aligned} \end{aligned}$$

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A.2 Proof [Theorem 2.2 (Updated Theorem 4.5 from [1])]

The proof for Theorem 2.2 is similar to that of Theorem 2.1, by replacing N by N / 2. Therefore, the number of ZF CPPs will be equal to:

$$\begin{aligned} C_{N, CPPs, ZF}&= \left( 2 \cdot 3^{n_{N,3}-1} \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \cdot (p_j - 1) \right) \nonumber \\&\quad \cdot \left( 3^{n_{N,3}-1} \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \right) \nonumber \\&\quad \cdot \left( 3^{\max \{0; n_{N,3}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \\&= 2 \cdot 3^{2 \cdot (n_{N,3}-1)} \cdot \prod _{j=3}^{s} \left( p_j^{2 \cdot (n_{N,p_j}-1)} \cdot (p_j - 1) \right) \nonumber \\&\quad \cdot \left( 3^{\max \{0; n_{N,3}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \nonumber \end{aligned}$$

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A.3 Proof [Theorem 2.3 (Updated Theorem 4.6 from [1])]

The number of possible coefficients $q_1$ is equal to $\displaystyle 2^{n_{N,2}-1} \cdot 2 \cdot 3^{n_{N,3}-1} \cdot \prod \nolimits _{j=3}^{s} \left( p_j^{n_{N,p_j}-1} \cdot (p_j - 1) \right) $, the number of coefficients $q_2$ is equal to $\displaystyle 2^{n_{N,2}-2} \cdot 3^{n_{N,3}-1} \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}-1}$, and the number of coefficients $q_3$ is equal to $\displaystyle 2^{n_{N,2}-2} \cdot 3^{\max \{0; n_{N,3}-2 \} } \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}-1}$, out of which one is zero. By removing the value $q_3 = 0$, the number of ZF CPPs will be equal to:

$$\begin{aligned} \begin{aligned}&C_{N, CPPs, ZF} \\&\quad = \left( 2^{n_{N,2}-1} \cdot 2 \cdot 3^{n_{N,3}-1} \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \cdot (p_j - 1) \right) \\&\qquad \cdot \left( 2^{n_{N,2}-2} \cdot 3^{n_{N,3}-1} \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \right) \\&\qquad \cdot \left( 2^{n_{N,2}-2} \cdot 3^{\max \{0; n_{N,3}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \\&\quad = 2^{2 \cdot (n_{N,2}-1)} \cdot 3^{2 \cdot (n_{N,3}-1)} \cdot \prod _{j=3}^{s} \left( p_j^{2 \cdot (n_{N,p_j}-1)} \cdot (p_j - 1) \right) \\&\qquad \cdot \left( 2^{n_{N,2}-2} \cdot 3^{\max \{0; n_{N,3}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \end{aligned} \end{aligned}$$

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Appendix B

B.1 Proof [Theorem 2.4 (Updated Theorem 4.9 from [1])]

The number of possible coefficients $q_1$ is equal to $\displaystyle 2 \cdot 3^{n_{N,3}-1} \cdot \prod \nolimits _{j=2}^{s} \left( p_j^{n_{N,p_j}-1} \cdot (p_j - 1) \right) $, the number of coefficients $q_2$ is equal to $\displaystyle 3^{n_{N,3}-1} \cdot \prod \nolimits _{j=2}^{s} p_j^{n_{N,p_j}-1}$, and the number of coefficients $q_3$ and $q_4$, respectively, is equal to $\displaystyle 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod \nolimits _{j=2}^{s} p_j^{n_{N,p_j}-1}$, out of which one value is zero. By removing the value $q_4 = 0$, the number of ZF 4-PPs will be equal to:

$$\begin{aligned} \begin{aligned} C_{N, 4-PPs, ZF}&= 2 \cdot 3^{n_{N,3}-1} \cdot \prod _{j=2}^{s} \left( p_j^{n_{N,p_j}-1} \cdot (p_j - 1) \right) \\&\quad \cdot \left( 3^{n_{N,3}-1} \cdot \prod _{j=2}^{s} p_j^{n_{N,p_j}-1} \right) \\&\quad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod _{j=2}^{s} p_j^{n_{N,p_j}-1} \right) \\&\quad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod _{j=2}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \\&= 2 \cdot 3^{2 \cdot (n_{N,3}-1) + \max \{ 0; n_{N,3}-2 \}} \\&\quad \cdot \prod _{j=2}^{s} \left( p_j^{3 \cdot (n_{N,p_j}-1)} \cdot (p_j - 1) \right) \\&\quad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod _{j=2}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \end{aligned} \end{aligned}$$

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B.2 Proof [Theorem 2.5 (Updated Theorem 4.11 from [1])]

The proof for Theorem 2.5 is similar to that of Theorem 2.4, by replacing N by N / 2. Thus, the number of ZF 4-PPs will be equal to:

$$\begin{aligned} C_{N, 4-PPs, ZF}&= 2 \cdot 3^{n_{N,3}-1} \cdot \prod _{j=3}^{s} \left( p_j^{n_{N,p_j}-1} \cdot (p_j - 1) \right) \nonumber \\&\quad \cdot \left( 3^{n_{N,3}-1} \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \right) \nonumber \\&\quad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \right) \\&\quad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \nonumber \\&= 2 \cdot 3^{2 \cdot (n_{N,3}-1) + \max \{ 0; n_{N,3}-2 \}} \nonumber \\&\quad \cdot \prod _{j=3}^{s} \left( p_j^{3 \cdot (n_{N,p_j}-1)} \cdot (p_j - 1) \right) \nonumber \\&\quad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \nonumber \end{aligned}$$

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B.3 Proof [Theorem 2.6 (Updated Theorem 4.13 from [1])]

The number of possible coefficients $q_1$ is equal to $\displaystyle 2 \cdot 2 \cdot 3^{n_{N,3}-1} \cdot \prod \nolimits _{j=3}^{s} \left( p_j^{n_{N,p_j}-1} \cdot (p_j - 1) \right) $, the number of coefficients $q_2$ is equal to $\displaystyle 3^{n_{N,3}-1} \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}-1}$, the number of coefficients $q_3$ is equal to $\displaystyle 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}-1}$, and the number of coefficients $q_4$ is equal to $\displaystyle 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}-1}$, out of which one is zero. Therefore, the number of ZF 4-PPs will be equal to:

$$\begin{aligned} C_{N, 4-PPs, ZF}= & {} 4 \cdot 3^{n_{N,3}-1} \cdot \prod _{j=3}^{s} \left( p_j^{n_{N,p_j}-1} \cdot (p_j - 1) \right) \nonumber \\&\quad \cdot \left( 3^{n_{N,3}-1} \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \right) \nonumber \\&\quad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \right) \\&\quad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \nonumber \\= & {} 4 \cdot 3^{2 \cdot (n_{N,3}-1) + \max \{ 0; n_{N,3}-2 \}} \nonumber \\&\quad \cdot \prod _{j=3}^{s} \left( p_j^{3 \cdot (n_{N,p_j}-1)} \cdot (p_j - 1) \right) \nonumber \\&\quad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \nonumber \end{aligned}$$

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B.4 Proof [Theorem 2.7 (Updated Theorem 4.14 from [1])]

The number of possible coefficients $q_1$ is equal to $\displaystyle 2^{n_{N,2}-1} \cdot 2 \cdot 3^{n_{N,3}-1} \cdot \prod \nolimits _{j=3}^{s} \left( p_j^{n_{N,p_j}-1} \cdot (p_j - 1) \right) $, the number of coefficients $q_3$ is equal to $\displaystyle 2^{n_{N,2}-2} \cdot 3^{ \max \{ 0; n_{N,3}-2\} } \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}-1}$, the number of even or odd coefficients $q_2$ is equal to $\displaystyle 2^{n_{N,2}-2} \cdot 3^{n_{N,3}-1} \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}-1}$, and the number of even or odd coefficients $q_4$ is equal to $\displaystyle 2^{n_{N,2}-4} \cdot 3^{\max \{ 0; n_{N,3}-2\}} \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}-1}$ if $n_{N,2} \ge 4$, out of which one is zero. If $n_{N,2} = 3$, there are $\displaystyle 2 \cdot 3^{\max \{ 0; n_{N,3}-2\}} \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}-1}$ coefficients $q_3$, $\displaystyle 3^{\max \{ 0; n_{N,3}-2\}} \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}-1}$ coefficients $q_4$ (all even), and $\displaystyle 2 \cdot 3^{n_{N,3}-1} \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}-1}$ coefficients $q_2$ (all even). Therefore, if $n_{N,2} \ge 4$, the number of ZF 4-PPs will be equal to:

$$\begin{aligned}&C_{N, 4-PPs, ZF} \nonumber \\&\quad = 2^{n_{N,2}-1} \cdot 2 \cdot 3^{n_{N,3}-1} \cdot \prod _{j=3}^{s} \left( p_j^{n_{N,p_j}-1} \cdot (p_j - 1) \right) \nonumber \\&\qquad \cdot \left( 2^{n_{N,2}-2} \cdot 3^{n_{N,3}-1} \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \right) \nonumber \\&\qquad \cdot \left( 2^{n_{N,2}-2} \cdot 3^{ \max \{ 0; n_{N,3}-2\} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \right) \nonumber \\&\qquad \cdot \left( 2^{n_{N,2}-4} \cdot 3^{ \max \{ 0; n_{N,3}-2\} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \nonumber \\&\qquad +\,2^{n_{N,2}-1} \cdot 2 \cdot 3^{n_{N,3}-1} \cdot \prod _{j=3}^{s} \left( p_j^{n_{N,p_j}-1} \cdot (p_j - 1) \right) \nonumber \\&\qquad \cdot \left( 2^{n_{N,2}-2} \cdot 3^{n_{N,3}-1} \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \right) \\&\qquad \cdot \left( 2^{n_{N,2}-2} \cdot 3^{ \max \{ 0; n_{N,3}-2\} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \right) \nonumber \\&\qquad \cdot \left( 2^{n_{N,2}-4} \cdot 3^{ \max \{ 0; n_{N,3}-2\} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \right) \nonumber \\&\quad = 2^{3 \cdot n_{N,2} - 4} \cdot 3^{2 \cdot (n_{N,3}-1) + \max \{ 0; n_{N,3}-2\} } \nonumber \\&\qquad \cdot \prod _{j=3}^{s} \left( p_j^{3 \cdot (n_{N,p_j}-1)} \cdot (p_j - 1) \right) \nonumber \\&\qquad \cdot \left( 2^{n_{N,2}-3} \cdot 3^{ \max \{ 0; n_{N,3}-2\} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \nonumber \end{aligned}$$

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We mention that, if $n_{N,2} = 3$, the number of ZF 4-PPs will be equal to:

$$\begin{aligned} C_{N, 4-PPs, ZF}= & {} 2^{3-1} \cdot 2 \cdot 3^{n_{N,3}-1} \nonumber \\&\cdot \prod _{j=3}^{s} \left( p_j^{n_{N,p_j}-1} \cdot (p_j - 1) \right) \nonumber \\&\cdot \left( 2 \cdot 3^{n_{N,3}-1} \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \right) \nonumber \\&\cdot \left( 2 \cdot 3^{ \max \{ 0; n_{N,3}-2\} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \right) \\&\cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \nonumber \\= & {} 32 \cdot 3^{2 \cdot (n_{N,3}-1) + \max \{ 0; n_{N,3}-2\} } \nonumber \\&\cdot \prod _{j=3}^{s} \left( p_j^{3 \cdot (n_{N,p_j}-1)} \cdot (p_j - 1) \right) \nonumber \\&\quad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \nonumber \end{aligned}$$

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so that relation (37) is also true.

Appendix C

C.1 Proof [Theorem 2.8 (Updated Theorem 4.17 from [1])]

The number of possible coefficients $q_1$ is equal to $\displaystyle 2 \cdot 3^{n_{N,3}-1} \cdot \prod \nolimits _{j=2}^{s} \left( p_j^{n_{N,p_j}-1} \cdot (p_j - 1) \right) $, the number of coefficients $q_2$ is equal to $\displaystyle 3^{n_{N,3}-1} \cdot \prod \nolimits _{j=2}^{s} p_j^{n_{N,p_j}-1}$, and the number of coefficients $q_3$, $q_4$ and $q_5$, respectively, is equal to $\displaystyle 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod \nolimits _{j=2}^{s} p_j^{n_{N,p_j}-1}$, out of which one value is zero. By removing the value $q_5 = 0$, the number of ZF 5-PPs will be equal to:

$$\begin{aligned} \begin{aligned} C_{N, 5-PPs, ZF}&= 2 \cdot 3^{n_{N,3}-1} \cdot \prod _{j=2}^{s} \left( p_j^{n_{N,p_j}-1} \cdot (p_j - 1) \right) \\&\quad \cdot \left( 3^{n_{N,3}-1} \cdot \prod _{j=2}^{s} p_j^{n_{N,p_j}-1} \right) \\&\quad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod _{j=2}^{s} p_j^{n_{N,p_j}-1} \right) \\&\quad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod _{j=2}^{s} p_j^{n_{N,p_j}-1} \right) \\&\quad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod _{j=2}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \\&= 2 \cdot 3^{2 \cdot (n_{N,3}-1) + 2 \cdot \max \{ 0; n_{N,3}-2 \}} \\&\quad \cdot \prod _{j=2}^{s} \left( p_j^{4 \cdot (n_{N,p_j}-1)} \cdot (p_j - 1) \right) \\&\quad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod _{j=2}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \end{aligned} \end{aligned}$$

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C.2 Proof [Theorem 2.9 (Updated Theorem 4.19 from [1])]

The number of possible coefficients $q_1$ is equal to $\displaystyle 4 \cdot 5^{n_{N,5}-1} \cdot \prod \nolimits _{j=2}^{s} \left( p_j^{n_{N,p_j}-1} \cdot (p_j - 1) \right) $, the number of coefficients $q_2$, $q_3$, and $q_4$, respectively, is equal to $\displaystyle 5^{n_{N,5}-1} \cdot \prod \nolimits _{j=2}^{s} p_j^{n_{N,p_j}-1}$, and the number of coefficients $q_5$ is equal to $\displaystyle 5^{\max \{ 0; n_{N,5}-2 \} } \cdot \prod \nolimits _{j=2}^{s} p_j^{n_{N,p_j}-1}$, out of which one value is zero. By removing the value $q_5 = 0$, the number of ZF 5-PPs will be equal to:

$$\begin{aligned} C_{N, 5-PPs, ZF}&= 4 \cdot 5^{n_{N,5}-1} \cdot \prod _{j=2}^{s} \left( p_j^{n_{N,p_j}-1} \cdot (p_j - 1) \right) \nonumber \\&\quad \cdot \left( 5^{n_{N,5}-1} \cdot \prod _{j=2}^{s} p_j^{n_{N,p_j}-1} \right) \nonumber \\&\quad \cdot \left( 5^{n_{N,5}-1} \cdot \prod _{j=2}^{s} p_j^{n_{N,p_j}-1} \right) \nonumber \\&\quad \cdot \left( 5^{n_{N,5}-1} \cdot \prod _{j=2}^{s} p_j^{n_{N,p_j}-1} \right) \\&\quad \cdot \left( 5^{\max \{ 0; n_{N,5}-2 \} } \cdot \prod _{j=2}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \nonumber \\&= 4 \cdot 5^{4 \cdot (n_{N,5}-1)} \nonumber \\&\quad \cdot \prod _{j=2}^{s} \left( p_j^{4 \cdot (n_{N,p_j}-1)} \cdot (p_j - 1) \right) \nonumber \\&\quad \cdot \left( 5^{\max \{ 0; n_{N,5}-2 \} } \cdot \prod _{j=2}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \nonumber \end{aligned}$$

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C.3 Proof [Theorem 2.10 (Updated Theorem 4.20 from [1])]

The proof for Theorem 2.10 is similar to that of Theorem 2.8, by replacing N by N / 2. Therefore, the number of ZF 5-PPs will be equal to:

$$\begin{aligned} \begin{aligned} C_{N, 5-PPs, ZF}&= 2 \cdot 3^{n_{N,3}-1} \cdot \prod _{j=3}^{s} \left( p_j^{n_{N,p_j}-1} \cdot (p_j - 1) \right) \\&\quad \cdot \left( 3^{n_{N,3}-1} \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \right) \\&\quad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \right) \\&\quad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \right) \\&\quad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \\&= 2 \cdot 3^{2 \cdot (n_{N,3}-1) + 2 \cdot \max \{ 0; n_{N,3}-2 \}} \\&\quad \cdot \prod _{j=3}^{s} \left( p_j^{4 \cdot (n_{N,p_j}-1)} \cdot (p_j - 1) \right) \\&\quad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \end{aligned} \end{aligned}$$

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C.4 Proof [Theorem 2.11 (Updated Theorem 4.22 from [1])]

The proof for Theorem 2.11 is similar to that of Theorem 2.9, by replacing N by N / 2. Therefore, the number of ZF 5-PPs will be equal to:

$$\begin{aligned} \begin{aligned} C_{N, 5-PPs, ZF}&= 4 \cdot 5^{n_{N,5}-1} \cdot \prod _{j=3}^{s} \left( p_j^{n_{N,p_j}-1} \cdot (p_j - 1) \right) \\&\quad \cdot \left( 5^{n_{N,5}-1} \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \right) \\&\quad \cdot \left( 5^{n_{N,5}-1} \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \right) \\&\quad \cdot \left( 5^{n_{N,5}-1} \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \right) \\&\quad \cdot \left( 5^{\max \{ 0; n_{N,5}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \\&= 4 \cdot 5^{4 \cdot (n_{N,5}-1)} \\&\quad \cdot \prod _{j=3}^{s} \left( p_j^{4 \cdot (n_{N,p_j}-1)} \cdot (p_j - 1) \right) \\&\quad \cdot \left( 5^{\max \{ 0; n_{N,5}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \end{aligned} \end{aligned}$$

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C.5 Proof [Theorem 2.12 (Updated Theorem 4.23 from [1])]

The number of possible coefficients $q_1$ is equal to $\displaystyle 2 \cdot 2 \cdot 3^{n_{N,3}-1} \cdot \prod \nolimits _{j=3}^{s} \left( p_j^{n_{N,p_j}-1} \cdot (p_j - 1) \right) $, the number of coefficients $q_2$ is equal to $\displaystyle 3^{n_{N,3}-1} \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}-1}$, the number of coefficients $q_3$, $q_4$ and $q_5$, respectively, is equal to $\displaystyle 3^{\max \{ 0; n_{N,3}-2 \}} \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}-1}$, out of which one is zero. Therefore, the number of ZF 5-PPs will be equal to:

$$\begin{aligned} C_{N, 5-PPs, ZF}&= 4 \cdot 3^{n_{N,3}-1} \cdot \prod _{j=3}^{s} \left( p_j^{n_{N,p_j}-1} \cdot (p_j - 1) \right) \nonumber \\&\quad \cdot \left( 3^{n_{N,3}-1} \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \right) \nonumber \\&\quad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \right) \nonumber \\&\quad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \right) \\&\quad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} }\cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \nonumber \\&= 4 \cdot 3^{2 \cdot (n_{N,3}-1) + 2 \cdot \max \{ 0; n_{N,3}-2 \}} \nonumber \\&\quad \cdot \prod _{j=3}^{s} \left( p_j^{4 \cdot (n_{N,p_j}-1)} \cdot (p_j - 1) \right) \nonumber \\&\quad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \nonumber \end{aligned}$$

(43)

C.6 Proof [Theorem 2.13 (Updated Theorem 4.24 from [1])]

The number of possible coefficients $q_1$ is equal to $\displaystyle 2 \cdot 4 \cdot 3^{n_{N,3}-1} \cdot 5^{n_{N,5}-1} \cdot \prod \nolimits _{j=3}^{s} \left( p_j^{n_{N,p_j}-1} \cdot (p_j - 1) \right) $, the number of coefficients $q_2$ is equal to $\displaystyle 3^{n_{N,3}-1} \cdot 5^{n_{N,5}-1} \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}-1}$, the number of coefficients $q_3$ and $q_4$, respectively, is equal to $\displaystyle 3^{\max \{ 0; n_{N,3}-2 \}} \cdot 5^{n_{N,5}-1} \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}-1}$. The number of coefficients $q_5$ is equal to $\displaystyle 3^{\max \{ 0; n_{N,3}-2 \}} \cdot 5^{\max \{ 0; n_{N,5}-2 \} } \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}-1}$, out of which one is zero. By removing the value $q_5 = 0$, the number of ZF 5-PPs will be equal to:

$$\begin{aligned}&C_{N, 5-PPs, ZF} \nonumber \\&\quad = 8 \cdot 3^{n_{N,3}-1} \cdot 5^{n_{N,5}-1} \cdot \prod _{j=3}^{s} \left( p_j^{n_{N,p_j}-1} \cdot (p_j - 1) \right) \nonumber \\&\qquad \cdot \left( 3^{n_{N,3}-1} \cdot 5^{n_{N,5}-1} \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \right) \nonumber \\&\qquad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot 5^{n_{N,5}-1} \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \right) \\&\qquad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot 5^{n_{N,5}-1} \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \right) \nonumber \\&\qquad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot 5^{\max \{ 0; n_{N,5}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \nonumber \\&\quad = 8 \cdot 3^{2 \cdot (n_{N,3}-1) + 2 \cdot \max \{ 0; n_{N,3}-2 \}} \cdot 5^{4 \cdot (n_{N,5}-1)}\nonumber \\&\qquad \cdot \prod _{j=3}^{s} \left( p_j^{4 \cdot (n_{N,p_j}-1)} \cdot (p_j - 1) \right) \nonumber \\&\qquad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot 5^{\max \{ 0; n_{N,5}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \nonumber \end{aligned}$$

(44)

C.7 Proof [Theorem 2.14 (Updated Theorem 4.25 from [1])]

The number of possible coefficients $q_1$ is equal to $\displaystyle 2 \cdot 4 \cdot 5^{n_{N,5}-1} \cdot \prod \nolimits _{j=3}^{s} \left( p_j^{n_{N,p_j}-1} \cdot (p_j - 1) \right) $, the number of coefficients $q_2$, $q_3$ and $q_4$, respectively, is equal to $\displaystyle 5^{n_{N,5}-1} \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}-1}$, and the number of coefficients $q_5$ is equal to $\displaystyle 5^{\max \{ 0; n_{N,5}-2 \}} \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}-1}$, out of which one is zero. Therefore, the number of ZF 5-PPs will be equal to:

$$\begin{aligned} C_{N, 5-PPs, ZF}&= 8 \cdot 5^{n_{N,5}-1} \cdot \prod _{j=3}^{s} \left( p_j^{n_{N,p_j}-1} \cdot (p_j - 1) \right) \nonumber \\&\quad \cdot \left( 5^{n_{N,5}-1} \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \right) \nonumber \\&\quad \cdot \left( 5^{n_{N,5}-1} \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \right) \\&\quad \cdot \left( 5^{n_{N,5}-1} \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \right) \nonumber \\&\quad \cdot \left( 5^{\max \{ 0; n_{N,5}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \nonumber \\&{=} 8 \cdot 5^{4 \cdot (n_{N,5}{-}1)} \cdot \prod _{j{=}3}^{s} \left( p_j^{4 \cdot (n_{N,p_j}{-}1)} \cdot (p_j {-} 1) \right) \nonumber \\&\quad \cdot \left( 5^{\max \{ 0; n_{N,5}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \nonumber \end{aligned}$$

(45)

C.8 Proof [Theorem 2.15 (Updated Theorem 4.26 from [1])]

The number of possible coefficients $q_1$ is equal to $\displaystyle 2^{n_{N,2}-1} \cdot 2 \cdot 3^{n_{N,3}-1} \cdot \prod \nolimits _{j=3}^{s} \left( p_j^{n_{N,p_j}-1} \cdot (p_j - 1) \right) $, the number of even or odd coefficients $q_2$ is equal to $\displaystyle 2^{n_{N,2}-2} \cdot 3^{n_{N,3}-1} \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}-1}$, the number of even or odd coefficients $q_3$ is equal to $\displaystyle 2^{n_{N,2}-2} \cdot 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}-1}$, and the number of even or odd coefficients $q_4$ and $q_5$, respectively, is equal to $\displaystyle 2^{n_{N,2}-4} \cdot 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}-1}$ if $n_{N,2} \ge 4$, out of which one is zero. If $n_{N,2} = 3$, there are $\displaystyle 2\cdot 3^{n_{N,3}-1} \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}-1}$ coefficients $q_2$, all of them even, $\displaystyle 2\cdot 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}-1}$ coefficients $q_3$, all of them even, and $\displaystyle 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}-1}$ coefficients $q_4$ and $q_5$, respectively, all of them even. Therefore, if $n_{N,2} \ge 4$, the number of ZF 5-PPs will be equal to:

$$\begin{aligned}&C_{N, 5-PPs, ZF} \nonumber \\&\quad = 2 \cdot 2^{n_{N,2}-1} \cdot 2 \cdot 3^{n_{N,3}-1} \nonumber \\&\qquad \cdot \prod _{j=3}^{s} \left( p_j^{n_{N,p_j}-1} \cdot (p_j - 1) \right) \nonumber \\&\qquad \cdot \left( 2^{n_{N,2}-2} \cdot 3^{n_{N,3}-1} \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \right) \nonumber \\&\qquad \cdot \left( 2^{n_{N,2}-4} \cdot 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1}\right) \nonumber \\&\qquad \cdot \left\{ \left( 2^{n_{N,2}-2} \cdot 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1}\right) \right. \nonumber \\&\quad \qquad \cdot \left( 2^{n_{N,2}-4} \cdot 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \\&\qquad + \left( 2^{n_{N,2}-2} \cdot 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \right) \nonumber \\&\qquad \left. \cdot \left( 2^{n_{N,2}-4} \cdot 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \right) \right\} \nonumber \\&\quad = 2^{4 \cdot n_{N,2} - 7} \cdot 3^{2 \cdot (n_{N,3}-1) + 2 \cdot \max \{ 0; n_{N,3}-2 \}} \nonumber \\&\qquad \cdot \prod _{j=3}^{s} \left( p_j^{4 \cdot (n_{N,p_j}-1)} \cdot (p_j - 1) \right) \nonumber \\&\qquad \cdot \left( 2^{n_{N,2}-3} \cdot 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \nonumber \end{aligned}$$

(46)

If $n_{N,2} = 3$, the number of ZF 5-PPs will be equal to:

$$\begin{aligned} C_{N, 5-PPs, ZF}&= 4 \cdot 2 \cdot 3^{n_{N,3}-1} \nonumber \\&\quad \cdot \prod _{j=3}^{s} \left( p_j^{n_{N,p_j}-1} \cdot (p_j - 1) \right) \nonumber \\&\quad \cdot \left( 2 \cdot 3^{n_{N,3}-1} \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \right) \nonumber \\&\quad \cdot \left( 2 \cdot 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1}\right) \\&\quad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1}\right) \nonumber \\&\quad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} -1 \right) \nonumber \\&= 32 \cdot 3^{2 \cdot (n_{N,3}-1)+ 2 \cdot \max \{ 0; n_{N,3}-2 \}} \nonumber \\&\quad \cdot \prod _{j=3}^{s} \left( p_j^{4 \cdot (n_{N,p_j}-1)} \cdot (p_j - 1) \right) \nonumber \\&\quad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} - 1 \right) ,\nonumber \end{aligned}$$

(47)

so that relation (46) is also true.

C.9 Proof [Theorem 2.16 (Updated Theorem 4.27 from [1])]

The proof for Theorem 2.16 is similar to that of Theorem 2.13, by replacing N by N / 2. Thus, the number of ZF 5-PPs will be equal to:

$$\begin{aligned} \begin{aligned}&C_{N, 5-PPs, ZF} \\&\quad = 8 \cdot 3^{n_{N,3}-1} \cdot 5^{n_{N,5}-1} \cdot \prod _{j=4}^{s} \left( p_j^{n_{N,p_j}-1} \cdot (p_j - 1) \right) \\&\qquad \cdot \left( 3^{n_{N,3}-1} \cdot 5^{n_{N,5}-1} \cdot \prod _{j=4}^{s} p_j^{n_{N,p_j}-1} \right) \\&\qquad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot 5^{n_{N,5}-1} \cdot \prod _{j=4}^{s} p_j^{n_{N,p_j}-1}\right) \\&\qquad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot 5^{n_{N,5}-1} \cdot \prod _{j=4}^{s} p_j^{n_{N,p_j}-1}\right) \\&\qquad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot 5^{\max \{ 0; n_{N,5}-2 \} } \cdot \prod _{j=4}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \\&\quad = 8 \cdot 3^{2 \cdot (n_{N,3}-1) + 2 \cdot \max \{ 0; n_{N,3}-2 \}} \cdot 5^{4 \cdot (n_{N,5}-1)} \\&\qquad \cdot \left( \prod _{j=4}^{s} p_j^{4 \cdot (n_{N,p_j}-1)} \cdot (p_j - 1) \right) \\&\qquad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot 5^{\max \{ 0; n_{N,5}-2 \} } \cdot \prod _{j=4}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \end{aligned} \end{aligned}$$

(48)

C.10 Proof [Theorem 2.17 (Updated Theorem 4.28 from [1])]

The number of possible coefficients $q_1$ is equal to $\displaystyle 2^{n_{N,2}-1} \cdot 4 \cdot 5^{n_{N,5}-1} \cdot \prod \nolimits _{j=3}^{s} \left( p_j^{n_{N,p_j}-1} \cdot (p_j - 1) \right) $, the number of even or odd coefficients $q_2$ and $q_3$, respectively, is equal to $\displaystyle 2^{n_{N,2}-2}\cdot 5^{n_{N,5}-1} \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}-1}$, the number of even or odd coefficients $q_4$ is equal to $\displaystyle 2^{n_{N,2}-4} \cdot 5^{n_{N,5}-1} \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}-1}$, and the number of even or odd coefficients $q_5$ is equal to $\displaystyle 2^{n_{N,2}-4} \cdot 5^{\max \{ 0; n_{N,5}-2 \} } \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}-1}$ if $n_{N,2} \ge 4$, out of which one value is zero. If $n_{N,2} = 3$, there are $\displaystyle 2 \cdot 5^{n_{N,5}-1} \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}-1}$ coefficients $q_2$ and $q_3$, respectively, all of them even, $\displaystyle 5^{n_{N,5}-1} \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}-1}$ coefficients $q_4$, all of them even, and $\displaystyle 5^{\max \{ 0; n_{N,5}-2 \} } \cdot \prod \nolimits _{j=3}^{s} p_j^{n_{N,p_j}-1}$ coefficients $q_5$, all of them even, out of which one is zero. Therefore, if $n_{N,2} \ge 4$, the number of ZF 5-PPs will be equal to:

$$\begin{aligned} \begin{aligned}&C_{N, 5-PPs, ZF} \\&\quad = 2 \cdot 2^{n_{N,2}-1} \cdot 4 \cdot 5^{n_{N,5}-1} \cdot \prod _{j=3}^{s} \left( p_j^{n_{N,p_j}-1} \cdot (p_j - 1) \right) \\&\qquad \cdot \left( 2^{n_{N,2}-2}\cdot 5^{n_{N,5}-1} \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \right) \\&\qquad \cdot \left( 2^{n_{N,2}-4}\cdot 5^{n_{N,5}-1} \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \right) \\&\qquad \cdot \left\{ \left( 2^{n_{N,2}-2}\cdot 5^{n_{N,5}-1} \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \right) \right. \\&\quad \qquad \cdot \left( 2^{n_{N,2}-4}\cdot 5^{\max \{ 0; n_{N,5}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1}- 1 \right) \\&\quad \qquad + \left( 2^{n_{N,2}-2}\cdot 5^{n_{N,5}-1} \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \right) \\&\quad \qquad \left. \cdot \left( 2^{n_{N,2}-4}\cdot 5^{\max \{ 0; n_{N,5}-2 \}} \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \right) \right\} \\&\quad = 2^{4 \cdot n_{N,2}-6} \cdot 5^{4 \cdot (n_{N,1}-1)} \\&\qquad \cdot \prod _{j=3}^{s} \left( p_j^{4 \cdot (n_{N,p_j}-1)} \cdot (p_j - 1) \right) \\&\qquad \cdot \left( 2^{n_{N,2}-3} \cdot 5^{\max \{ 0; n_{N,5}-2 \}} \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \end{aligned} \end{aligned}$$

(49)

If $n_{N,2} = 3$, the number of ZF 5-PPs will be equal to:

$$\begin{aligned} C_{N, 5-PPs, ZF}&= 4 \cdot 4 \cdot 5^{n_{N,5}-1} \nonumber \\&\quad \cdot \prod _{j=3}^{s} \left( p_j^{n_{N,p_j}-1} \cdot (p_j - 1) \right) \nonumber \\&\quad \cdot \left( 2 \cdot 5^{n_{N,5}-1} \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \right) \nonumber \\&\quad \cdot \left( 2 \cdot 5^{n_{N,5}-1} \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \right) \nonumber \\&\quad \cdot \left( 5^{n_{N,5}-1} \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} \right) \\&\quad \cdot \left( 5^{\max \{ 0; n_{N,5}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \nonumber \\&= 64 \cdot 5^{4 \cdot (n_{N,1}-1)} \nonumber \\&\quad \cdot \prod _{j=3}^{s} \left( p_j^{4 \cdot (n_{N,p_j}-1)} \cdot (p_j - 1) \right) \nonumber \\&\quad \cdot \left( 5^{\max \{ 0; n_{N,5}-2 \} } \cdot \prod _{j=3}^{s} p_j^{n_{N,p_j}-1} - 1 \right) ,\nonumber \end{aligned}$$

(50)

so that relation (49) is also true.

C.11 Proof [Theorem 2.18 (Updated Theorem 4.29 from [1])]

The proof for Theorem 2.18 is similar to that of Theorem 2.13, by replacing N by N / 4 and multiplying the number of coefficients $q_1$ by 2. The number of ZF 5-PPs will be equal to:

$$\begin{aligned}&C_{N, 5-PPs, ZF} \nonumber \\&\quad = 16 \cdot 3^{n_{N,3}-1} \cdot 5^{n_{N,5}-1} \cdot \prod _{j=4}^{s} \left( p_j^{n_{N,p_j}-1} \cdot (p_j - 1) \right) \nonumber \\&\qquad \cdot \left( 3^{n_{N,3}-1} \cdot 5^{n_{N,5}-1} \cdot \prod _{j=4}^{s} p_j^{n_{N,p_j}-1} \right) \nonumber \\&\qquad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot 5^{n_{N,5}-1} \cdot \prod _{j=4}^{s} p_j^{n_{N,p_j}-1} \right) \nonumber \\&\qquad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot 5^{n_{N,5}-1} \cdot \prod _{j=4}^{s} p_j^{n_{N,p_j}-1} \right) \\&\qquad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot 5^{\max \{ 0; n_{N,5}-2 \} } \cdot \prod _{j=4}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \nonumber \\&\quad = 16 \cdot 3^{2 \cdot (n_{N,3}-1) + 2 \cdot \max \{ 0; n_{N,3}-2 \}} \cdot 5^{4 \cdot (n_{N,5}-1)}\nonumber \\&\qquad \cdot \prod _{j=4}^{s} \left( p_j^{4 \cdot (n_{N,p_j}-1)} \cdot (p_j - 1) \right) \nonumber \\&\qquad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot 5^{\max \{ 0; n_{N,5}-2 \} } \cdot \prod _{j=4}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \nonumber \end{aligned}$$

(51)

C.12 Proof [Theorem 2.19 (Updated Theorem 4.30 from [1])]

The number of possible coefficients $q_1$ is equal to $\displaystyle 2^{n_{N,2}-1} \cdot 2 \cdot 3^{n_{N,3}-1} \cdot 4 \cdot 5^{n_{N,5}-1} \cdot \prod \nolimits _{j=4}^{s} \left( p_j^{n_{N,p_j}-1} \cdot (p_j - 1) \right) $, the number of even or odd coefficients $q_2$ is equal to $\displaystyle 2^{n_{N,2}-2} \cdot 3^{n_{N,3}-1} \cdot 5^{n_{N,5}-1} \cdot \prod \nolimits _{j=4}^{s} p_j^{n_{N,p_j}-1}$, and the number of even or odd coefficients $q_3$ is equal to $\displaystyle 2^{n_{N,2}-2} \cdot 3^{\max \{ 0; n_{N,3}-2 \} } \cdot 5^{n_{N,5}-1} \cdot \prod \nolimits _{j=4}^{s} p_j^{n_{N,p_j}-1}$. The number of even or odd coefficients $q_4$ is equal to $\displaystyle 2^{n_{N,2}-4} \cdot 3^{\max \{ 0; n_{N,3}-2 \} } \cdot 5^{n_{N,5}-1} \cdot \prod \nolimits _{j=4}^{s} p_j^{n_{N,p_j}-1}$ if $n_{N,2} \ge 4$. If $n_{N,2} = 3$, the number of coefficients $q_4$ is equal to $\displaystyle 3^{\max \{ 0; n_{N,3}-2 \} } \cdot 5^{n_{N,5}-1} \cdot \prod \nolimits _{j=4}^{s} p_j^{n_{N,p_j}-1}$, all of them even. The number of even or odd coefficients $q_5$ is equal to $\displaystyle 2^{n_{N,2}-4} \cdot 3^{\max \{ 0; n_{N,3}-2 \} } \cdot 5^{\max \{ 0; n_{N,5}-2 \} } \cdot \prod \nolimits _{j=4}^{s} p_j^{n_{N,p_j}-1}$ if $n_{N,2} \ge 4$. If $n_{N,2} = 3$, the number of coefficients $q_5$ is equal to $\displaystyle 3^{\max \{ 0; n_{N,3}-2 \} } \cdot 5^{\max \{ 0; n_{N,5}-2 \} } \cdot \prod \nolimits _{j=4}^{s} p_j^{n_{N,p_j}-1}$, all of them even, out of which one is zero. By removing the value $q_5 = 0$, if $n_{N,2} \ge 4$, the number of ZF 5-PPs will be equal to:

$$\begin{aligned}&C_{N, 5-PPs, ZF} \nonumber \\&\quad = 2 \cdot 8 \cdot 2^{n_{N,2}-1} \cdot 3^{n_{N,3}-1} \cdot 5^{n_{N,5}-1} \nonumber \\&\qquad \cdot \prod _{j=4}^{s} \left( p_j^{n_{N,p_j}-1} \cdot (p_j - 1) \right) \nonumber \\&\qquad \cdot \left( 2^{n_{N,2}-2} \cdot 3^{n_{N,3}-1} \cdot 5^{n_{N,5}-1} \cdot \prod _{j=4}^{s} p_j^{n_{N,p_j}-1} \right) \nonumber \\&\qquad \cdot \left( 2^{n_{N,2}-4} \cdot 3^{\max \{ 0; n_{N,3}-2 \} } \cdot 5^{n_{N,5}-1} \cdot \prod _{j=4}^{s} p_j^{n_{N,p_j}-1}\right) \nonumber \\&\qquad \cdot \left\{ \left( 2^{n_{N,2}-2} \cdot 3^{\max \{ 0; n_{N,3}-2 \} } \cdot 5^{n_{N,5}-1} \cdot \prod _{j=4}^{s} p_j^{n_{N,p_j}-1}\right) \right. \nonumber \\&\quad \qquad \cdot \left( 2^{n_{N,2}-4} \cdot 3^{\max \{ 0; n_{N,3}-2 \} } \cdot 5^{\max \{ 0; n_{N,5}-2 \} } \cdot \prod _{j=4}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \\&\quad \qquad + \left( 2^{n_{N,2}-2} \cdot 3^{\max \{ 0; n_{N,3}-2 \} } \cdot 5^{n_{N,5}-1} \cdot \prod _{j=4}^{s} p_j^{n_{N,p_j}-1} \right) \nonumber \\&\quad \qquad \left. \cdot \left( 2^{n_{N,2}-4} \cdot 3^{\max \{ 0; n_{N,3}-2 \} } \cdot 5^{\max \{ 0; n_{N,5}-2 \} } \cdot \prod _{j=4}^{s} p_j^{n_{N,p_j}-1} \right) \right\} \nonumber \\&\quad = 2^{4 \cdot n_{N,2} - 5} \cdot 3^{2 \cdot (n_{N,3}-1) + 2 \cdot \max \{ 0; n_{N,3}-2 \}} \cdot 5^{4 \cdot (n_{N,5}-1)} \nonumber \\&\qquad \cdot \prod _{j=4}^{s} \left( p_j^{4 \cdot (n_{N,p_j}-1)} \cdot (p_j - 1) \right) \nonumber \\&\qquad \cdot \left( 2^{ n_{N,2}- 3} \cdot 3^{\max \{ 0; n_{N,3}-2 \} } \cdot 5^{\max \{ 0; n_{N,5}-2 \} } \cdot \prod _{j=4}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \nonumber \end{aligned}$$

(52)

If $n_{N,2} = 3$, the number of ZF 5-PPs will be equal to:

$$\begin{aligned}&C_{N, 5-PPs, ZF} \nonumber \\&\quad = 8 \cdot 4 \cdot 3^{n_{N,3}-1} \cdot 5^{n_{N,5}-1} \cdot \prod _{j=4}^{s} \left( p_j^{n_{N,p_j}-1} \cdot (p_j - 1) \right) \nonumber \\&\qquad \cdot \left( 2 \cdot 3^{n_{N,3}-1} \cdot 5^{n_{N,5}-1} \cdot \prod _{j=4}^{s} p_j^{n_{N,p_j}-1} \right) \nonumber \\&\qquad \cdot \left( 2 \cdot 3^{\max \{ 0; n_{N,3}-2 \} } \cdot 5^{n_{N,5}-1} \cdot \prod _{j=4}^{s} p_j^{n_{N,p_j}-1}\right) \nonumber \\&\qquad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot 5^{n_{N,5}-1} \cdot \prod _{j=4}^{s} p_j^{n_{N,p_j}-1}\right) \\&\qquad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot 5^{\max \{ 0; n_{N,5}-2 \} } \cdot \prod _{j=4}^{s} p_j^{n_{N,p_j}-1} - 1 \right) \nonumber \\&\quad = 128 \cdot 3^{2 \cdot (n_{N,3}-1) + 2 \cdot \max \{ 0; n_{N,3}-2 \}} \cdot 5^{4 \cdot (n_{N,5}-1)} \nonumber \\&\qquad \cdot \prod _{j=4}^{s} \left( p_j^{4 \cdot (n_{N,p_j}-1)} \cdot (p_j - 1) \right) \nonumber \\&\qquad \cdot \left( 3^{\max \{ 0; n_{N,3}-2 \} } \cdot 5^{\max \{ 0; n_{N,5}-2 \} } \cdot \prod _{j=4}^{s} p_j^{n_{N,p_j}-1} - 1 \right) ,\nonumber \end{aligned}$$

(53)

so that relation (52) is also true.

Appendix D

D.1 Example for Theorem 2.3 (Updated example for Theorem 4.6 from [1])

Let the interleaver length be $N = 2100 = 2^2 \cdot 3 \cdot 5^2 \cdot 7$. According to (11), the number of ZF CPPs will be equal to:

$$\begin{aligned} \begin{aligned} C_{2100, CPPs, ZF}&= 2^{2 \cdot (2-1)} \cdot 3^{2 \cdot (1-1)} \cdot 5^{2 \cdot (2-1)} \cdot (5 - 1) \\&\quad \cdot 7^{2 \cdot (1-1)} \cdot (7 - 1) \\&\quad \cdot \left( 2^{2-2} \cdot 3^{\max \{0; 1-2 \} } \cdot 5^{2-1} \cdot 7^{1-1} - 1 \right) \\&= 4 \cdot 1 \cdot 25 \cdot 4 \cdot 1 \cdot 6 \cdot \left( 1 \cdot 3^0 \cdot 5 \cdot 1 - 1 \right) \\&= 2400 \cdot (5 - 1) = 9600. \end{aligned} \end{aligned}$$

Further, we detail the coefficients of the 9600 true different ZF CPPs. As $6 \mid N$ and $9 \not \mid N$, it is required that $q_3 < 3 \cdot N/6 = N/2 = 1050$ and $q_2 < N/2 = 1050$. The coefficients $q_2$ and $q_3$ have to be multiples of $2 \cdot 3 \cdot 5 \cdot 7 = 210$, and coefficient $q_1$ has to be relatively prime with 2100. Therefore, coefficient $q_3$ can take 4 values: 210, 420, 630 and 840, $q_2$ can take 5 values: 0, 210, 420, 630 and 840, and coefficient $q_1$ can take $\displaystyle \Phi (2100) = 2100 \cdot \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{4}{5} \cdot \frac{6}{7} = 480$ values. The total number of ZF CPPs is equal to $4 \cdot 5 \cdot 480 = 9600$, just the number found by the formula we determined. We note that the number of all true different CPP is also 9600. However, the equivalence conditions for CPPs, in this case, are: $q_3 < N/6 =350$ and $q_2 < N/2 = 1050$. We note that every ZF CPP having coefficient $q_3 > 350$ (i.e., 420, 630, or 840) has an equivalent CPP having coefficient $q_3 < 350$ and coefficient $q_2 < 1050$. For example, ZF CPP $\pi _{CPP, ZF}(x) = 1 \cdot x + 0 \cdot x^2 + 420 \cdot x^3 \ (\mathrm {mod}\ 2100)$ is equivalent to CPP $\pi _{CPP}(x) = 351 \cdot x + 0 \cdot x^2 + 70 \cdot x^3 \ (\mathrm {mod}\ 2100)$, since $\pi _{CPP}(x) = \pi _{CPP, ZF}(x) + z(x) \ (\mathrm {mod}\ 2100)$ only for NP $z(x) = 350 \cdot x + 0 \cdot x^2 + 1750 \cdot x^3 \ (\mathrm {mod}\ 2100)$. This NP is not a ZF NP since $3 \not \mid 1750$.

D.2 Example 2 for Theorem 2.19 (Updated example 2 for Theorem 4.30 from [1])

Let the interleaver length be $N = 2400 = 2^5 \cdot 3 \cdot 5^2$. According to (29), the number of ZF 5-PPs will be equal to:

$$\begin{aligned} \begin{aligned} C_{2400, 5-PPs, ZF}&= 2^{4 \cdot 5 - 5} \cdot 3^{2 \cdot (1-1) + 2 \cdot \max \{ 0; 1-2 \}} \cdot 5^{4 \cdot (2-1)} \\&\quad \cdot \left( 2^{ 5- 3} \cdot 3^{\max \{ 0; 1-2 \}} \cdot 5^{\max \{ 0; 2-2 \}} - 1 \right) \\&= 32768 \cdot 1 \cdot 625 \cdot ( 4 \cdot 1 \cdot 1 - 1 ) \\&= 61440000. \end{aligned} \end{aligned}$$

In this case, since $9 \not \mid N$ and $600 \mid N$, it is required that $q_5 < N/40 = 60$, $q_4 < N/8 = 300$, $q_3 < N/2 = 1200$ and $q_2 < N/2 = 1200$. Coefficients $q_5$, $q_4$, $q_3$, and $q_2$ have to be multiples of $3 \cdot 5 = 15$. The other conditions coefficients $q_5$, $q_4$, $q_3$ and $q_2$ have to meet are the same as those for $N = 25200$ (in Example 1 for Theorem 4.30 from [1]), and coefficient $q_1$ has to be relatively prime with 2400. Therefore, the pair of coefficients $(q_5, q_3)$ can take the values (15, 15), (15, 45), (15, 75), $\dots $, (15, 1185), (45, 15), (45, 45), (45, 75), $\dots $, (45, 1185) ($2 \cdot 40 = 80$ pairs with both coefficients odd) and (30, 0), (30, 30), (30, 60), $\dots $ (30, 1170) (40 pairs with both coefficients even), i.e., a total of 120 pairs. The pair of coefficients $(q_4, q_2)$ can take the values (15, 15), (15, 45), (15, 75), $\dots $, (15, 1185), (45, 15), (45, 45), (45, 75), $\dots $, (45, 1185), (75, 15), (75, 45), (75, 75), $\dots $, (75, 1185), $\dots $, (285, 15), (285, 45), (285, 75), $\dots $, (285, 1185) ($10 \cdot 40 = 400$ pairs with both coefficients odd) and (0, 0), (0, 30), (0, 60), $\dots $, (0, 1170), (30, 0), (30, 30), (30, 60), $\dots $, (30, 1170), (60, 0), (60, 30), (60, 60), $\dots $, (60, 1170), (90, 0), (90, 30), (90, 60), $\dots $, (90, 1170), $\dots $, (270, 0), (270, 30), (270, 60), $\dots $, (270, 1170) ($10 \cdot 40 = 400$ pairs with both coefficients even), i.e., a total of 800 pairs. Coefficient $q_1$ can take $\displaystyle \Phi (2400) = 2400 \cdot \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{4}{5} = 640$ values. The total number of ZF 5-PPs is equal to $120 \cdot 800 \cdot 640 = 61440000$, just the number we found by the formula we determined.

We note that the equivalence conditions for 5-PPs in this case are: $q_5 < N/120 = 20$, $q_4 < N/24 = 100$, $q_3 < N/6 = 400$ and $q_2 < N/2 = 1200$. We note that every ZF 5-PP having coefficient $q_5 > 20$ (i.e., 30 or 45) and/or coefficient $q_4 > 100$ and/or coefficient $q_3 > 400$ has an equivalent 5-PP having coefficients $q_5 < 20$, $q_4 < 100$, $q_3 < 400$, and $q_2 < 1200$. For example, ZF 5-PP $\pi _{5-PP, ZF}(x) = 1 \cdot x + 0 \cdot x^2 + 1185 \cdot x^3 + 270 \cdot x^4 + 45 \cdot x^5 \ (\mathrm {mod}\ 2400)$ is equivalent to 5-PP $\pi _{5-PP}(x) = 1041 \cdot x + 200 \cdot x^2 + 185 \cdot x^3 + 70 \cdot x^4 + 5 \cdot x^5 \ (\mathrm {mod}\ 2400)$ since $\pi _{5-PP}(x) = \pi _{5-PP, ZF}(x) + z(x) \ (\mathrm {mod}\ 2400)$ only for NP $z(x) = 1040 \cdot x + 200 \cdot x^2 + 1400 \cdot x^3 + 2200 \cdot x^4 + 2360 \cdot x^5 \ (\mathrm {mod}\ 2400)$. This NP is not a ZF NP since $3 \not \mid 2360$, $3 \not \mid 2200$, $3 \not \mid 1400$, and $3 \not \mid 200$.

References

1.
Trifina, L., & Tarniceriu, D. (2016). The number of different true permutation polynomial based interleavers under Zhao and Fan sufficient conditions. Telecommunication Systems, 63(4), 593–623.CrossRefGoogle Scholar
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Zhao, H., & Fan, P. (2007). Simple method for generating $m$th-order permutation polynomials over integer rings. IEE Electronics Letters, 43(8), 449–451.CrossRefGoogle Scholar
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Ryu, J. (2007). Permutation polynomial based interleavers for turbo codes over integer rings, Ph.D. thesis, [Online] Available https://etd.ohiolink.edu/rws_etd/document/get/osu1181139404/inline.
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3GPP TS 36.212 V8.3.0, 3rd Generation partnership project, multiplexing and channel coding (Release 8), 2008. [Online] http://www.etsi.org.
5.
Trifina, L., & Tarniceriu, D. (2018). Determining the number of true different permutation polynomials of degrees up to five by Weng and Dong algorithm. Telecommunication Systems, 67(2), 211–215.CrossRefGoogle Scholar

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Authors and Affiliations

Lucian Trifina
- 1
Email authorView author's OrcID profile
Daniela Tarniceriu
- 1

1.Department of Telecommunications and Information Technologies, Faculty of Electronics, Telecommunications and Information Technology“Gheorghe Asachi” Technical University of IasiIasiRomania

1.a	\(p = 2\)	\(n_{N,2}=1\)	\((q_1 + q_2 + \dots + q_n) \ne 0 \ (\mathrm {mod}\ 2)\)
1.b		\(n_{N,2}>1\)	\(q_1 \ne 0\), \(q_2+q_4+q_6+ \dots = 0 \ (\mathrm {mod}\ 2)\) and \(q_3+q_5+q_7+ \dots = 0 \ (\mathrm {mod}\ 2)\)
2	\(p > 2\)	\(n_{N,p} \ge 1\)	\(q_1 \ne 0, q_2 = q_3 = \dots = q_n = 0 \ (\mathrm {mod}\ p)\)

Correction to: The number of different true permutation polynomial based interleavers under Zhao and Fan sufficient conditions

1 Correction to: Telecommun Syst (2016) 63:593–623 https://doi.org/10.1007/s11235-016-0144-8

2 Null polynomials under Zhao and Fan sufficient conditions

3 Corrections for formulas in Sects. 4.1, 4.2, and 4.3 from [1] considering null polynomials under Zhao and Fan sufficient conditions

3.1 Corrections for formulas in Sect. 4.1 from [1]

Theorem 2.1

Proof

Theorem 2.2

Proof

Theorem 2.3

Proof

3.2 Corrections for formulas in Sect. 4.2 from [1]

Theorem 2.4

Proof

Theorem 2.5

Proof

Theorem 2.6

Proof

Theorem 2.7

Proof

3.3 Corrections for formulas in Sect. 4.3 from [1]

Theorem 2.8

Proof

Theorem 2.9

Proof

Theorem 2.10

Proof

Theorem 2.11

Proof

Theorem 2.12

Proof

Theorem 2.13

Proof

Theorem 2.14

Proof

Theorem 2.15

Proof

Theorem 2.16

Proof

Theorem 2.17

Proof

Theorem 2.18

Proof

Theorem 2.19

Proof

References

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