## Abstract

Romanov proved that the proportion of positive integers which can be represented as a sum of a prime and a power of 2 is positive. We establish similar results for integers of the form \(n=p+2^{2^k}+m!\) and \(n=p+2^{2^k}+2^q\) where \(m,k \in \mathbb {N}\) and *p*, *q* are primes. In the opposite direction, Erdős constructed a full arithmetic progression of odd integers none of which is the sum of a prime and a power of two. While we also exhibit in both cases full arithmetic progressions which do not contain any integers of the two forms, respectively, we prove a much better result for the proportion of integers not of these forms: (1) The proportion of positive integers not of the form \(p+2^{2^k}+m!\) is larger than \(\frac{3}{4}\). (2) The proportion of positive integers not of the form \(p+2^{2^k}+2^q\) is at least \(\frac{2}{3}\).

## Keywords

Romanov’s theorem Smooth numbers Diophantine equation Sumsets## Mathematics Subject Classification

Primary 11P32 Secondary 11A07 11A41 11N36## 1 Introduction

*p*is a prime and \(g \ge 2\) is a positive integer. As there are about Open image in new window primes \(p \le x\) and Open image in new window powers \(g^k \le x\), this result implicitly gives some information about the number

*r*(

*n*) of representations of \(n=p+g^k\). There are not too many integers \(n \le x\) with a very large number of representations and on average

*r*(

*n*) is bounded. The most prominent special case of Romanov’s result is the one concerning sums of primes and powers of 2. Euler [9] observed in a letter to Goldbach that 959 can not be written as the sum of a prime and a power of two. Euler’s letter was also mentioned by de Polignac [3] and provides a counter example to a conjecture of de Polignac himself, stating that any odd positive integer is the sum of a prime and a power of 2. In 1950, Erdős [5] and van der Corput [18] independently proved that also the lower density of odd integers not of the form \(p+2^k\) is positive. Here and in the following the lower density of a set \(\mathcal {A} \subset \mathbb {N}\) is defined to beReplacing \(\liminf \) with \(\limsup \) leads to what we call upper density and if lower and upper density coincide we speak of the density of the set \(\mathcal {A}\).

Concerning Romanov’s theorem one may ask how this result can be generalized. One way would be by replacing the sequence of powers of *g* with another sequence \((a_n)_{n \ge 1}\). Generalizing a result of Lee [13] who replaced the powers of *g* by the Fibonacci sequence, Ballot and Luca [1] proved an analogue of Romanov’s theorem for the case when \((a_n)_{n \ge 1}\) is a linearly recurrent sequence with certain additional properties. For certain quadratic recurrences \((a_n)_{n\ge 1}\) this was done by Dubickas [4].

*c*, one can write a positive proportion of integers \(n \le x\) as \(n=p+a\),

*p*prime and \(a \in \mathcal {A}\). In this paper we study sets \(\mathcal {A}\) with \(|\mathcal {A}\cap [1,x]| \sim c_{\mathcal {A}} \log x\) but of a quite different nature compared to previous ones. In particular, we study

### Theorem 1

The lower density of integers of the form \(p+2^{2^k}+m!\) for \(k,m \in \mathbb {N}_0\) and *p* prime is positive.

### Theorem 2

The lower density of integers of the form \(p+2^{2^k}+2^q\) for \(k \in \mathbb {N}_0\) and *p*, *q* prime is positive.

Concerning integers not of the form \(p+2^{2^k}+m!\) we consider two different questions: The first one is finding a large set, in the sense of lower density, of odd positive integers not of this form.

The second question is if there is a full arithmetic progression of odd positive integers not of the form \(p+2^{2^k}+m!\). The positive answer to this question is given in Theorem 4. Note that, the density of the set constructed in the proof of Theorem 4 is considerably less than the density of the set used in the proof of Theorem 3.

### Theorem 3

The lower density of odd positive integers not of the form \(p+2^{2^k}+m!\) for \(k,m \in \mathbb {N}_0\) and *p* prime is at least Open image in new window . The lower density of all positive integers without a representation of the form \(p+2^{2^k}+m!\) is therefore larger than Open image in new window .

### Theorem 4

There exists a full arithmetic progression of odd positive integers not of the form \(p+2^{2^k}+m!\) for \(k,m \in \mathbb {N}_0\) and *p* prime.

Finally, we prove analogous results for integers not of the form \(p+2^{2^k}+2^q\).

### Theorem 5

There exists a subset of the odd positive integers not of the form \(p+2^{2^k}+2^q\), for \(k \in \mathbb {N}\) and *p*, *q* prime, with lower density Open image in new window . The lower density of all positive integers without a representation of the form \(p+2^{2^k}+2^q\) is therefore larger than Open image in new window .

Furthermore, there exists a full arithmetic progression of odd positive integers not of the form \(p+2^{2^k}+2^q\).

Concerning the last result, we recall that Erdős conjectured that the lower density of the set of positive odd integers not of the form \(p+2^k+2^m\) is positive for \(k,m \in \mathbb {N}_0\), *p* prime (see for example [10, Sect. A19]).

*n*in the form \(p+2^{2^k}+m!\), and \(r_2(n)\) counts the number of representations of

*n*in the form \(p+2^{2^k}+2^q\). Note that the first sum on the left-hand side of Eq. (1) equals the number of integers less than

*x*having a representation of the required form. It thus suffices to check that

## 2 Notation

Let \(\mathbb {N}\), as usual, denote the set of positive integers, \(\mathbb {N}_0\) the set of non-negative integers and let \(\mathbb {P}\) denote the set of primes. The variables *p* and *q* will always denote prime numbers. For any prime \(p \in \mathbb {P}\) and any positive integer \(n \in \mathbb {N}\), let \(\nu _p(n)\) denote the *p*-adic valuation of *n*, i.e. \(\nu _p(n)=k\) where \(p^k\) is the highest power of *p* dividing *n*. For an integer *n*, *P*(*n*) denotes its largest prime factor. For any set \(S \subset \mathbb {N}\) let \(S(x) = |S \cap [1,x]|\) denote the counting function of *S*. As usual \(\varphi \) denotes Euler’s totient function and \(\mu \) the Möbius function. Furthermore, for an odd positive integer *n* we denote by *t*(*n*) the order of \(2 \bmod n\). We use the symbols \(\ll \), \(\gg \), \(\mathcal {O}\) and *o* within the context of the well-known Vinogradov and Landau notation.

## 3 Integers of the form \(p+2^{2^k}+m!\)

Before proving Lemmas 3 and 4, we establish and collect several results needed in due course. The following is a classical result due to Legendre (see for example Theorems 2.6.1 and 2.6.4 in [14]).

### Lemma A

*p*digits of

*n*, then

### Theorem 6

### Proof

### Theorem 7

If we exclude solutions arising from interchanging \((x_1,y_1)\) and \((x_2,y_2)\), the equation \(2^{x_1}+y_1!=2^{x_2}+y_2!\) has only four non-negative integer solutions \((x_1,y_1,x_2,y_2)\) with \((x_1,y_1) \ne (x_2,y_2)\) and \((y_1,y_2) \not \in \{(1,0),(0,1)\}\) if \(x_1=x_2\). These are the solutions presented in Theorem 6.

### Proof

We compare the 2-adic and 3-adic valuation of both sides of equivalent forms of the equation \(2^{x_1}+y_1!=2^{x_2}+y_2!\) to get information about the size of the parameters \(x_1,x_2,y_1\) and \(y_2\).

*p*. Using the above bound for

*k*and the fact that \(y_1 \le 2x_2\), we getNext we find an upper bound for \(x_1\) in terms of \(x_2\). Consider the equation

### Lemma 1

### Proof

### Lemma 2

*n*, let

*t*(

*n*) be the order of \(2 \bmod n\) and \(t(n)=2^{a(n)}b(n)\) such that

*b*(

*n*) is odd. Then the series

### Proof

*P*(

*n*) denotes the largest prime factor of

*n*and observe that if

*u*|

*v*then

*t*(

*u*)|

*t*(

*v*), thus

*b*(

*u*)|

*b*(

*v*) and further

*t*(

*b*(

*u*))|

*t*(

*b*(

*v*)). From this and Mertens’ formula in the weak form

*y*and

*u*tend to infinity. For \(p \in \mathcal {P}_2\) we may suppose that Open image in new window since there are at most Open image in new window primes in \(\mathcal {P}_2\) less than \(\sqrt{x}\). If Open image in new window , then Open image in new window for sufficiently large

*x*, and hence for Open image in new window in \({\mathcal P}_2\), we havePut Open image in new window . Thus, \(p-1\) is a number which is at most

*x*, having a divisor Open image in new window , whose largest prime factor is at most

*y*. It follows that \(p-1\le x\) is a multiple of some number Open image in new window with \(P(d)\le y\). For a fixed

*d*, the number of such

*p*is at most Open image in new window . Summing over

*d*, we get thatPutting Open image in new window , we get that Open image in new window for all Open image in new window , and

*x*. Using (14) and (15), we thus get that

*p*such that \(p-1\) is divisible by some prime Open image in new window but \(q \in \mathcal {P}_1\). We may assume again that Open image in new window , then Open image in new window , where as before Open image in new window . Fixing

*q*, the number of primes \(p \le x\) such that \(p-1\) is a multiple of

*q*is at most Open image in new window . Summing up over \(q \in \mathcal {P}_1\) and using (13), we get thatFinally, choose \(p_0\) such that for \(p>p_0\) we have that Open image in new window and get

*q*|

*b*(

*p*) for large

*p*. Furthermore, \(q \not \in \mathcal {P}_1\) so Open image in new window . By the choice of the constant \(p_0\) in the definition of \(\mathcal {P}_4\) this implies that \(t(b(p))\ge t(q) >(\log p)^3\). Finally, this implies that

### Lemma 3

### Proof

*x*. This implies thatThe bounds in (16), (17) and (18) show that

### Lemma 4

### Proof

^{1}. The number of choices of the form \((p_1,p_2,k_1,k_2,m_1,m_2)\) in this case is

*h*is odd, then one of the primes \(p_1\) and \(p_2\) equals 2 and any choice of \((k_1,k_2,m_1,m_2)\) fixes the other prime. There are

*d*, the cardinality of the set

*d*is odd this implies that \(2^{2^{k_1}-2^{k_2}}\equiv 1\pmod d\). Recall that

*t*(

*d*) is the order of 2 modulo

*d*. The above congruence makes \(2^{k_1}\equiv 2^{k_2}\pmod {t(d)}\). As above we write \(t(d)=2^{a(d)} b(d)\), where

*b*(

*d*) is odd and

*a*(

*d*) is some non-negative integer. This implies that \(2^{k_1-k_2}\equiv 1\pmod {b(d)}\). The above cancellation again is justified since

*b*(

*d*) is odd. Hence, for \(k_2\) fixed, \(k_1\) is in a fixed arithmetic progression modulo

*t*(

*b*(

*d*)). The number of such \(k_1\) with \(2^{2^{k_1}}\le x\) is of order (up to a constant) at most

*d*in the range of summation into two different sets

*A*and

*B*. We set

*A*we thus collect all

*d*for which all solutions in \(S_d\backslash S_{1,d}\) give the same

*h*and the set

*B*contains all other

*d*. For \(d \in A\) we fix \(k_1\) and \(k_2\) for solutions in \(S_d\backslash S_{1,d}\) and get

*d*and thus \(d|h-h'\). Every prime factor of

*d*(in particular the ones larger than \(\log x\)) divideswhere the product is taken over all \(m_i\) with \(m_i!\le x\) and all \(k_i\) with \(2^{2^{k_i}}\le x\) for \(i=1,2,3,4\). The dash indicates that the product is to be taken over the non-zero factors only. Since each factor in this product is of size \(\mathcal {O}(x)\) any of these factors has at most \(\mathcal {O}(\log x)\) prime factors. Furthermore, for the octuple \((k_1,k_2,k_3,k_4,m_1,m_2,m_3,m_4)\) we have Open image in new window choices and altogether we have that \(|\mathcal {P}|=\mathcal {O}((\log x)^5)\). Write \(d=u_dv_d\), where \(u_d\) is divisible by primes \(p \le \log x\) only. Hence the factor \(v_d\) is divisible only by primes in \(\mathcal {P}\). Then

*t*(

*b*(

*u*)). The number of \(k_1\) with \(2^{2^{k_1}}\le x\) in this progression is of order Open image in new window . Thus, we have

### Proof of Theorem 3

*a*is a residue class \(\bmod \, 2^{2^6}-1\) with \((a,2^{2^6}-1)>1\), then \((n-2^{2^k}-m!,2^{2^6}-1)>1\) which leaves only finitely many choices for the prime \(p=n-2^{2^k}-m!\). This implies that the proportion of such

*n*with a representation of the form \(n=p+2^{2^k}+m!\) is zero. We have \(2^{2^6}-1-\varphi (2^{2^6}-1)\) choices for the residue class

*a*and half of the integers in these residue classes are odd which yields a density of

### Proof of Theorem 4

*n*satisfying the following system of congruences is of the form \(p+2^{2^k}+m!:\)

*n*be an element of this progression and suppose that \(n=p+2^{2^k}+m!\).

If \(m \ge 3\), then \(n=p+2^{2^k}+m! \equiv p+2^{2^k} \bmod 3\). All primes except for 3 are in the residue classes \(1,2 \bmod 3\) and \(2^{2^k} \equiv 1 \bmod 3\) for \(k \ge 1\). Thus, for \(m \ge 3\) and \(k \ge 1\) we have that \(n = p+2^{2^k}+m! \equiv 1 \bmod 3\); hence, the only possible choice for *p* is \(p=3\).

Next, we show that if \(p=3\), then \(m<5\). To do so, we use that \(2^{2^k}\equiv 1 \bmod 5\) for \(k\ge 2\); hence for \(m \ge 5\) we are left with \(n=3+2^{2^k}+m! \equiv \{0,2,4\} \bmod 5\), a contradiction to \(n \equiv 3 \bmod 5\).

In the case that \(k=0\), we will show that \(m \ge 3\) implies \(m<7\). Let \(n=p+2+m!\) and \(m \ge 3\). Then \(n \equiv 1 \bmod 3\) implies that \(p \equiv 2 \bmod 3\). If additionally \(m \ge 7\), then \(n=p+2+m! \equiv p+2 \bmod 7\). Since \(n \equiv 2 \bmod 7\), the only possible choice for *p* is \(p=7\), which contradicts \(p \equiv 2 \bmod 3\).

Using the above observations, the only cases we need to consider are those of \(m=0\), \(m=1\), \(m=2\), \(m=3,4\) and \(k=0\) or \(p=3\) and \(m=5,6\) and \(k=0\).

If \(m\in \{0,1\}\) and we additionally have that *p* is odd, then \(n=p+2^{2^k}+1\) is even, a contradiction to \(n \equiv 1 \bmod 2\). It remains to deal with the case when \(p=2\). Then we have \(n=2+2^{2^k}+1\) and we get a contradiction from \(n \equiv 3 \bmod 5\) which would imply that \(2^{2^k} \equiv 0 \bmod 5\).

For the case \(m=2\), we use that \(2^{2^k} \equiv 1 \bmod 17\) for \(k\ge 3\). Hence, for \(m=2\) and \(k \ge 3\), we have that \(n=p+2^{2^k}+2 \equiv p+3 \bmod 17\) which together with \(n \equiv 3 \bmod 17\) leaves us with \(p=17\). We use that \(n=17+2^{2^k}+2 \equiv 2 \bmod 3\) to get a contradiction to \(n \equiv 1 \bmod 3\). Since \(m=2\) and \(k=0\) imply \(n=p+4 \equiv p+1 \bmod 3\), the only possible choice for *p* in this case is \(p=3\) but \(n=7 \not \equiv 3 \bmod 5\). If \(m=2\) and \(k=1,\) then \(n=p+6\) and \(n \equiv 6 \bmod 11\) implies that \(p=11\). This contradicts \(n \equiv 1 \bmod 3\). Last we need to deal with \(m=2\) and \(k=2\). In this case, \(n=p+18 \equiv p+3 \bmod 5\), and hence, \(n \equiv 3 \bmod 5\) implies that \(p=5\). Now \(n=23\) does not satisfy the congruence \(n \equiv 1 \bmod 3\).

If \(m=3\) and \(p=3\) we have that \(n=9+2^{2^k} \equiv {8,10,11,13} \bmod 17\) contradicting \(n \equiv 3 \bmod 17\). On the other hand, if \(m=3\) and \(k=0\), then \(n=p+8 \equiv p+3 \bmod 5\) and we get a contradiction as shown above.

For \(m=4\) and \(p=3\) we get \(n=27+2^{2^k} \equiv \{9,11,12,14\} \bmod 17\), a contradiction to \(n \equiv 3 \bmod 17\). If \(m=4\) and \(k=0\), it follows that \(n=p+26 \equiv p+7 \bmod 19\) which implies \(p=19\) and \(n=45\). This contradicts \(n \equiv 3 \bmod 5\).

In the case when \(m=5\) and \(k=0\), we have that \(n=p+122 \equiv p+3 \bmod 17\). Together with \(n \equiv 3 \bmod 17\) this only leaves \(p=17\) which contradicts \(n \equiv 3 \bmod 5\).

Finally, if \(m=6\) and \(k=0\), then \(n=p+722 \equiv p+9 \bmod 23\). Together with \(n \equiv 9 \bmod 23\), this only leaves \(p=23\) which yields a contradiction to \(n \equiv 3 \bmod 5\). \(\square \)

## 4 Integers of the form \(p+2^{2^k}+2^q\)

### Lemma 5

### Proof

### Lemma 6

### Proof

*h*is even, we may directly use the sieve bound from [15, Theorem 7.3] which, after summing over all

*h*, yields an upper bound of orderfor the sum in the lemma, where the dash indicates that \((k_1,q_1) \ne (k_2,q_2)\). Noting that the contribution of the prime 2 is just a constant factor, we disregard it. Furthermore \(h \le x\) by definition, and a very crude upper bound for the number of prime factors of

*h*, in particular for those larger than \(\log x\), is given by Open image in new window . We thus getIf we fix \(k_1,q_1\) and \(k_2\), then the fact that \(d\mid h\) implies

*l*is a fixed residue class \(\bmod \ d\). This puts \(q_2\) in a fixed residue class \(\bmod \ t(d)\). Since we are counting representations of integers \(n \le x\), we have Open image in new window . Hence if \(t(d) > \log x\) there are at most two choices for \(q_2\). If \(t(d) \le \log x\), the Brun–Titchmarsh inequality yields an upper bound offor the number of choices of \(q_2\). We thus get an upper bound of the following order for (23)As earlier, by Mertens’ formula

### Proof of Theorem 5

We prove the theorem by showing that the subset of positive integers in the residue class \(3 \bmod 6\) having a representation of the form \(p+2^{2^k}+2^q\) has density 0.

*n*is in none of the sets

*n*has a representation of the form \(n=p+2^{2^k}+2^q\), then

*n*is in one of the residue classes

*n*satisfying the congruences

*n*is in the arithmetic progression constructed above. We use that except for \(q\in \{2,3\}\), we have that \(q \equiv \{1,5,7,11\} \bmod 12\) and that for any \(l\in \mathbb {N}_0\) we have that

*p*is \(p=7\). Then \(9+2^{12l+7} \equiv 2 \bmod 5\), a contradiction to \(n \equiv 4 \bmod 5\). Finally if \(q=12l+11\), then \(n = p+2+2^{12l+11} \equiv p+9 \bmod 13\) and from \(n \equiv 9 \bmod 13\) we get \(p=13\). Since \(n=15+2^{12l+11} \equiv 3 \bmod 5\), we again get a contradiction to \(n \equiv 4 \bmod 5\). To finish off the integers in the set \(S_1\), it remains to deal with \(q \in \{2,3\}\). If \(q=2\) we have \(n=p+6 \equiv p \bmod 6\). Since \(n \equiv 3 \bmod 6\), we are left with \(p=3\) and \(n=9\) which contradicts to \(n \equiv 4 \bmod 7\). If \(q=3\) then \(n=p+10\) and from \(n \equiv 10 \bmod 37\), we need to have that \(p=37\), and hence \(n=47\). This is impossible since it contradicts to \(n \equiv 4 \bmod 5\).

Next, we deal with the integers in \(S_2\) and we use that \(2^{2^k} \equiv 1 \bmod 17\) for \(k \ge 3\). Thus, for \(k \ge 3\) and \(n=p+2^{2^k}+4 \in S_2\) we have that \(n=p+2^{2^k}+4 \equiv p + 5 \bmod 17\). From \(n \equiv 5 \bmod 17\), we see that the only admissible choice for *p* is \(p=17\), and hence, \(n=21+2^{2^k}\). As above we use that \(2^{2^k} \equiv \{2,4\} \bmod 6\) and thus \(21+2^{2^k} \equiv \{1,5\} \bmod 6\) a contradiction to \(n \equiv 3 \bmod 6\). We are left with \(k \in \{0,1,2\}\). For \(k=0\), we get \(n=p+6\) which was ruled out when we dealt with the integers in \(S_1\). If \(k=1\) we have \(n=p+8\) and from \(n\equiv 8 \bmod 19\), the only possible choice for *p* is \(p=19\) and thus \(n=27\). This contradicts to \(n \equiv 4 \bmod 5\). Finally, if \(k=2\) we have \(n=p+20\) and from \(n \equiv 20 \bmod 23\) we again are left with a single possible choice for *p*, namely \(p=23\). Now \(n=43\), contradicting to \(n \equiv 4 \bmod 5\).

For integers *n* in the set \(S_3\), we have \(n=2+2^{2^k}+2^q\). If \(q=2\) we have \(n \equiv 2^{2^k} \bmod 6\) and again using that \(2^{2^k} \in \{2,4\} \bmod 6\), we get a contradiction to \(n \equiv 3 \bmod 6\). If *q* is odd, then \(2^q \equiv 2 \bmod 6\). If furthermore \(k=0\), then \(n = 4+2^q \equiv 0 \bmod 6\), and if \(k=1\), we get \(n=6+2^q \equiv 2 \bmod 6\). In both cases this yields a contradiction to \(n \equiv 3 \bmod 6\). For \(k \ge 2\) and *q* odd, we have that \(2^{2^k} \equiv \{16,24,25\} \bmod 29\) and \(2^{q} \equiv \{2,3,8,10,11,12,14,15,17,18,19,21,26,27\} \bmod 29\). For \(k \ge 2\) and *q* odd, it is thus true that \(2^{2^k} +2^q \not \equiv 0 \bmod 29\) and thus \(n=2+2^{2^k}+2^q \equiv 2 \bmod 29\) yields a contradiction in this case.

Finally, for integers in the set \(S_4\) we apply a similar argument as for integers in the set \(S_3\). For any prime *q* we have that \(2^{q} \equiv \{1,2,4,8,16\} \bmod 31\), and for all \(k \in \mathbb {N}_0\) we get \(2^{2^k}\equiv \{2,4,8,16\} \bmod 31\). Again \(2^{2^k}+2^q \not \equiv 0 \bmod 31\) for any prime *q* and any non-negative integer *k*. Thus, \(n=3+2^{2^k}+2^q \equiv 3 \bmod 31\) yields a contradiction. \(\square \)

## Footnotes

## Notes

### Acknowledgements

Open access funding provided by Austrian Science Fund (FWF). Parts of this research work were done when the second author was visiting the Institute of Analysis and Number Theory at Graz University of Technology and the Max Planck Institute for Mathematics in Bonn, Germany. He wants to thank these institutions and Professor Robert Tichy. Furthermore, we would like to thank the referee for carefully reading the manuscript.

## References

- 1.Ballot, C., Luca, F.: On the sumset of the primes and a linear recurrence. Acta Arith.
**161**(1), 33–46 (2013)MathSciNetCrossRefGoogle Scholar - 2.Canfield, E.R., Erdős, P., Pomerance, C.: On a problem of Oppenheim concerning “factorisatio numerorum”. J. Number Theory
**17**(1), 1–28 (1983)MathSciNetCrossRefGoogle Scholar - 3.de Polignac, A.: Recherches nouvelles sur les nombres premiers. C. R. Hebd. Séances Acad. Sci.
**29**, 397–401 and 738–739 (1849)Google Scholar - 4.Dubickas, A.: Sums of primes and quadratic linear recurrence sequences. Acta Math. Sin. (Engl. Ser.)
**29**(12), 2251–2260 (2013)MathSciNetCrossRefGoogle Scholar - 5.Erdős, P.: On integers of the form \(2^k+p\) and some related problems. Summa Brasil. Math.
**2**, 113–123 (1950)MathSciNetGoogle Scholar - 6.Erdős, P., Murty, M.R.: On the order of \(a\,(\text{ mod } \, p)\). CRM Proceedings & Lecture Notes, vol. 19. American Mathematical Society, Providence, RI (1999)zbMATHGoogle Scholar
- 7.Erdős, P., Turán, P.: Ein zahlentheoretischer Satz. Mitt. Forsch.-Inst. Math. Mech. Univ. Tomsk
**1**, 101–103 (1935)zbMATHGoogle Scholar - 8.Erdős, P., Turán, P.: Über die Vereinfachung eines Landauschen Satzes. Mitt. Forsch.-Inst. Math. Mech. Univ. Tomsk
**1**, 144–147 (1935)zbMATHGoogle Scholar - 9.Euler, L.: Letter to Goldbach, 16.12.1752. http://eulerarchive.maa.org/correspondence/letters/OO0879.pdf
- 10.Guy, R.K.: Unsolved Problems in Number Theory. Problem Books in Mathematics, 3rd edn. Springer, New York (2004)CrossRefGoogle Scholar
- 11.Habsieger, L., Roblot, X.F.: On integers of the form \(p+2^k\). Acta Arith.
**122**(1), 45–50 (2006)MathSciNetCrossRefGoogle Scholar - 12.Jacobi, C.G.J.: Canon arithmeticus sive tabulae quibus exhibentur pro singulis numeris primis vel primorum potestatibus infra 1000 numeri ad datos indices et indices ad datos numeros pertinentes. Berolinum (1839)Google Scholar
- 13.Lee, K.S.E.: On the sum of a prime and a Fibonacci number. Int. J. Number Theory
**6**(7), 1669–1676 (2010)MathSciNetCrossRefGoogle Scholar - 14.Moll, V.H.: Numbers and Functions: From a Classical-Experimental Mathematician’s Point of View. American Mathematical Society, Providence, RI (2012)CrossRefGoogle Scholar
- 15.Nathanson, M.B.: Additive Number Theory—The Classical Bases. Graduate Texts in Mathematics, vol. 164. Springer, New York (1996)CrossRefGoogle Scholar
- 16.Romanoff, N.P.: Über einige Sätze der additiven Zahlentheorie. Math. Ann.
**109**(1), 668–678 (1934)MathSciNetCrossRefGoogle Scholar - 17.Rosser, J.B., Schoenfeld, L.: Approximate formulas for some functions of prime numbers. Ill. J. Math.
**6**, 64–94 (1962)MathSciNetzbMATHGoogle Scholar - 18.van der Corput, J.G.: Over het vermoeden van de Polignac. Simon Stevin
**27**, 99–105 (1950)MathSciNetGoogle Scholar

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