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The Ramanujan Journal

, Volume 47, Issue 2, pp 267–289 | Cite as

Romanov type problems

  • Christian Elsholtz
  • Florian Luca
  • Stefan Planitzer
Open Access
Article
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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 pq 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

An old result of Romanov [16] states that a positive proportion of the positive integers can be written in the form \(p+g^k\), where 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].

We would expect that for many sets \(\mathcal {A} \subset \mathbb {N}\), with \(|\mathcal {A}\cap [1,x]| \ge c\log x\) for some positive constant 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
$$\begin{aligned} \mathcal {A}_1&=\{2^{2^k}+m!:k,m \in \mathbb {N}_0\}, \\ \mathcal {A}_2&=\{2^{2^k}+2^q: k \in \mathbb {N}_0, q \text { prime}\}. \end{aligned}$$
Using the machinery of Romanov [16], we prove the following two theorems.

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 pq 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 pq 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]).

For the proofs of Theorems 1 and 2, we apply the method of Romanov [16]. This means that we start with the Cauchy–Schwarz inequality in the form
$$\begin{aligned} \left( \sum _{\begin{array}{c} n \le x \\ r_i(n)>0 \end{array}}1\right) \left( \sum _{n \le x}r_i(n)^2 \right) \ge \left( \sum _{n \le x}r_i(n)\right) ^2 \end{aligned}$$
(1)
for \(i \in \{1,2\}\), where \(r_1(n)\) denotes the number of representations of 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
$$\begin{aligned} \sum _{n \le x}r_i(n) \gg x \text { and } \sum _{n \le x}r_i(n)^2 \ll x \end{aligned}$$
for both \(i=1,~2\) in order to get positive lower density for the sets of those integers. The estimates \(\sum _{n \le x}r_1(n) \gg x\) and \(\sum _{n \le x}r_1(n)^2 \ll x\) are proved in Sect. 3, Lemmas 3 and 4, respectively. The analogous results for \(r_2(n)\) are proved in Sect. 4, Lemmas 5 and 6, respectively. Theorems 3 and 4 are proved at the end of Sect. 3 and Theorem 5 at the end of Sect. 4.

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

For any prime \(p \in \mathbb {P}\) and any positive integer \(n \in \mathbb {N}\), we have that
$$\begin{aligned} \nu _p(n!) = \sum _{k=1}^{\infty }\left\lfloor \frac{n}{p^k}\right\rfloor . \end{aligned}$$
Furthermore, if \(\sigma _p(n)\) denotes the sum of base p digits of n, then
$$\begin{aligned} \nu _p(n!) = \frac{n-\sigma _p(n)}{p-1}. \end{aligned}$$

Theorem 6

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 \(x_1>x_2\) where either \(x_2 \le 52\) or \(y_2 \le 8\). These solutions are
$$\begin{aligned} (x_1,y_1,x_2,y_2) \in \{(1,0,0,2),(1,1,0,2),(3,2,2,3),(7,4,5,5)\}. \end{aligned}$$

Proof

Suppose that \(x_2 \le 52\) and note that \(y_1=0\) either implies that \(y_2\in \{0,1\}\) if \(x_2>0\), which leads to a solution where \(x_1=x_2\), which is excluded, or implies that \(x_2=0\), whence \(x_1=1\) and \(y_2=2\). Hence, the only solution where \(y_1=0\) is \((x_1,y_1,x_2,y_2)=(1,0,0,2)\). From now on, we may suppose that \(y_1 \ge 1\). In this case, from Lemma A, we get that \(\nu _2(y_1!) \ge \frac{y_1}{2}-1\). This yields \(\frac{y_1}{2}-1 \le x_2\) and thus \(y_1 \le 106\). Since
$$\begin{aligned} 2^{x_2}-y_1!=2^{x_1}-y_2!, \end{aligned}$$
we have \(\nu _2(2^{x_2}-y_1!)=\nu _2(2^{x_1}-y_2!)\). Certainly \(|2^{x_2}-y_1!| \le 2^{52}+106!\) which implies that \(\nu _2(2^{x_2}-y_1!) \le \frac{\log (2^{52}+106!)}{\log 2}<816\). If \(x_1\ge 816\) and \(y_2 \ge 822\), then \(\nu _2(2^{x_1}-y_2!)\ge 816\), a contradiction. The cases where either \(x_1 \le 815\) or \(y_2 \le 821\) can be checked by a computer search which leads to the solutions
$$\begin{aligned} (x_1,y_1,x_2,y_2) \in \{(1,0,0,2),(1,1,0,2),(3,2,2,3),(7,4,5,5)\}. \end{aligned}$$
Now suppose that \(y_2 \le 8\) and consider
$$\begin{aligned} 0<2^{x_1}-2^{x_2}=y_2!-y_1!, \end{aligned}$$
which implies that \(y_1 \le y_2 \le 8\). In particular, \(|y_2!-y_1!|\le 2\cdot 8!\) and thus
$$\begin{aligned} \nu _2(y_2!-y_1!) \le \frac{\log (2\cdot 8!)}{\log 2} < 17. \end{aligned}$$
Since \(\nu _2(2^{x_1}-2^{x_2})=x_2\), we have that \(x_2 < 17\) which is included in the case \(x_2 \le 52\) treated above. \(\square \)

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 \(y_2\).

If \(x_1=x_2\) we have that \(y_1!=y_2!\) and hence either \(y_1=y_2\) or \((y_1,y_2) \in \{(1,0),(0,1)\}\) which leads to the excluded trivial solutions. Therefore, w.l.o.g., we may suppose that \(x_1>x_2\) and write
$$\begin{aligned} 2^{x_2}(2^{x_1-x_2}-1)=y_1!((y_1+1) \cdots y_2-1). \end{aligned}$$
(2)
Next we compute the 2-adic valuation of both sides of the last equality. For the left-hand side we simply have \(\nu _2(2^{x_2}(2^{x_1-x_2}-1))=x_2\) while for the right-hand side we use that the factor \(((y_1+1)\cdots y_2-1)\) is odd as soon as \(y_2 \ge y_1+2\) which yields
$$\begin{aligned} \nu _2(y_1!((y_1+1)\cdots y_2-1)) ={\left\{ \begin{array}{ll} \nu _2(y_1!), &{}\text { if } y_2 \ge y_1+2, \\ \nu _2(y_1!)+\nu _2(y_1), &{}\text { if } y_2=y_1+1. \end{array}\right. } \end{aligned}$$
From this, Lemma A and the fact that \(1 \le \sigma _2(y_1) \le \frac{\log y_1}{\log 2}+1\) (note that as in the proof of Theorem 6, \(y_1 \in \{0,1\}\) leads to a single non-trivial solution listed there), we get the following two inequalities:
$$\begin{aligned} x_2=\nu _2(2^{x_2}(2^{x_1-x_2}-1))&=\nu _2(y_1!((y_1+1)\cdots y_2-1)) \le \nu _2(y_1!)+\nu _2(y_1) \nonumber \\&<y_1+\frac{\log y_1}{\log 2}, \end{aligned}$$
(3)
$$\begin{aligned} x_2=\nu _2(2^{x_2}(2^{x_1-x_2}-1))&=\nu _2(y_1!((y_1+1)\cdots y_2-1)) \ge \nu _2(y_1!) \nonumber \\&\ge y_1-\left( \frac{\log y_1}{\log 2}+1\right) . \end{aligned}$$
(4)
By Theorem 6, we may suppose that \(x_2 \ge 5\) without loosing solutions. In this case, the last inequality implies \(y_1 \le 2x_2\).
Next, we look at
$$\begin{aligned} 2^{x_1}=2^{x_2}+y_2!-y_1!. \end{aligned}$$
Since \(2^{x_2} \le 2^{x_1-1}=\frac{2^{x_1}}{2}\), we have that \(y_2!>\frac{2^{x_1}}{2}\), whence we get
$$\begin{aligned} y_2^{y_2} \ge y_2!>\frac{2^{x_1}}{2}, \end{aligned}$$
and thus
$$\begin{aligned} y_2\log y_2> (x_1-1)\log 2 \text { and } y_2>\frac{(x_1-1)\log 2}{\log y_2}. \end{aligned}$$
To get the last inequality we used that by Theorem 6 we may suppose that \(y_2 \ge 9\) whence \(\log y_2 >0\). Now \(x_2 \ge 5\) implies that \(x_1 \ge 6\). If we would have that \(y_2 \le x_1\) the last inequality would imply that
$$\begin{aligned} y_2> \frac{\log 2}{2}\left( \frac{x_1}{\log y_2}\right) >\frac{1}{4}\left( \frac{x_1}{\log x_1}\right) . \end{aligned}$$
(5)
In order to prove (5), it therefore suffices to prove that \(y_2 \le x_1\) for \(x_1 \ge 6\). In order to do so, we consider the equation
$$\begin{aligned} 2^{x_1}=y_1!((y_1+1)\cdots y_2-1)+2^{x_2} \end{aligned}$$
from which we readily deduce that \(y_1! <2^{x_1}\). This together with \(2^{x_1}=y_2!-y_1!+2^{x_2}\) implies that
$$\begin{aligned} y_2!<2\cdot 2^{x_1}. \end{aligned}$$
This implies that \(y_2 \le x_1\), since otherwise \((x_1+1)! \le 2^{x_1+1}\) which is true for \(x_1 \le 2\) only. By Theorem 6 again, we may suppose that \(y_2 \ge 9\). In this case, applying Lemma A, we obtain
$$\begin{aligned} \nu _3(y_2!) \ge \left\lfloor \frac{y_2}{3}\right\rfloor +\left\lfloor \frac{y_2}{9} \right\rfloor \ge \frac{y_2}{3} > \frac{1}{12}\left( \frac{x_1}{\log x_1}\right) , \end{aligned}$$
(6)
where the last inequality follows by (5). Now we compute the 3-adic valuation of both sides of Eq. (2). By inequality (3) and Lemma A for the right-hand side, we get
$$\begin{aligned} k=\nu _3(y_1!((y_1+1)\cdots y_2-1))&\ge \nu _3(y_1!)=\frac{y_1-\sigma _3(y_1)}{2}\ge \frac{y_1}{2}-\frac{\log y_1}{\log 3}-1 \\&\ge \frac{x_2}{2}-\log (y_1)\left( \frac{1}{2\log 2}+\frac{1}{\log 3}\right) -1. \end{aligned}$$
Since for the left-hand side of (2) we have \(3^k|2^{x_1-x_2}-1\), we deduce that \(\varphi (3^k)=2\cdot 3^{k-1}|x_1-x_2\). Here we used that 2 is a primitive root modulo any power of 3. This is a direct consequence of Jacobi’s observation [12, p. XXXV] that a primitive root modulo \(p^2\) is also a primitive root modulo any higher power of 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
$$\begin{aligned} 2^{x_1}-y_2!=2^{x_2}-y_1!. \end{aligned}$$
Equation (5) yields that \(y_2> \frac{1}{4}\frac{x_1}{\log x_1}>\frac{1}{4} \sqrt{x_1}.\) Thus, by Lemma A, \(\nu _2(y_2!)>\frac{\sqrt{x_1}}{8}-1\) and hence \(\nu _2(2^{x_1}-y_2!)\ge \frac{\sqrt{x_1}}{8}-1\).
On the other hand, \(|2^{x_2}-y_1!|\le 2^{x_2}+y_1! \le 2^{x_2}+(2x_2)^{2x_2}\le 2\cdot (2x_2)^{2x_2}\). Now \(\nu _2(2^{x_2}-y_1!)\) is certainly bounded from above by the highest power of 2 less than \(2\cdot (2x_2)^{2x_2}\):
$$\begin{aligned} 2^a \le 2\cdot (2x_2)^{2x_2} \Leftrightarrow a \le \frac{2x_2\log (2x_2)}{\log 2}+1. \end{aligned}$$
We therefore have that \(\nu _2(2^{x_2}-y_1!) \le 4x_2\log (2x_2)+1\) and putting everything together, we get
$$\begin{aligned} \frac{\sqrt{x_1}}{8}-1 \le \nu _2(2^{x_1}-y_2!)=\nu _2(2^{x_2}-y_1!)\le 4x_2\log (2x_2)+1, \end{aligned}$$
which implies that \(x_1 \le (32x_2\log (2x_2)+16)^2\). Combining this with (7), we finally arrive atThis inequality is valid only for \(x_2 \le 52\) and the solutions satisfying this restriction are given in Theorem 6. \(\square \)

Lemma 1

Let \(m_1,m_2,m_3,m_4 \in \mathbb {N}\) such that \(m_1>m_2\), \(m_3>m_4\) and
$$\begin{aligned} m_1!-m_2!=m_3!-m_4!. \end{aligned}$$
(8)
Then \((m_1,m_2)=(m_3,m_4)\) or \(m_1=m_3\) and \((m_2,m_4) \in \{(0,1),(1,0)\}\).

Proof

We start with the case where either \(m_1=m_2+1\) or \(m_3=m_4+1\) and w.l.o.g suppose that \(m_1=m_2+1\). If furthermore \(m_2 \le m_4\), we get from Eq. (8)
$$\begin{aligned} m_2!\, m_2=m_4!((m_4+1)\cdots m_3-1) \ge m_2!m_4, \end{aligned}$$
which leads to \(m_2 \ge m_4\) and thus \(m_2=m_4\) which implies \(m_1=m_3\). On the other hand, if \(m_1=m_2+1\) and \(m_2>m_4\) Eq. (8) implies that
$$\begin{aligned} m_2(m_4+1)\cdots m_2=(m_4+1) \cdots m_3-1, \end{aligned}$$
(9)
and therefore \(m_4+1|1\) if \(m_3>m_4+1\) and \(m_4+1|m_4\) otherwise, whence \(m_4=0\) in both cases. Now \(m_3=1\) implies that \((m_1,m_2)=(1,0)\) and we are done. Otherwise, if \(m_3 \ne 1\), then the right-hand side of (9) is odd. In order for the left-hand side to be odd we need \(m_2=1\), which implies that \(m_1=m_3\).
It remains to consider the case where \(m_1\ge m_2+2\) and \(m_3 \ge m_4+2\) and w.l.o.g. we suppose that \(m_2 > m_4\). We look at Eq.  (8) in the form
$$\begin{aligned} m_2!((m_2+1)\cdots m_1-1)=m_4!((m_4+1)\cdots m_3-1). \end{aligned}$$
(10)
By assumption, we have that \(\nu _2(m_2!)=\nu _2(m_4!)\) which implies that \(m_4\) is even and \(m_2=m_4+1\). We hence may rewrite Eq.  (10) to get
$$\begin{aligned} (m_4+1)\cdots m_1 -m_4=(m_4+1)\cdots m_3. \end{aligned}$$
It follows that \(m_4+1|m_4\) which implies that \(m_4=0\). This leads to \(m_2=1\) and \(m_1=m_3\). \(\square \)

Lemma 2

For odd positive 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
$$\begin{aligned} \sum _{\begin{array}{c} 2 \not \mid n \\ \mu ^2(n)=1 \end{array}} \frac{1}{nt(b(n))} \end{aligned}$$
converges.

Proof

Recall that 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
$$\begin{aligned} \prod _{p \le x}\left( 1+\frac{1}{p}\right) \ll \log x, \end{aligned}$$
we get
$$\begin{aligned} \sum _{\begin{array}{c} 2 \not \mid n \\ \mu ^2(n)=1 \end{array} }\frac{1}{nt(b(n))}&\le \sum _{\begin{array}{c} p \ge 3 \\ p \in \mathbb {P} \end{array}}\frac{1}{pt(b(p))}\sum _{\begin{array}{c} 2 \not \mid m \\ \mu (m)^2=1 \\ P(m)<p \end{array}}\frac{1}{m}=\sum _{\begin{array}{c} p \ge 3 \\ p \in \mathbb {P} \end{array}}\frac{1}{pt(b(p))}\prod _{\begin{array}{c} q<p \\ q \in \mathbb {P} \end{array}}\left( 1+\frac{1}{q}\right) \nonumber \\&\ll \sum _{\begin{array}{c} p \ge 3 \\ p \in \mathbb {P} \end{array}}\frac{\log p}{pt(b(p))}. \end{aligned}$$
(11)
We split the primes into two subsets \(\mathcal {P}\) and \(\mathcal {Q}\) and consider the contribution of these sets separately. We set \(\mathcal {P}=\mathcal {P}_1 \cup \mathcal {P}_2 \cup \mathcal {P}_3 \cup \mathcal {P}_4\) wherefor some fixed \(p_0\) to be chosen later. The set \(\mathcal {Q}\) is then defined to be \(\mathbb {P}\backslash (\mathcal {P}\cup \{2\})\). We start by showing that
$$\begin{aligned} \mathcal {P}(x) \ll \frac{x}{(\log x)^3}. \end{aligned}$$
(12)
For \(\mathcal {P}_1\), applying an idea of Erdős and Murty [6], we use that \(p|2^k-1\) where \(k=t(p)\), whence we have thatFrom this, we getwhich shows thatTo deal with the contribution of the set \(\mathcal {P}_2\), we set
$$\begin{aligned} {\varPsi }(x,y):=|\{n \le x: P(n) \le y\}|. \end{aligned}$$
By known results on smooth numbers (in particular, a result of Canfield, Erdős and Pomerance from [2, Corollary p.15]), we have for \(y>(\log x)^2\),
$$\begin{aligned} {\varPsi }(x,y)=\frac{x}{\exp ((1+o(1))u\log u)},\quad \text { where }\quad u=\frac{\log x}{\log y}, \end{aligned}$$
(14)
as both 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
$$\begin{aligned} (1+o(1))u_0\log u_0=\left( \frac{1}{12}+o(1)\right) \log \log x \log \log \log x>4\log \log x \end{aligned}$$
(15)
for large x. Using (14) and (15), we thus get that
$$\begin{aligned} {\mathcal P}_2(x)\ll \frac{x+x\log x}{\exp ((1+o(1))u_0\log u_0)}\ll \frac{x}{(\log x)^3}. \end{aligned}$$
Next we consider the contribution of \(\mathcal {P}_3\). This set contains primes 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
$$\begin{aligned} \mathcal {P}_3(x) \ll 1 \ll \frac{x}{(\log x)^3}. \end{aligned}$$
We are now ready to prove that the sum on the right-hand side of (11) converges. For the contribution of primes \(p \in \mathcal {P}\), we use the Abel summation formula as well as (12) and get
$$\begin{aligned} \sum _{\begin{array}{c} p \le x \\ p \in \mathcal {P} \end{array}}\frac{\log p}{pt(b(p))}&\le \sum _{\begin{array}{c} p \le x \\ p \in \mathcal {P} \end{array}} \frac{\log p}{p} =\int _3^x\frac{\log t}{t} \mathrm {d}\mathcal {P}(t) \\&= \frac{\mathcal {P}(t)\log t}{t}\Big |_{t=3}^x-\int _3^x\frac{1-\log t}{t^2}\mathcal {P}(t)\mathrm {d}t \\&\ll 1+ \int _3^x\frac{\log t}{t^2}\frac{t}{(\log t)^3} \mathrm {d}t =1+\int _3^x\frac{\mathrm {d}t}{t(\log t)^2} \ll 1. \end{aligned}$$
By the definition of \(\mathcal {Q}\) for \(p \in \mathcal {Q}\) we have that Open image in new window which implies that 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
$$\begin{aligned} \sum _{p \in Q}\frac{\log p}{pt(b(p))} \le \sum _{n \in \mathbb {N}}\frac{1}{n(\log n)^2} \ll 1, \end{aligned}$$
which finishes the proof of the lemma. \(\square \)

Lemma 3

The following estimate holds:
$$\begin{aligned} \sum _{n \le x}r_1(n) \gg x. \end{aligned}$$

Proof

We certainly have thatBy the Prime Number Theoremand Open image in new window implies that Open image in new window and henceWe use that \(m! \le m^m\) and that Open image in new window for Open image in new window and sufficiently large x. This implies thatThe bounds in (16), (17) and (18) show that
$$\begin{aligned} \sum _{n \le x}r_1(n) \gg x. \end{aligned}$$
\(\square \)

Lemma 4

The following estimate holds:
$$\begin{aligned} \sum _{n \le x}r_1(n)^2 \ll x. \end{aligned}$$

Proof

We begin with the observation that the sum counts exactly the number of solutions of the equation
$$\begin{aligned} p_1+2^{2^{k_1}}+m_1!=p_2+2^{2^{k_2}}+m_2! \end{aligned}$$
in \(p_1,p_2,k_1,k_2,m_1\) and \(m_2\) where \(p_1+2^{2^{k_1}}+m_1! \le x\). For fixed \(k_1,k_2,m_1\) and \(m_2\) this amounts to counting pairs of primes \((p_1,p_2)\) such that \(p_2=p_1+h\), where
$$\begin{aligned} h:=2^{2^{k_1}}+m_1!-2^{2^{k_2}}-m_2!. \end{aligned}$$
If \(h=0\), then we apply Theorem 7 to get that either \((k_1,m_1)=(k_2,m_2)\) or \(k_1=k_2\) and \((m_1,m_2) \in \{(1,0),(0,1)\}\)1. The number of choices of the form \((p_1,p_2,k_1,k_2,m_1,m_2)\) in this case is
$$\begin{aligned} \mathcal {O}\left( \frac{x}{\log x}\left( \log \log x \frac{\log x}{\log \log x}+\log \log x \right) \right) =\mathcal {O}(x). \end{aligned}$$
If 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
$$\begin{aligned} \mathcal {O}\left( (\log \log x)^2 \left( \frac{\log x}{\log \log x}\right) ^2\right) =o(x) \end{aligned}$$
choices for \((p_1,p_2,k_1,k_2,m_1,m_2)\) in this case. To deal with the remaining even \(h \ne 0\), we use a classical sieve bound (cf. for example [15, Theorem 7.3]). In this case, the number of pairs \((p_1,p_2)\) of primes such that \(p_2=p_1+h\) is
$$\begin{aligned} \mathcal {O}\left( \frac{x}{(\log x)^2}\prod _{p|h}\left( 1+\frac{1}{p}\right) \right) . \end{aligned}$$
Summing over all choices \((k_1,k_2,m_1,m_2)\) such that \(h \ne 0\) is even (this range of summation is indicated by the dash in the superscript of the sum below), we hence need to show thatObserving that the prime \(p=2\) contributes just a constant factor, this amounts to showing thatwhich we do in what follows. We now rewrite the left-hand side of the last inequality above asTherefore we need to study, for a given odd square-free d, the cardinality of the set
$$\begin{aligned} S_d:=\{(k_1,k_2,m_1,m_2):d|h, h\ne 0, 2\not \mid h\}. \end{aligned}$$
We start with the subset \(S_{1,d} \subset S_d\) where
$$\begin{aligned} S_{1,d}:=\{(k_1,k_2,m_1,m_2)\in S_d: m_1=m_2~{\text {or}}~\{m_1,m_2\}=\{0,1\}\}. \end{aligned}$$
(20)
We thus first deal withBy (20), \((m_1,m_2)\) is chosen in at most Open image in new window ways. As for \((k_1,k_2)\), we have \(2^{2^{k_1}}\equiv 2^{2^{k_2}}\pmod d\). Since 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
$$\begin{aligned} \left\lfloor \frac{\log \log x}{t(b(d))}\right\rfloor +1. \end{aligned}$$
Since \(k_2\) is chosen in \(\mathcal {O}(\log \log x)\) ways, we havewhere we used Lemma 2 and the fact that
$$\begin{aligned} \sum _{\begin{array}{c} d \le x \\ d \text { odd} \\ \mu (d)^2=1 \end{array}}\frac{1}{d} \ll \log x. \end{aligned}$$
From now on, we deal with \(S_d\backslash S_{1,d}\). Any quadruple \((k_1,k_2,m_1,m_2)\) in the above set gives \(m_1!-m_2!\ne 0\) and we assume that \(m_1>m_2\). We partition the numbers d in the range of summation into two different sets A and B. We set
$$\begin{aligned}&A:= \\&\quad \left\{ d \in \mathbb {N} :\begin{gathered} 2 \not \mid d, \mu (d)^2=1,\forall \{(k_1,k_2,m_1,m_2),(k_3,k_4,m_3,m_4)\} \in (S_d\backslash S_{1,d})^2: \\ 2^{2^{k_1}}+m_1!-2^{2^{k_2}}-m_2!=2^{2^{k_3}}+m_3!-2^{2^{k_4}}-m_4!=h \end{gathered} \right\} \!, \\&B:=\\&\quad \left\{ d \in \mathbb {N} :\begin{gathered} 2\not \mid d, \mu (d)^2=1,\exists \{(k_1,k_2,m_1,m_2),(k_3,k_4,m_3,m_4)\} \in (S_d\backslash S_{1,d})^2: \\ 2^{2^{k_1}}+m_1!-2^{2^{k_2}}-m_2!\ne 2^{2^{k_3}}+m_3!-2^{2^{k_4}}-m_4! \end{gathered} \right\} \!. \end{aligned}$$
In the 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
$$\begin{aligned} m_1!-m_2!=h-2^{2^{k_1}}+2^{2^{k_2}}. \end{aligned}$$
The existence of some other element \((k_1,k_2,m_3,m_4) \in S_d\backslash S_{1,d}\) with \(m_3>m_4\) would imply that \(m_1!-m_2!=m_3!-m_4!\) which by Lemma 1 leads to \((m_1,m_2)=(m_3,m_4)\). Hence, for \(d\in A\) and for \((k_1,k_2,m_1,m_2) \in S_d\backslash S_{1,d}\) with \(m_1>m_2\), the last two coordinates are uniquely determined by the first two whence for \(d \in A\) we have
$$\begin{aligned} | (S_d\backslash S_{1,d})|\ll (\log \log x)^2. \end{aligned}$$
We thus get that
$$\begin{aligned} \sum _{d\in A} \frac{|(S_d\backslash S_{1,d})|}{d} \ll (\log \log x)^2 \sum _{d\le x} \frac{1}{d}\ll (\log x)(\log \log x)^2=o((\log x)^2). \end{aligned}$$
Finally, we deal with the contribution of \(d \in B\). By definition, we may find two quadruples \((k_1,k_2,m_1,m_2)\) with \(m_1>m_2\) and \((k_3,k_4,m_3,m_4)\) with \(m_3>m_4\) both in \(S_d\backslash S_{1,d}\) such that
$$\begin{aligned} h:=2^{2^{k_1}}+m_1!-2^{2^{k_2}}-m_2! \ne 2^{2^{k_3}}+m_3!-2^{2^{k_4}}-m_4!=:h'. \end{aligned}$$
(21)
Let \(\mathcal {P}\) be the set of possible prime factors of \(d \in B\) which exceed \(\log x\). We shall prove that \(|\mathcal {P}| = \mathcal {O}((\log x)^5)\). For \(h,h'\) in (21) we have that they are both divisible by 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
$$\begin{aligned} \sum _{d\in B} \frac{|(S_d\backslash S_{1,d})|}{d}\le \left( \sum _{\begin{array}{c} u \text { odd} \\ \mu (u)^2=1 \\ P(u)<\log x \end{array}} \frac{|(S_u\backslash S_{1,u})|}{u} \right) \left( \sum _{\begin{array}{c} v \text { odd} \\ \mu (v)^2=1 \\ p\mid v\Rightarrow p\in \mathcal {P} \end{array}} \frac{1}{v}\right) , \end{aligned}$$
where we used that \(S_d\backslash S_{1,d}\subset S_u\backslash S_{1,u}\) if \(u\mid d\). For the second sum we have
$$\begin{aligned} \sum _{\begin{array}{c} v \text { odd} \\ \mu (v)^2=1 \\ p\mid v\Rightarrow p\in \mathcal {P} \end{array}} \frac{1}{v}=\prod _{p \in \mathcal {P}}\left( 1+\frac{1}{p}\right) =\mathcal {O}(1), \end{aligned}$$
which follows from partial summation and the fact that \(\mathcal {P}\) has \(\mathcal {O}((\log x)^5)\) elements all larger than \(\log x\). It thus remains to bound
$$\begin{aligned} \sum _{\begin{array}{c} u \text { odd} \\ \mu (u)^2=1 \\ P(u)<\log x \end{array}} \frac{|(S_u\backslash S_{1,u})|}{u}. \end{aligned}$$
For this, we fix \((m_1,m_2)\) with \(m_1>m_2\) not both in \(\{0,1\}\). Then putting \(M_{1,2}=m_2!-m_1!\), we need to count the number of \((k_1,k_2)\) such that \(2^{2^{k_1}}-2^{2^{k_2}}\equiv M_{1,2} \pmod u\). Analogously as before, for fixed \(k_2\), this puts \(k_1\) into a fixed arithmetic progression modulo 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
$$\begin{aligned}&\sum _{\begin{array}{c} u \text { odd} \\ \mu (u)^2=1 \\ P(u)<\log x \end{array}} \frac{|(S_u\backslash S_{1,u})|}{u} \ll \left( \frac{\log x}{\log \log x}\right) ^2(\log \log x) \\&\qquad \times \left( \log \log x \sum _{\begin{array}{c} u \text { odd} \\ \mu (u)^2=1 \\ P(u)<\log x \end{array}}\frac{1}{ut(b(u))}+\sum _{\begin{array}{c} u \text { odd} \\ \mu (u)^2=1 \\ P(u)<\log x \end{array}}\frac{1}{u}\right) \ll (\log x)^2. \end{aligned}$$
Here, we used Lemma 2 and Mertens’ formula, which yields
$$\begin{aligned} \sum _{\begin{array}{c} u \text { odd} \\ \mu (u)^2=1 \\ P(u)<\log x \end{array}}\frac{1}{u} = \prod _{3 \le p \le \log x} \left( 1+\frac{1}{p}\right) \ll \log \log x. \end{aligned}$$
\(\square \)

Proof of Theorem 3

Since the density of integers of the form \(p+2^{2^k}+m!\), \(p \in \mathbb {P}\), \(m,k \in \mathbb {N}\) and \(m < 2^{2^6}-1\) is zero, we may suppose that \(m \ge 2^{2^6}-1\). In this case, we have \(m! \equiv 0 \bmod 2^{2^6}-1\), and for \(k \ge 6\), we have that \(2^{2^k} \equiv 1 \bmod 2^{2^6}-1\). If \(n \equiv a+1 \bmod 2^{2^6}-1\), where 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
$$\begin{aligned} \frac{2^{2^6}-1-\varphi (2^{2^6}-1)}{2 \cdot ( 2^{2^6}-1)} = \frac{615850829669273873}{2459565876494606882}. \end{aligned}$$
We note that a more refined version of the above argument was used by Habsieger and Roblot [11, Sect. 3] to prove an upper bound on the proportion of odd integers not of the form \(p+2^k\). \(\square \)

Proof of Theorem 4

We will show that none of the integers n satisfying the following system of congruences is of the form \(p+2^{2^k}+m!:\)
$$\begin{aligned} 1&\bmod 2&1&\bmod 3&3&\bmod 5 \\ 2&\bmod 7&6&\bmod 11&3&\bmod 17 \\ 7&\bmod 19&9&\bmod 23. \end{aligned}$$
By the Chinese Remainder Theorem, the arithmetic progressions above intersect in a unique arithmetic progression. Let 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

The following estimate holds:
$$\begin{aligned} \sum _{n \le x}r_2(n) \gg x. \end{aligned}$$

Proof

The lemma follows fromBy the Prime Number Theorem, we haveTogether withthis finishes the proof of the lemma. \(\square \)

Lemma 6

The following estimate holds:
$$\begin{aligned} \sum _{n \le x}r_2(n)^2 \ll x. \end{aligned}$$

Proof

Again \(r_2(n)^2\) counts the number of solutions of the equation
$$\begin{aligned} p_1+2^{2^{k_1}}+2^{q_1}=p_2+2^{2^{k_2}}+2^{q_2} \end{aligned}$$
in \(p_1,p_2,k_1,k_2,q_1\) and \(q_2\) where \(p_1+2^{2^{k_1}}+2^{q_1} \le x\). This means counting pairs of primes \((p_1,p_2)\) such that \(p_2=p_1+h\), where
$$\begin{aligned} h:=2^{2^{k_1}}+2^{q_1}-2^{2^{k_2}}-2^{q_2}. \end{aligned}$$
If \(h=0\) then either \((k_1,q_1)=(k_2,q_2)\) or w.l.o.g. \(k_1>k_2\) and
$$\begin{aligned} 2^{2^{k_2}}\left( 2^{2^{k_1}-2^{k_2}}-1\right) =2^{q_1}\left( 2^{q_2-q_1}-1\right) . \end{aligned}$$
Since \(2^{2^{k_1}-2^{k_2}}-1\) and \(2^{q_2-q_1}-1\) are odd, we have that \(2^{k_2} = q_1\) and hence \(k_2=1\) and \(q_1=2\). This leads to \(2^{k_1} = q_2\) and hence to \(k_1=1\) and \(q_2=2\) a contradiction to \(k_1>k_2\). If \(h=0\) we thus have that \((k_1,q_1)=(k_2,q_2)\) and \(p_2\) is fixed by a choice of \(p_1,k_1\) and \(q_1\). The last three parameters may be chosen in \(\mathcal {O}(x)\) ways and we can deal with the contribution of solutions of the equation \(p_2=p_1+h\) where \(h \ne 0\). Since 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
$$\begin{aligned} 2^{q_2} \equiv 2^{2^{k_1}}+2^{q_1}-2^{2^{k_2}}=:l \bmod d, \end{aligned}$$
where 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
$$\begin{aligned} \sum _{\begin{array}{c} d \text { odd} \\ P(d)\le \log x \end{array}}\frac{\mu (d)^2}{d} \ll \log \log x. \end{aligned}$$
To deal with the first sum in (24), we use Open image in new window (see [17, Theorem 15]) and split the range of summation in two parts and getBy a result of Erdős and Turán [7, 8], the sums
$$\begin{aligned} \sum _{d \text { odd}} \frac{\log \log t(d)}{dt(d)} \text { and } \sum _{d \text { odd }}\frac{\log \log t(d)}{d\sqrt{t(d)}} \end{aligned}$$
converge which altogether proves an upper bound of order \(\mathcal {O}((\log x)^2)\) for (23) and hence an upper bound of order \(\mathcal {O}(x)\) for (22). \(\square \)

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.

If \(k>0\), then \(2^{2^k}=4^{2^{k-1}}\). The fact that \(4^2 \equiv 4 \bmod 6\) puts the term \(2^{2^k}\) into the residue class \(4 \bmod 6\) if \(k>0\). Using the same fact again, we get for \(q=2l+1\)
$$\begin{aligned} 2^q =2^{2l+1}=2\cdot 4^l \equiv 2 \bmod 6. \end{aligned}$$
Furthermore, all primes except 2 and 3 are in the residue classes \(\{1,5\} \bmod 6\). Thus if n is in none of the sets
$$\begin{aligned}&S_1:=\{p+2+2^q: p,q\in \mathbb {P}\}, \\&S_2:=\{p+2^{2^k}+4: p\in \mathbb {P}, k \in \mathbb {N}\},\\&S_3:=\{2+2^{2^k}+2^q: k \in \mathbb {N},q\in \mathbb {P}\},\\&S_4:=\{3+2^{2^k}+2^q: k \in \mathbb {N},q\in \mathbb {P}\}, \end{aligned}$$
all of which have density 0, and if n has a representation of the form \(n=p+2^{2^k}+2^q\), then n is in one of the residue classes
$$\begin{aligned} \{1,5\}+\{4\}+\{2\} =\{1,5\} \bmod 6. \end{aligned}$$
The set
$$\begin{aligned} S=\{n \in \mathbb {N}: n \equiv 3 \bmod 6\} \backslash (S_1\cup S_2 \cup S_3 \cup S_4) \end{aligned}$$
has density Open image in new window , consists of odd integers only and none of its members is of the form \(p+2^{2^k}+2^q\). This proves the first part of the Theorem.
To find a full arithmetic progression of integers not of the form \(p+2^{2^k}+2^q\), we will add additional congruences ruling out the integers in the sets \(S_1\), \(S_2\), \(S_3\) and \(S_4\). We claim that none of the integers n satisfying the congruences
$$\begin{aligned} 3&\bmod 6&4&\bmod 5&4&\bmod 7 \\ 9&\bmod 13&5&\bmod 17&8&\bmod 19 \\ 20&\bmod 23&2&\bmod 29&3&\bmod 31 \\ 10&\bmod 37 \end{aligned}$$
is of the form \(p+2^{2^k}+2^q\). By the above considerations, it suffices to check that none of the integers in the sets \(S_1\), \(S_2\), \(S_3\) and \(S_4\) is contained in this arithmetic progression.
We start with the integers in \(S_1\). Take \(n=p+2+2^q \in S_1\) and suppose that 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
$$\begin{aligned} 2^{12l+1} \equiv 2^{12l+5} \equiv 2 \bmod 5 \text {, } 2^{12l+7} \equiv 2 \bmod 7 \text {, } 2^{12l+11} \equiv 7 \bmod 13. \end{aligned}$$
If \(q \equiv \{1,5\} \bmod 12\), then \(n=p+2+2^q \equiv p+4 \bmod 5\). Since \(n \equiv 4 \bmod 5\), this implies that \(p=5\). Now \(7+2^{12l+1} \equiv 2 \bmod 7\) and \(7+2^{12l+5} \equiv 0 \bmod 13\), contradiction to \(n \equiv 4 \bmod 7\) and \(n \equiv 9 \bmod 13\). In the case of \(q=12l+7\), we get \(n=p+2+2^{12l+7} \equiv p+ 4 \bmod 7\) and the only possible choice for 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

  1. 1.

    Note that \(x_1\) and \(x_2\) in the non-trivial solutions in Theorem 7 are never both powers of 2.

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. 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. 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. 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. 4.
    Dubickas, A.: Sums of primes and quadratic linear recurrence sequences. Acta Math. Sin. (Engl. Ser.) 29(12), 2251–2260 (2013)MathSciNetCrossRefGoogle Scholar
  5. 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. 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. 7.
    Erdős, P., Turán, P.: Ein zahlentheoretischer Satz. Mitt. Forsch.-Inst. Math. Mech. Univ. Tomsk 1, 101–103 (1935)zbMATHGoogle Scholar
  8. 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. 9.
    Euler, L.: Letter to Goldbach, 16.12.1752. http://eulerarchive.maa.org/correspondence/letters/OO0879.pdf
  10. 10.
    Guy, R.K.: Unsolved Problems in Number Theory. Problem Books in Mathematics, 3rd edn. Springer, New York (2004)CrossRefGoogle Scholar
  11. 11.
    Habsieger, L., Roblot, X.F.: On integers of the form \(p+2^k\). Acta Arith. 122(1), 45–50 (2006)MathSciNetCrossRefGoogle Scholar
  12. 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. 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. 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. 15.
    Nathanson, M.B.: Additive Number Theory—The Classical Bases. Graduate Texts in Mathematics, vol. 164. Springer, New York (1996)CrossRefGoogle Scholar
  16. 16.
    Romanoff, N.P.: Über einige Sätze der additiven Zahlentheorie. Math. Ann. 109(1), 668–678 (1934)MathSciNetCrossRefGoogle Scholar
  17. 17.
    Rosser, J.B., Schoenfeld, L.: Approximate formulas for some functions of prime numbers. Ill. J. Math. 6, 64–94 (1962)MathSciNetzbMATHGoogle Scholar
  18. 18.
    van der Corput, J.G.: Over het vermoeden van de Polignac. Simon Stevin 27, 99–105 (1950)MathSciNetGoogle Scholar

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© The Author(s) 2018

Open AccessThis article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors and Affiliations

  • Christian Elsholtz
    • 1
  • Florian Luca
    • 2
    • 3
  • Stefan Planitzer
    • 1
  1. 1.Institute of Analysis and Number TheoryGraz University of TechnologyGrazAustria
  2. 2.School of MathematicsWits UniversityJohannesburgSouth Africa
  3. 3.Department of Mathematics, Faculty of SciencesUniversity of OstravaOstrava 1Czech Republic

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