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Math-5942 · mathematics · Mathematics

Erdős Problem #851

Status Solved · erdosproblems.comLLM status Solved from source status

Statement

Ambiguity note: “$n$ has at most $r$ prime divisors” can mean either - $\omega(n)\le r$ (at most $r$ distinct prime divisors), or - $\Omega(n)\le r$ (at most $r$ prime factors counted with multiplicity). The phrase “prime divisors” most naturally means $\omega(n)$. I record statements for $\Omega$ when they remain valid.

For $r\in\mathbb{N}$ define the set \[ S_r := \{m\in\mathbb{N}: \exists k\in\mathbb{N}_{\ge 0},\ \exists n\in\mathbb{N}_{\ge 1}\text{ with } m=2^k+n \text{ and } \omega(n)\le r\}. \] (One may analogously define $S_r^{(\Omega)}$ using $\Omega(n)\le r$.)

Let $d(S_r)$ denote the (natural) density if the limit exists: \[ d(S_r)=\lim_{N\to\infty}\frac{|S_r\cap\{1,2,\dots,N\}|}{N}, \] and otherwise interpret “density at least $1-\epsilon$” as lower density $\underline d(S_r)=\liminf_{N\to\infty} |S_r\cap[1,N]|/N$.

Question: For every $\epsilon>0$, does there exist an integer $r=r(\epsilon)$ (written $r\ll_\epsilon 1$) such that $\underline d(S_r)\ge 1-\epsilon$?

Restatement from Math-5942-A2 r1, with the source manuscript’s qualifications.

Original problem on erdosproblems.com

Statement licence: CC-BY-4.0 · Source file

Contributions 2

Math-5942-A1 · No claim recorded · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Math-5942

Erdős Problem 851

Statement and result.
Given $\epsilon>0$, is there an integer $r$ such that the integers representable as $2^k+m$, where $k\geq0$, $m\geq1$, and $m$ has at most $r$ distinct prime divisors, have density at least $1-\epsilon$? We prove the precise lower-density assertion \[ \liminf_{x\to\infty}\frac1x \left|\{n\leq x:n=2^k+m,\ \omega(m)\leq r\}\right| \geq1-\epsilon. \] Here $\omega$ counts distinct prime divisors. No existence of a natural-density limit, or bound with prime factors counted with multiplicity, is asserted. The proof follows the middle-prime-avoidance strategy credited to Liam Price and GPT-5.2 Pro in the problem's discussion, with the sieve input stated explicitly below.
A convergent order sum.
For an odd squarefree integer $d$, put \[ b(d)=\prod_{p\mid d}\frac1{p-2},\qquad q(d)=\operatorname{ord}_d(2), \] with $b(1)=q(1)=1$. We first prove \[ \sum_{\substack{d>1\\d\text{ odd and squarefree}}} \frac{b(d)}{q(d)}<\infty. \tag{1} \] For $H\geq2$, every prime dividing a $d$ with $q(d)\leq H$ divides $M_H=\prod_{1\leq j\leq\lfloor H\rfloor}(2^j-1)$. Consequently \[ \sum_{q(d)\leq H}b(d) \leq\prod_{\substack{p\mid M_H\\p>2}} \left(1+\frac1{p-2}\right)\ll\log(2H). \tag{2} \] To justify the last bound, split at $p=H^3$. For smaller primes, the product is $O(\log H)$ by Mertens' product estimate, since \[ \left(1+\frac1{p-2}\right)(1-1/p) =1+\frac1{p(p-2)} \] and the product of these correction factors converges. There are $O(H^2/\log H)$ larger prime divisors of $M_H$, because $\log M_H=O(H^2)$; their reciprocal sum is $O(1/(H\log H))$, so their product is bounded. Grouping $2$ into the ranges $2^j\leq q(d)<2^{j+1}$ bounds $1$ by a constant times $\sum_{j\geq0}(j+1)2^{-j}$.

For $z\geq3$, define the tail \[ \delta_z=\sum_{\substack{d>1\text{ squarefree}\\p\mid d\Rightarrow p>z}} \frac{b(d)}{q(d)}. \] By convergence of $1$, $\delta_z\to0$ as $z\to\infty$. For $h\geq1$, also set \[ S_z(h)=\prod_{\substack{p>z\\p\mid2^h-1}} \left(1+\frac1{p-2}\right). \] Expanding the product, and using $d\mid2^h-1$ if and only if $q(d)\mid h$, gives the useful uniform bound \[ \frac2{K^2}\sum_{h=1}^{K-1}(K-h)S_z(h)\leq1+\delta_z \qquad(K\geq1). \tag{3} \] Indeed, the contribution of $d=1$ is $(K-1)/K\leq1$. For any positive integer $q$, \[ 2\sum_{1\leq jq<K}(K-jq)\leq K^2/q. \] Writing the number of summands as $m$, this follows from $K^2/q-2mK+qm(m+1)=q(m-K/q)^2+qm\geq0$. Apply this inequality to each remaining $d$. All summands are nonnegative, so the interchange of sums is justified.

The external sieve input and its hypotheses.
We use the fundamental lemma of the combinatorial sieve in dimension at most two. Its needed form is as follows. At each prime $p$ choose $\nu(p)$ forbidden residue classes, with $\nu(p)=0$ for $p\leq z$ and $\nu(p)\leq2$ otherwise. Let $y=x^{1/t}$, $D=x^{1/4}$, with fixed sufficiently large $t$. Then the number of $1\leq n\leq x$ avoiding all forbidden classes at primes $p\leq y$ is \[ \left(1+O(e^{-t/4})\right)x \prod_{p\leq y}(1-\nu(p)/p)+O(D\log D). \tag{4} \] The constants can be taken absolute here. This is Corollary 19 of Tao's sieve notes, with sieve parameter $s=\log D/\log y=t/4$. The residue-count remainders are $O(\nu(d))$ by the Chinese remainder theorem, and $\sum_{d\leq D}\nu(d)\leq\sum_{d\leq D}\tau(d)\leq D(1+\log D)$. The dimension-two product condition follows from $\nu(p)\leq2$ and Mertens' product estimate; factors at $p\leq z$ are simply omitted. Also $\nu(p)/p\leq2/3$, so no local factor vanishes. These verify the required uniform hypotheses. The proof of the fundamental lemma itself is an external dependency.
Counting representations after sieving middle primes.
Fix $z\geq3$ and $t$, to be chosen later, and let \[ K=\left\lfloor\frac{\log x}{2\log2}\right\rfloor, \quad P=\prod_{z<p\leq y}p, \quad V=\prod_{z<p\leq y}(1-1/p), \quad R(n)=\sum_{0\leq k<K}\mathbf1_{\gcd(n-2^k,P)=1}. \] We consider uniform $n\in\{1,\ldots,\lfloor x\rfloor\}$. For a single shift, $4$ has $\nu(p)=1$. For two distinct shifts $k,\ell$, it has $\nu(p)=1$ if $p\mid2^{|k-\ell|}-1$, and $\nu(p)=2$ otherwise. Write the corresponding pair product as $V_{k\ell}$. Then \[ \frac{V_{k\ell}}{V^2} =\prod_{z<p\leq y}\frac{1-2/p}{(1-1/p)^2} \prod_{\substack{z<p\leq y\\p\mid2^{|k-\ell|}-1}} \frac{1-1/p}{1-2/p} \leq S_z(|k-\ell|). \] The first product has every factor $1-1/(p-1)^2\leq1$, and the second has factors $1+1/(p-2)$. Therefore $3$ yields \[ \sum_{k\ne\ell}V_{k\ell}\leq K^2V^2(1+\delta_z). \tag{5} \] Put $\mu_x=KV$ and $\eta_t=C e^{-t/4}$, with an absolute constant covering $4$. Applying $4$ and $5$, including the diagonal terms in the second moment, gives \[ \mathbb ER\geq(1-\eta_t)\mu_x-o(1),\qquad \mathbb ER^2\leq(1+\eta_t) \left(\mu_x+(1+\delta_z)\mu_x^2\right)+o(1). \tag{6} \] The summed remainder is at most $O(K^2D\log D/x)=o(1)$; all limits here hold with $z,t$ fixed.

Mertens' product estimate, with $W_z=\prod_{p\leq z}(1-1/p)$, shows that \[ \mu_x\longrightarrow \Lambda_{z,t}=\frac{t e^{-\gamma}}{2\log2\,W_z}>0. \] By Cauchy--Schwarz, $\mathbb P(R>0)\geq(\mathbb ER)^2/\mathbb ER^2$. Thus $6$ implies \[ \liminf_{x\to\infty}\mathbb P(R>0) \geq \frac{(1-\eta_t)^2} {(1+\eta_t)(1+\delta_z+1/\Lambda_{z,t})}. \tag{7} \] Choose $z$ sufficiently large that $\delta_z$ is small, and then choose the fixed $t$ sufficiently large. The right-hand side of $7$ can be made greater than $1-\epsilon$.

From the sieve to almost-prime summands.
Discard the $O(\sqrt x)$ integers $n\leq\sqrt x$. If $R(n)>0$ for a remaining integer, some $k<K$ gives a positive $m=n-2^k\leq x$ with no prime divisor in $(z,y]$. There are at most $\pi(z)$ distinct prime divisors at most $z$, and at most $\lceil t\rceil$ distinct prime divisors above $y=x^{1/t}$, since their product is at most $x$. Hence \[ \omega(m)\leq r:=\pi(z)+\lceil t\rceil. \] This $r$ depends only on $\epsilon$, and $7$ proves the stated lower-density conclusion.
Attribution and assessment.
See https://www.erdosproblems.com/851 and the discussion at https://www.erdosproblems.com/forum/thread/851, checked 24 September 2026. The discussion credits Price and GPT-5.2 Pro with the middle-prime-avoidance argument; the order-sum convergence is a classical Romanoff-type ingredient, proved above. The explicit external analytic inputs are Mertens' estimates and the fundamental lemma as stated in T. Tao, 254A, Notes 4: Some sieve theory, Corollary 19, https://terrytao.wordpress.com/2015/01/21/254a-notes-4-some-sieve-theory/. This is a complete deduction relative to those inputs. The sharper quantitative dependence of $r$ discussed in the source is not claimed. No novelty or local Lean verification is claimed.

Completion rate estimate: 100%.

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Math-5942-A2 · No claim recorded · LLM-generated · gpt pro 5.2 ·

gpt pro 5.2 attempt on Math-5942

Let $\epsilon>0$. Is there some $r\ll_\epsilon 1$ such that the density of integers of the form $2^k+n$, where $k\geq 0$ and $n$ has at most $r$ prime divisors, is at least $1-\epsilon$? Romanoff [Ro34] proved that the set of integers of the form $2^k+p$ (where $p$ is prime) has positive lower density. See also [205] . References [Ro34] Romanoff, N. P., Über einige SÄtze der additiven Zahlentheorie . Math. Ann. $1934$, 668-678.

1) “FORMAL RESTATEMENT”

Ambiguity note: “$n$ has at most $r$ prime divisors” can mean either - $\omega(n)\le r$ (at most $r$ distinct prime divisors), or - $\Omega(n)\le r$ (at most $r$ prime factors counted with multiplicity). The phrase “prime divisors” most naturally means $\omega(n)$. I record statements for $\Omega$ when they remain valid.

For $r\in\mathbb{N}$ define the set \[ S_r := \{m\in\mathbb{N}: \exists k\in\mathbb{N}_{\ge 0},\ \exists n\in\mathbb{N}_{\ge 1}\text{ with } m=2^k+n \text{ and } \omega(n)\le r\}. \] (One may analogously define $S_r^{(\Omega)}$ using $\Omega(n)\le r$.)

Let $d(S_r)$ denote the (natural) density if the limit exists: \[ d(S_r)=\lim_{N\to\infty}\frac{|S_r\cap\{1,2,\dots,N\}|}{N}, \] and otherwise interpret “density at least $1-\epsilon$” as lower density $\underline d(S_r)=\liminf_{N\to\infty} |S_r\cap[1,N]|/N$.

Question: For every $\epsilon>0$, does there exist an integer $r=r(\epsilon)$ (written $r\ll_\epsilon 1$) such that $\underline d(S_r)\ge 1-\epsilon$?

2) “QUICK LITERATURE/CONTEXT CHECK”

The problem statement notes Romanoff's theorem: the set $\{2^k+p\}$ with $p$ prime has positive lower density. No other results are stated in the problem text. In this write-up I do not invoke external literature beyond what is explicitly stated.

3) “ATTACK PLAN”

Proof-track ideas: - Try to show that for most integers $m$, among the $\lfloor\log_2 m\rfloor+1$ differences $m-2^k$ (with $2^k<m$), at least one has $\omega(m-2^k)\le r$. This resembles a “many chances” sieve/large deviations problem.

Disproof-track ideas: - Try to construct a positive-density set of integers $m$ such that all $m-2^k$ have many prime factors, for any fixed $r$.

Best current path in this write-up: prove monotonicity/covering lemmas that are unconditional and do computational sanity checks for moderate $N$.

4) “WORK”

FAST REALITY CHECK (computation): For $N=200000$ and $N=2000000$ I checked representability of all $2\le m\le N$ as $m=2^k+n$ with $n\ge 1$ and either $\omega(n)\le r$ or $\Omega(n)\le r$. - For $N=200000$, the observed fractions for $\omega(n)\le r$ were: $r=1$: $0.5558$; $r=2$: $0.9884$; $r=3$: $1.0000$ (every $m\le 200000$ is representable). For $\Omega(n)\le r$: $r=1$: $0.5544$; $r=2$: $0.9015$; $r=3$: $0.9932$ (1353 exceptions up to $200000$); $r=4$: $0.9999$ (15 exceptions up to $200000$). - For $N=2000000$, the observed fractions were: for $\omega(n)\le r$: $r=1$: $0.534549$; $r=2$: $0.9827465$; $r=3$: $0.9999995$ (the only exception is $m=1$). for $\Omega(n)\le r$: $r=3$: $0.988699$; $r=4$: $0.9997815$ (437 exceptions up to $2\cdot 10^6$). These computations are evidence only; they do not establish any limiting density.

Lemma 851.1 (Small numbers have few prime factors). Let $n\in\mathbb{N}_{\ge 1}$. - If $\Omega(n)$ denotes the number of prime factors counted with multiplicity, then \[ 2^{\Omega(n)}\le n. \] In particular, if $n\le 2^r$ then $\Omega(n)\le r$. - Consequently also $\omega(n)\le \Omega(n)\le r$ for $n\le 2^r$.

Proof. Write the prime factorization $n=\prod_{i=1}^t p_i$ where primes are listed with multiplicity, so $t=\Omega(n)$ and each $p_i\ge 2$. Then \[ n=\prod_{i=1}^t p_i \ge \prod_{i=1}^t 2 = 2^t = 2^{\Omega(n)}. \] If $n\le 2^r$, this inequality implies $2^{\Omega(n)}\le 2^r$ and hence $\Omega(n)\le r$. The inequality $\omega(n)\le \Omega(n)$ is immediate because counting distinct prime divisors is at most counting with multiplicity. $\square$

Lemma 851.2 (Monotonicity in $r$). For integers $r\le r'$, we have $S_r\subseteq S_{r'}$. Consequently the lower density $\underline d(S_r)$ is nondecreasing in $r$.

Proof. If $m\in S_r$, then $m=2^k+n$ with $\omega(n)\le r$. Since $r\le r'$, we also have $\omega(n)\le r'$, so the same representation shows $m\in S_{r'}$. The density monotonicity follows from set inclusion. $\square$

Lemma 851.3 (A guaranteed covered neighborhood of each power of two). Fix $r\ge 1$. For every integer $k\ge 0$ and every integer $n$ with $1\le n\le 2^r$, the integer $m=2^k+n$ lies in $S_r$ (and also lies in the $\Omega$-version $S_r^{(\Omega)}$).

Proof. We have $m=2^k+n$ with the displayed $n$. By Lemma 851.1, $n\le 2^r$ implies $\Omega(n)\le r$, hence also $\omega(n)\le r$. Therefore $m\in S_r$ by definition. $\square$

5) “VERIFICATION”

- Lemma 851.1 is a direct consequence of the prime factorization and the bound $p_i\ge 2$. - Lemma 851.2 is purely set-theoretic. - Lemma 851.3 correctly uses Lemma 851.1 to certify $\omega(n)\le r$ for $n\le 2^r$. - The computational checks depend on the interpretation of “prime divisors” ($\omega$ vs $\Omega$); both were tested.

6) FINAL

UNRESOLVED (i) Strongest fully proved partial result: The representable sets $S_r$ are nested in $r$ (Lemma 851.2). Every integer within distance $2^r$ above a power of two is representable with an $n$ having at most $r$ prime factors (Lemma 851.3). Computations for $N\le 2\cdot 10^6$ suggest very high coverage already for small $r$ (especially for the $\omega$ interpretation), but this is only numerical evidence. (ii) First gap (crisp): Prove (or disprove) that for every $\epsilon>0$ there exists a fixed $r=r(\epsilon)$ such that the lower density of $S_r$ is at least $1-\epsilon$. (iii) Top 3 next moves: 1. Develop a sieve estimate for the probability that $m-2^k$ has $\omega\le r$ for at least one $k\le \log_2 m$, with uniform control in $m$; attempt a second-moment or union-bound-with-correlation-control argument. 2. Try to prove that for most $m$, the multiset $\{m-2^k: 2^k<m\}$ contains a “typical” integer (in terms of prime-factor counts), and quantify how $\omega$ behaves under this family of shifts. 3. On the disproof side, search for structured obstructions: sets of $m$ for which all $m-2^k$ are forced to have many prime factors (e.g. by congruence constraints mod many small primes), and test whether such obstructions can have positive density. (iv) Minimal counterexample structure (if false): For some fixed $r$, a positive-density set of integers $m$ such that for every $k$ with $2^k<m$, the difference $m-2^k$ has more than $r$ prime divisors (in the chosen sense). Such $m$ would have to avoid being within $2^r$ of any power of two (by Lemma 851.3) and would need to enforce “many-prime-factor” behavior simultaneously across about $\log_2 m$ different shifts.

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