Erdős Problem #234

Erdős Problem #234 (Round 2)

1) ROUND-2 OBJECTIVE

Path (C): obstruction/gap-closure. Round 1 left two distinct gaps: existence of the densities \[ F_N(c):=\frac{1}{N}\#\left\{2\le n\le N:\ \frac{p_{n+1}-p_n}{\log n}<c\right\} \] for fixed $c$, and continuity of the limiting function if the limit exists. The most promising strict advance is to remove one of those obstacles unconditionally: I will prove a quantitative upper bound \[ F_N(c)\ll c \] for fixed $c>0$. This shows that any putative limiting distribution is automatically continuous at $0$, so the only genuinely open part that remains is existence of the densities and continuity away from the origin.

2) Round-1 FOUNDATION USED

I use the following Round-1 results without re-proving them.

• Round-1 Lemma 234.1: the normalized gaps \[ a_n:=\frac{p_{n+1}-p_n}{\log n} \] have mean value $1+o(1)$, i.e. \[ \frac{1}{N}\sum_{n=2}^{N} a_n \to 1. \]
• Round-1 Corollary 234.2: for every fixed $c>0$, \[ \limsup_{N\to\infty}\frac{1}{N}\#\{2\le n\le N: a_n\ge c\}\le \frac{1}{c}, \] so any limiting law would be tight on the right.

3) NEW INSIGHT / TOOL (ROUND-2)

The new tool is a pair-counting upper bound averaged over short gap lengths. The key quantitative input is a standard Selberg-sieve prime-pair estimate together with the elementary average-order identity \[ \sum_{h\le H}\frac{h}{\phi(h)}\ll H. \] This converts upper bounds for prime pairs into an upper bound for the frequency of small consecutive prime gaps.

4) ATTACK PLAN (ROUND-2)

After Round 1, the exact unresolved issue was: even if one believes the densities should exist, no rigorous argument had been given for continuity of the limiting function. I now target the specific sub-gap at $c=0$.

The precise claim to prove is:

Claim 234-R2. There exists an absolute constant $C>0$ such that for every fixed $c>0$ and all sufficiently large $N$, \[ F_N(c)\le Cc+O\!\left(\frac{1}{N}\right). \] In particular, \[ \limsup_{N\to\infty}F_N(c)\le Cc. \]

This overcomes one concrete obstacle from Round 1: continuity at the origin is no longer part of the open problem.

5) WORK (ROUND-2)

Let \[ d_n:=p_{n+1}-p_n \qquad (n\ge 1). \]

I will use the following standard external theorem.

Prime-pair upper bound (Selberg sieve). There exists an absolute constant $A>0$ such that for all integers $h\ge 2$ and all $X\ge h+2$, \[ \Pi(X;h):=\#\{p\le X:\ p\text{ and }p+h\text{ are both prime}\} \le A\,\frac{h}{\phi(h)}\,\frac{X}{(\log X)^2}. \] For odd $h>1$, the left-hand side is actually $0$.

The only other ingredient needed is the following elementary mean-value lemma.

Lemma 234.2.1. For all $H\ge 1$, \[ \sum_{h\le H}\frac{h}{\phi(h)}\ll H. \]

Proof. Use the divisor identity \[ \frac{n}{\phi(n)}=\sum_{d\mid n}\frac{\mu^2(d)}{\phi(d)}. \] Indeed, both sides are multiplicative, and on prime powers $p^k$ they equal \[ \frac{p}{p-1}=1+\frac{1}{p-1}. \] Therefore \[ \sum_{h\le H}\frac{h}{\phi(h)} = \sum_{h\le H}\sum_{d\mid h}\frac{\mu^2(d)}{\phi(d)} = \sum_{d\le H}\frac{\mu^2(d)}{\phi(d)}\lfloor\frac{H}{d}\rfloor. \] Hence \[ \sum_{h\le H}\frac{h}{\phi(h)} \le H\sum_{d=1}^{\infty}\frac{\mu^2(d)}{d\phi(d)}. \] The Euler product for the last series is \[ \prod_{p}\left(1+\frac{1}{p(p-1)}\right), \] which converges absolutely. So the whole sum is $O(H)$. $\square$

Now prove the announced bound on small normalized gaps.

Theorem 234.2.2. There exists an absolute constant $C>0$ such that for every fixed $c>0$ and all sufficiently large $N$, \[ F_N(c)\le Cc+O\!\left(\frac{1}{N}\right). \] Consequently, \[ \limsup_{N\to\infty}F_N(c)\le Cc. \]

Proof. Set \[ X:=p_{N+1}. \] If $2\le n\le N$ and $d_n/\log n<c$, then certainly $d_n<c\log N$ because $\log n\le \log N$. Therefore \[ N F_N(c) \le \#\{2\le n\le N:\ d_n<c\log N\}. \] Now all prime gaps are even except the unique gap $p_2-p_1=1$. Hence \[ \#\{2\le n\le N:\ d_n<c\log N\} \le 1+\sum_{2\le h<c\log N}\#\{2\le n\le N:\ d_n=h\}. \] If $d_n=h$, then $(p_n,p_n+h)$ is a prime pair with $p_n\le p_N<X$. Discarding the consecutiveness condition can only enlarge the count, so \[ \#\{2\le n\le N:\ d_n=h\}\le \Pi(X;h). \] Hence \[ N F_N(c) \le 1+\sum_{2\le h<c\log N}\Pi(X;h) \le 1+A\frac{X}{(\log X)^2}\sum_{2\le h<c\log N}\frac{h}{\phi(h)}. \] By Lemma 234.2.1 with $H=\lfloor c\log N\rfloor$, \[ \sum_{2\le h<c\log N}\frac{h}{\phi(h)}\ll c\log N+1. \] Also, by the prime number theorem, \[ X=p_{N+1}\asymp N\log N, \qquad \log X\asymp \log N. \] Therefore \[ N F_N(c) \ll 1+\frac{N\log N}{(\log N)^2}(c\log N+1) \ll 1+cN. \] Dividing by $N$ yields \[ F_N(c)\le Cc+O\!\left(\frac{1}{N}\right) \] for some absolute $C$. $\square$

Corollary 234.2.3. If the densities \[ f(c)=\lim_{N\to\infty}F_N(c) \] exist for all $c\ge 0$, then automatically \[ f(c)\ll c \qquad (c\downarrow 0), \] so $f$ is continuous at $0$.

Proof. Since $f(0)=0$ and Theorem 234.2.2 gives $f(c)\le Cc$ for every fixed $c>0$, one has $f(c)\to 0$ as $c\downarrow 0$. $\square$

6) ADVERSARIAL VERIFICATION

• Did I silently use consecutiveness where only prime pairs were counted? No. I only used the inequality \[ \#\{n:d_n=h\}\le \Pi(X;h), \] which is valid because every consecutive prime pair is a prime pair, but not conversely.

• What about odd gaps? For $n\ge 2$, prime gaps are even. The exceptional gap $1$ between $2$ and $3$ contributes the harmless additive term $1$.

• Are the sieve hypotheses satisfied? Yes. Here $h<c\log N$ and $X=p_{N+1}\asymp N\log N$, so $h\le X$ for all large $N$.

• Could the bound depend on $c$ in a hidden way? The theorem is uniform in fixed $c$ because the only $c$-dependence enters through the upper limit $H\asymp c\log N$ in Lemma 234.2.1.

• Did this prove too much? No. The argument only bounds the upper density of small gaps. It does not control oscillation of $F_N(c)$ in $N$, so it does not prove the existence of the density.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

The new rigorous advance over Round 1 is:

• the continuity issue at $c=0$ is now settled conditionally on existence of the densities, because any putative limit must satisfy $f(c)\ll c$ as $c\downarrow 0$;
• equivalently, the only genuinely open part remaining is existence of the densities and continuity for positive $c$.

The exact unresolved gap is still the same hard one identified in Round 1: proving that $F_N(c)$ has a limit for each fixed $c$, or producing a value of $c$ for which the densities fail to exist.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 45%

References

Only references actually used or checked in this round are listed.

1. T. F. Bloom, Erdős Problem #234, erdosproblems.com, accessed 2026-03-07.
2. T. F. Bloom, Erdős Problem #236, erdosproblems.com, accessed 2026-03-07.
3. T. F. Bloom, Erdős Problem #238, erdosproblems.com, accessed 2026-03-07.
4. H. Halberstam and H.-E. Richert, Sieve Methods, Academic Press, 1974.
5. H. L. Montgomery and R. C. Vaughan, Multiplicative Number Theory I: Classical Theory, Cambridge University Press, 2007.

Erdős Problem #234: independent continuation

1) OBJECTIVE AND SOURCE AUDIT

2) WORK: WHAT COMPACTNESS DOES AND DOES NOT GIVE

Every resulting probability measure $\mu$ satisfies \[ \int_0^\infty t\,d\mu(t)\le1. \] For proof, apply weak convergence to $\min(t,R)$, whose average is at most $1+o(1)$, and let $R$ increase. Equality of the limiting mean is not justified without uniform integrability: a small amount of mass can escape to large values while preserving total first moment.

If one additionally retains the source's small-gap estimate $\limsup_N\mu_N([0,c))\le Cc$, then the open-set inequality for weak convergence gives $\mu([0,c))\le Cc$. Letting $c$ decrease to zero shows $\mu(\{0\})=0$. This last conclusion uses the stated sieve input; the compactness and moment arguments above do not require it.

3) ADVERSARIAL VERIFICATION

7) FINAL AND REFERENCE

8) COMPLETION ESTIMATE

Erdős Problem #234

1. FORMAL RESTATEMENT.

   Let $p_n$ denote the $n$th prime ($p_1=2,p_2=3,\dots$) and let \[d_n := p_{n+1}-p_n \qquad (n\ge 1)\] be the $n$th prime gap. Throughout, $\log$ denotes the natural logarithm.

   For $c\ge 0$, define the set \[A_c := \left\{n\in\mathbb{N} : n\ge 2\ \text{and}\ \frac{d_n}{\log n} < c\right\}. \] (We restrict to $n\ge 2$ so that $\log n$ is defined.)

   Define the natural density of $A_c$ (if it exists) by \[ f(c) := \lim_{N\to\infty} \frac{1}{N}\,\left|A_c\cap\{1,2,\dots,N\}\right|. \]

   Claim/conjecture. For every $c\ge 0$, the limit defining $f(c)$ exists, and the resulting function $f:[0,\infty)\to[0,1]$ is continuous.

   Edge cases. For $c=0$, $A_0=\varnothing$ (since $d_n>0$), so $f(0)=0$ if the above definition is used.

2. QUICK LITERATURE/CONTEXT CHECK.

   This is an "existence of limiting distribution" problem for the normalized gaps $a_n:=d_n/\log n$. Heuristically, in Cramér/Hardy--Littlewood models one expects a limiting distribution (often Poisson/exponential for suitably normalized gaps), which would in particular imply the existence of $f(c)$ and its continuity.

   Unconditionally, one has very strong information on extremes of prime gaps (liminf $=0$ in the $d_n/\log p_n$ normalization, and arbitrarily large multiples of the average gap), but no unconditional theorem gives the full limiting distribution of the normalized gaps.

3. ATTACK PLAN.

   Proof strategies.

   • $P1$ Try to show convergence of the empirical distribution functions $F_N(c):=\frac1N|\{2\le n\le N: d_n/\log n<c\}|$ to a limit using deep distributional results on primes in short intervals / correlations of primes.
   • $P2$ Prove existence of the density for all $c$ by deriving asymptotics for counts of gaps $\le c\log n$ from (a strong form of) the Hardy--Littlewood prime tuples conjecture.

   Disproof/construction strategies.

   • $D1$ Attempt to use known irregularity phenomena for primes in short intervals (à la Maier) to construct a $c$ for which $F_N(c)$ has distinct limit points.
   • $D2$ Attempt to show $f$ cannot be continuous by forcing a nontrivial point mass in any putative limiting distribution.

   Best path chosen here. In this write-up, the proof track $P1$--$P2$ and the disproof track $D1$--$D2$ both stall at the point where one would need substantially finer distributional control of prime gaps than is established here. Accordingly, I record rigorous partial results that any full solution must be consistent with: (i) the normalized gaps have mean $1+o(1)$; (ii) this yields quantitative bounds on upper/lower densities for large $c$.

4. WORK.

   Lemma 234.1 (Mean of normalized gaps).

   Let $a_n:=\dfrac{p_{n+1}-p_n}{\log n}$ for $n\ge 2$. Assuming the Prime Number Theorem, one has \[ \lim_{N\to\infty}\frac{1}{N}\sum_{n=2}^{N} a_n = 1. \]
   Proof: Write $d_n:=p_{n+1}-p_n$. For $N\ge 2$ consider \[ S(N):=\sum_{n=2}^{N}\frac{d_n}{\log n}. \] For each $n\ge 2$ we have \[ \frac{d_n}{\log n}=\int_{p_n}^{p_{n+1}} \frac{dt}{\log n}. \] On the interval $[p_n,p_{n+1})$ the prime-counting function satisfies $\pi(t)=n$, hence $\log n = \log \pi(t)$ there. Summing over $n=2,\dots,N$ and concatenating the integrals gives the identity \[ S(N)=\int_{p_2}^{p_{N+1}}\frac{dt}{\log\pi(t)}. \] By the Prime Number Theorem, $\pi(t)\sim t/\log t$, so for sufficiently large $t$ we have \[ \log\pi(t)=\log t + O(\log\log t). \] For large $t$, the error term satisfies $|O(\log\log t)|\le \tfrac12\log t$, so the function $u\mapsto 1/(\log t+u)$ can be expanded to first order, yielding \[ \frac{1}{\log\pi(t)} = \frac{1}{\log t} + O\!\left(\frac{\log\log t}{(\log t)^2}\right). \] Therefore, with $X:=p_{N+1}$, \[ S(N)=\int_{p_2}^{X}\frac{dt}{\log t} \;+ O\!\left(\int_{p_2}^{X}\frac{\log\log t}{(\log t)^2}\,dt\right). \] The main integral is the logarithmic integral $\operatorname{li}(X)$ up to an additive constant. By integration by parts, \[ \int_{2}^{X}\frac{dt}{\log t}=\frac{X}{\log X}+O\!\left(\frac{X}{(\log X)^2}\right). \] For the error integral, since the integrand is eventually increasing in $t$ and bounded above by its value at $X$, one has \[ \int_{2}^{X}\frac{\log\log t}{(\log t)^2}\,dt \ll \frac{X\log\log X}{(\log X)^2}. \] Combining these estimates gives \[ S(N)=\frac{X}{\log X}+O\!\left(\frac{X\log\log X}{(\log X)^2}\right). \] Finally, by the Prime Number Theorem in the equivalent form $p_{N+1}\sim (N+1)\log(N+1)$, we have $X\sim N\log N$ and $\log X = \log N + O(\log\log N)$. Hence \[ \frac{X}{\log X}=N\,(1+o(1)), \qquad \frac{X\log\log X}{(\log X)^2}=N\,o(1). \] Therefore $S(N)=N+o(N)$, and dividing by $N$ yields \[ \frac{1}{N}\sum_{n=2}^{N}a_n = \frac{S(N)}{N} \to 1. \] ∎

   Corollary 234.2 (A Markov-type bound on large normalized gaps).

   Assume Lemma 234.1. Fix $c>0$. Then \[ \limsup_{N\to\infty}\frac{1}{N}\left|\{2\le n\le N: a_n\ge c\}\right|\le \frac{1}{c}. \] In particular, for $c>1$, \[ \liminf_{N\to\infty}\frac{1}{N}\left|\{2\le n\le N: a_n< c\}\right|\ge 1-\frac{1}{c}. \]
   Proof: For each $N$ and $c>0$, \[ \sum_{n=2}^{N}a_n \ge \sum_{\substack{2\le n\le N\\ a_n\ge c}} a_n \ge c\,\left|\{2\le n\le N: a_n\ge c\}\right|. \] Divide by $N$ and take $\limsup$; Lemma 234.1 gives $\lim_{N\to\infty}\frac1N\sum_{n=2}^N a_n =1$, hence the displayed bound. The second inequality is the complement. ∎

   What this does not give. These arguments control only the first moment of $a_n$ and hence give only coarse bounds on the distribution of $a_n$; they do not address existence of the density $f(c)$ for a fixed $c$, nor continuity of $f$.

5. VERIFICATION.
   • Quantifiers. Lemma 234.1 uses only $N\to\infty$ and PNT; the conversion $S(N)=\int dt/\log\pi(t)$ is exact.
   • Edge cases. The definition starts at $n\ge 2$ so $\log n$ is defined. Any finite number of initial terms does not affect densities.
   • Error control. The step $1/\log\pi(t)=1/\log t + O(\log\log t/(\log t)^2)$ requires $\log\pi(t)=\log t + O(\log\log t)$ and that $\log t$ dominates $\log\log t$; this holds for all sufficiently large $t$.

6. FINAL.

   UNRESOLVED.

   • (i) Strongest fully proved partial result obtained here. Assuming PNT, the mean of the normalized gaps $a_n=(p_{n+1}-p_n)/\log n$ tends to $1$, and hence for every fixed $c>0$, $\limsup_{N\to\infty}\frac1N|\{n\le N: a_n\ge c\}|\le 1/c$.
   • (ii) First gap. No unconditional (or even PNT-level) method in this write-up proves that the limit defining $f(c)$ exists for an arbitrary fixed $c$, nor continuity of the putative limit function.
   • (iii) Top 3 next moves.
     • Target an "asymptotic for the count of gaps $\le c\log n$" from a strong Hardy--Littlewood prime tuples input.
     • Try to convert known irregularity results for primes in short intervals into oscillations of $F_N(c)$ to attack existence of the density.
     • Study whether any currently known moment information on prime gaps (even conditional) implies continuity/no-atom properties of a limiting law.
   • (iv) What a minimal counterexample would likely look like. A value $c_0$ and two sequences $N_j\to\infty$, $M_j\to\infty$ such that $F_{N_j}(c_0)$ and $F_{M_j}(c_0)$ converge to different limits, i.e. a persistent oscillation in the frequency of "$d_n< c_0\log n$".

7. COMPLETION ESTIMATE (MANDATORY).
   COMPLETION: 35%.