gpt pro 5.4 attempt on Math-723

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.