GPT 6 Astra Ultra attempt on Math-5522
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Erdős Problem 235: gaps between reduced residues of a primorial
1) Statement and scope
Let $N=\prod_{j\leq k}p_j$, and list its $\phi(N)$ reduced residues in $[1,N)$ in increasing order. The question asks for a continuous limiting distribution of successive gaps divided by $N/\phi(N)$. Hooley proved that the limit is $1-e^{-c}$ at $c\geq0$. Here we derive exact finite formulas, tightness, and disappearance of bounded unscaled gaps. The exponential limit is unresolved in this attempt.2) Cyclic normalization
Write $a_{\phi+1}=a_1+N$ and $d_i=a_{i+1}-a_i$ for $1\leq i\leq\phi=\phi(N)$. Then \[ \sum_i d_i=N,\qquad \frac1\phi\sum_i\frac{d_i}{N/\phi}=1. \] For each $c>0$, Markov's inequality gives \[ \frac1\phi\#\{i:d_i>cN/\phi\}\leq\frac1c. \] Thus the normalized cyclic gap laws are tight. The empirical distribution in the problem omits just the wraparound gap and uses the same denominator $\phi$, so it differs from the cyclic one by at most $1/\phi$, which tends to zero.3) An exact formula for the zero count
Choose a reduced residue $a$ uniformly modulo $N$. For $0\leq H<N$ put \[ X_H(a)=\sum_{h=1}^H1_{\gcd(a+h,N)=1}. \] The next cyclic gap exceeds $H$ if and only if $X_H(a)=0$. For a subset $S\subseteq\{1,\ldots,H\}$, let $\nu_p(S)$ denote the number of distinct residues represented by $\{0\}\cup S$ modulo $p$. The Chinese remainder theorem gives \[ \mathbb P\left(\gcd(a+h,N)=1\text{ for every }h\in S\right) =\frac{\prod_{p\mid N}(p-\nu_p(S))}{\phi(N)}. \] Consequently, for each integer $r\geq0$, \[ \mathbb E\binom{X_H}{r} =\frac1{\phi(N)} \sum_{\substack{S\subseteq[1,H]\\|S|=r}} \prod_{p\mid N}(p-\nu_p(S)), \] and inclusion--exclusion gives the exact identity \[ \mathbb P(X_H=0) =\sum_{r=0}^H(-1)^r\mathbb E\binom{X_H}{r}. \] No independence of the different shifts has been assumed.4) Fixed, unscaled gaps disappear
Fix $h\geq1$. If $p\mid h$, its factor in the conditional probability $\mathbb P(\gcd(a+h,N)=1)$ is one; otherwise its factor is $(p-2)/(p-1)$. In particular the probability is zero for odd $h$, once $2\mid N$. For any fixed even $h$, all primes greater than $h$ contribute $(p-2)/(p-1)$, and \[ \prod_{h<p\leq p_k}\left(1-\frac1{p-1}\right)\longrightarrow0 \] by divergence of the sum of prime reciprocals. The union bound over $1\leq h\leq H$ therefore proves $\mathbb P(d_i\leq H)\to0$ for every fixed $H$. This is consistent with the mean scale $N/\phi(N)\to\infty$.5) The precise missing estimate
For fixed $c>0$, take $H=\lfloor cN/\phi(N)\rfloor$. It would suffice to prove, for every fixed $r$, \[ \mathbb E\binom{X_H}{r}\longrightarrow\frac{c^r}{r!}. \] Indeed the odd and even Bonferroni partial sums then sandwich $\mathbb P(X_H=0)$ between the corresponding partial sums for $e^{-c}$. Letting the truncation order tend to infinity gives the desired limit. The case $c=0$ is immediate because all gaps are positive. The displayed moment asymptotic is not proved here; neither the exact mean nor tightness can replace it.Sources: https://www.erdosproblems.com/235; C. Hooley, On the difference between consecutive numbers prime to $n$. II, Publ. Math. Debrecen 12 $1965$, 39--49. The historical resolution is credited to Hooley; no new resolution or formal verification is claimed.