GPT 6 Astra Ultra attempt on Math-5847
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Contents
Erdős Problem #728: balanced factorial divisibility
Full logarithmic-excess construction by digit counting, 24 September 2026.Statement and a nontrivial target
For every fixed $C>0$ and $0<\varepsilon<1/2$, we prove that infinitely many triples satisfy \[ \varepsilon n\leq a,b\leq(1-\varepsilon)n,\qquad a+b>n+C\log n,\qquad a!b!\mid n!(a+b-n)!. \] Thus the proof addresses the intended balanced version, as well as the printed version that permits trivial choices $a=b=n$. We reconstruct the carry-versus-large-valuation method from the cited contemporary proofs, with simpler nonoptimal constants.Valuations and a useful interval bound
Put $B_m=\binom{2m}{m}$ and $\kappa_p(m)=v_p(B_m)$. Counting multiples of each prime power in factorials gives \[ \kappa_p(m)=\sum_{j\geq1} \left(\left\lfloor\frac{2m}{p^j}\right\rfloor -2\left\lfloor\frac m{p^j}\right\rfloor\right). \tag{1} \] Every summand is zero or one. In particular each base-$p$ digit of $m$ that is at least $\lceil p/2\rceil$ contributes one at the corresponding digit level.For any $K\geq1$ let $J_p=\max_{1\leq i\leq K}v_p(m+i)$. Then \[ v_p\binom{m+K}{K}\leq J_p. \tag{2} \] To prove this, select $m+i_0$ attaining the maximum. For each $j\leq J_p$, the number of other multiples of $p^j$ in an interval of $K$ consecutive integers is at most $\lceil K/p^j\rceil-1\leq\lfloor K/p^j\rfloor$. For $j>J_p$ there are no such multiples. Thus \[ v_p((m+1)\cdots(m+K))\leq J_p+v_p(K!), \] which proves $2$.
Good integers in a dyadic interval
Fix $D>0$, let $M$ be a large integer, $L=\log M$, and $K=\lceil DL\rceil$. We show that all but $o_D(M)$ integers $m\in[M,2M)$ simultaneously satisfy, for every prime $p\leq2K$, \[ \kappa_p(m)\geq\frac L{18\log p}, \qquad J_p\leq T_p:=\left\lceil\frac{4\log L}{\log p}\right\rceil. \tag{3} \]For the first assertion put $d=\lfloor L/(2\log p)\rfloor$. Modulo $p^d$ the $d$ low digits are independent uniform digits. The probability that any one is at least $\lceil p/2\rceil$ is $\lfloor p/2\rfloor/p\geq1/3$. The elementary binomial lower-tail estimate gives probability at most $e^{-d/24}$ that fewer than $d/6$ digits have this property. One may obtain this estimate directly from exponential Markov inequality; equivalently, if the mean is $\mu\geq d/3$, the bound $\mathbb P(Y\leq\mu/2)\leq e^{-\mu/8}$ suffices.
The modulus $p^d\leq\sqrt M$. Counting complete residue periods and a remaining incomplete one shows that the number of such bad $m$ in $[M,2M)$ is at most $2M e^{-d/24}$. Uniformly for $p\leq2K$ and large $M$, we have $d\geq L/(3\log p)$. By $1$, the good digits imply the first inequality in $3$. Summing the exceptional count over at most $2K$ primes gives \[ 4KM\exp\left(-\frac L{72\log(2K)}\right)=o_D(M). \tag{4} \]
For the second assertion, $v_p(m+i)>T_p$ forces $p^{T_p+1}\mid m+i$, where $p^{T_p+1}\geq L^4$. For a fixed $p,i$, at most $M/L^4+1$ integers in the interval can do this. A union bound over at most $2K$ primes and $K$ offsets bounds this exceptional count by \[ 2K^2(M/L^4+1)=O_D(M/L^2+L^2)=o_D(M). \tag{5} \] Equations $4$--$5$ prove $3$. No independence between different primes is assumed.
Every prime is controlled
Choose a good $m$. For $p\leq2K$, \[ T_p\log p\leq4\log L+\log(2K)<L/18 \] once $M$ is large. Equations $2$--$3$ then give $v_p\binom{m+K}{K}\leq\kappa_p(m)$.For $p>2K$, let $J=J_p$. If $J=0$ there is nothing to prove. Otherwise choose $i\leq K$ with $p^J\mid m+i$. For every $1\leq j\leq J$ we have \[ m\bmod p^j=p^j-i>p^j/2, \] since $i\leq K<p/2$. Thus each of the first $J$ terms of $1$ is one, and $\kappa_p(m)\geq J$. Applying $2$ proves the same valuation inequality for these primes too. We conclude that \[ \binom{m+K}{K}\mid\binom{2m}{m}. \tag{6} \]
Returning to the requested parameters
Take $D=C+1$ and set $n=2m$, $a=m+K$, $b=m$. The ratio in $6$ is exactly \[ \frac{\binom{2m}{m}}{\binom{m+K}{K}} =\frac{(2m)!K!}{m!(m+K)!}, \] so it is the desired divisibility. Since $m\in[M,2M)$, $K\geq(C+1)\log M>C\log(2m)$ for large $M$. Also $K/m\to0$, giving both balance inequalities for every fixed $\varepsilon<1/2$. Taking $M$ through increasing powers of two supplies infinitely many distinct $n$.More precisely, if $0<C_1<C_2$, choose $D$ strictly between them. Since $\log(2m)/\log M\to1$ uniformly in the interval, the construction also gives $C_1\log n<a+b-n<C_2\log n$.