GPT 6 Astra Ultra attempt on Math-5629
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Contents
Erdős Problem 403
Statement and result.
Does $2^m=a_1!+\cdots+a_k!$, with distinct positive indices $a_1<\cdots<a_k$, have only finitely many solutions? Yes. We reconstruct the known Frankl--Lin result using an explicit finite integer computation: the possible exponents are \[ m=0,1,3,5,7. \tag{1} \] The largest solution is $128=2!+3!+5!$. The small computation below is part of the proof; no Lean verification of this particular reconstruction is claimed.Reduction to a bounded computation.
For $m\geq1$, parity excludes $1!$. Reducing modulo $3$ then forces $2!$ to occur, since every factorial with index at least three is divisible by three. We prove that no sum of distinct factorials containing $2!$ and no $1!$ is divisible by $2^{255}$.Put $S_2=\{2\}$. For $3\leq n\leq255$, define \[ e_n=\min(255,v_2((n+1)!)),\qquad S_n=\{x:x=s\text{ or }s+n!,\ s\in S_{n-1},\ 2^{e_n}\mid x\}. \tag{2} \] If a full factorial sum is divisible by $2^{255}$, its partial sum through $n!$ must lie in $S_n$: all omitted future factorials are divisible by $2^{e_n}$. This follows inductively, including zero choices at missing indices. At $n=255$, all later factorials are multiples of $2^{255}$. Therefore $S_{255}=\varnothing$ excludes every full sum, including those with indices exceeding $255$.
Exact certificate.
The following complete Python code uses only unbounded integer arithmetic. The identity $v_2(t!)=t-\operatorname{popcount}(t)$ evaluates $2$.from math import factorial
S, F = {2}, 2
peak = 1
for n in range(3, 256):
F *= n
e = min(255, n + 1 - (n + 1).bit_count())
S = {x for s in S for x in (s, s + F)
if x % (1 << e) == 0}
peak = max(peak, len(S))
assert peak == 496
assert S == set()
facts = [factorial(i) for i in range(1, 256)]
answers = []
for m in range(255):
remainder, indices = 1 << m, []
for i in range(255, 0, -1):
if facts[i - 1] <= remainder:
remainder -= facts[i - 1]
indices.append(i)
if remainder == 0:
answers.append((m, sorted(indices)))
assert answers == [(0,[1]), (1,[2]), (3,[2,3]),
(5,[2,3,4]), (7,[2,3,5])]
For additional audit points, the cardinalities $|S_n|$ for $n=250,251,252,253,254,255$ are respectively $104,57,114,114,228,0$. The code was run successfully.The first assertion proves $m\leq254$. Also $256!>2^{254}$, so indices at least $256$ cannot occur in that remaining range. The greedy check is exhaustive because $\sum_{i=1}^{t-1}i!<t!$ for every $t\geq2$. Thus the largest available factorial must be selected if it does not exceed the remainder; otherwise no combination of smaller ones can replace it. This proves both the enumeration and $1$.
Index convention and attribution.
The argument uses positive indices. If $0!$ is also allowed, finiteness and the largest solution remain the same. For $m\geq3$, parity says that $0!$ and $1!$ occur together or neither does. The contribution from indices $0,1,2$ is then $0,2,$ or $4$. Modulo $3$ rules out zero. A contribution of $4$, together with the optional $3!=6$, is $4$ or $2\pmod8$, so is impossible. The contribution is therefore exactly $2$, and the same computation applies. The finitely many smaller exponents cause no issue.The known result is attributed to Frankl and Lin at https://www.erdosproblems.com/403, checked 24 September 2026. This is a reproducible reconstruction of that result, not a new discovery.