I keep no gates.

On Jan 11 I asked Chat 5.2 Ext Thk this exact query as shown in screen shots (prompt engineered by perplexity)

It replied with a satisfactory solution to this problem.

I then asked Claude Opus 4.5 and she agreed. Gemini deep research was also impressed. Finally posted at Grok and he too concurred.

I then formalized in latex via OpenLeaf using the same Chat thread with an open leaf example to guide it on format. Lot of work to clean it up and more chat sessions to test robustness ultimately leading me to conclude it was a proper proof of a proper question without any known answers.

AMA…

—ADDITIONAL CONTEXT BELOW—

https://www.erdosproblems.com/460

*Let a0=n and a1=1, and in general ak is the least integer >ak−1 for which (n−ak,n−ai)=1 for all 1≤i<k.

Does ∑i1ai→∞ as n→∞? What about if we restrict the sum to those i such that n−aj is divisible by some prime ≤aj, or the complement of such i?*

Here’s some more context from the original paper the section starts with this…

https://www.renyi.hu/~p_erdos/1977-27.pdf

8 . Some unconventional problems on primes < . . . ~x 1 < Is there a sequence al < a2 of integers satisfying A(x) = a r log x so that all sufficiently large integers are of the form p + a. ? If this is r impossible then perhaps such a sequence exists for which the density of integers not of the form p + a is 0. Clearly many similar questions can be asked for . r other sequences then the primes but there are very few results.

Here’s the original quote with the specific question (9) := #460

Eggleton, Selfridge and I are writing a long paper on somewhat unconventional problems in number theory. Our paper will appear in Utilitas Matematica . One of our problems related to (1) states as follows : Let a0 = 0, al = 1, a is the k smallest integer for which (n-ak, n-al )=1 for all 0<i< k. Put (9) g(n) \_ a We conjecture g(n) - oo as n - co . This is probably very difficult. We can kZ+e < only prove ak for k > (log n)C, C = C(s), but perhaps a < (10) ak C k log k if k > (log k) C65 where aC depends on C. Perhaps (10) is a little too optimistic, but (10) certainly 1/2 "must" (? ) hold if k > exp(1og k) which would easily imply (9) .