Wikipedia:Reference desk/Archives/Mathematics/2024 July 21

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July 21

If the sum of the first m factorial numbers is equal to the sum of the first n positive integers

If the sum of the first m factorial numbers is equal to the sum of the first n positive integers, i.e. 1! + 2! + 3! + … + m! = 1 + 2 + 3 + … + n, then (m,n) = (0,0), (1,1), (2,2), (5,17), right? 220.132.216.52 (talk) 20:30, 21 July 2024 (UTC)

The triangular numbers, modulo 19, are reduced to one of 10 possibilities: 0, 1, 2, 3, 6, 7, 9, 10, 15 and 17. The sum 1! + 2! + 3! + ... + m!, for m > 17, modulo 19, is reduced to 8. Therefore no further factorial sums are triangular.  --Lambiam 22:19, 21 July 2024 (UTC)

I looked at mod 7 with about the same result. The left hand side is 5 for m≥6 and the right hand side can never be 5. So you only have to check m from 1 to 5. (Btw, I would count 0! = 1 as a factorial number, so the sums of factorials would be 1, 2, 4, 10, 34, ... . (sequence A003422 in the OEIS)) --RDBury (talk) 22:28, 21 July 2024 (UTC)

Wrote up a quick MATLAB script to find numbers which can be used as modulos to show that the list is finite, it starts: $7,13,14,17,19,21,25,26,28,34,35,38,\ldots$. Obviously if a number appears in the list then all its positive multiples do too. The list of nontrivial numbers starts $7,13,17,19,25,59,67,73,79,83,94,97,\ldots$ GalacticShoe (talk) 03:01, 22 July 2024 (UTC)

Submit to OEIS! —Tamfang (talk) 20:31, 28 July 2024 (UTC)

Have you submitted this sequence to OEIS? 220.132.216.52 (talk) 12:28, 2 August 2024 (UTC)

I like that idea, although I feel that I should probably collect more information about the sequence before I do. GalacticShoe (talk) 16:08, 2 August 2024 (UTC)