cp's OEIS Frontend

A109657 Numbers n such that the sum of the digits of Sum_{k=1..n} (k!) is divisible by n.

%I A109657 #19 Apr 16 2023 12:41:48
%S A109657 1,3,6,9,12,18,54,117,272,294,296,320,783,1125,2088,3375,16164,16407,
%T A109657 26286,26777,26784,27516,27568,45945,74970,124236,125589,208116,
%U A109657 348705,583746,586218,586353,586368,586536,588567,2712944,2714655,2714912,2720288,2720399
%N A109657 Numbers n such that the sum of the digits of Sum_{k=1..n} (k!) is divisible by n.
%C A109657 Most, but not all, of the terms in this sequence are divisible by 3; is this a coincidence?
%C A109657 In general, terms should be more likely to occur in regions where the number of digits in the sum of the first n factorials is close to an integer multiple of 2*n/9. This happens, e.g., around n = 268, 449, 752, 1257, 2100, 3506, 5851, 9763, 16290, 27177, 45337, 75631, 126165, etc. - _Jon E. Schoenfield_, Jun 16 2010
%C A109657 Numbers n such that A349403(n) (mod n) == 0. - _Kevin P. Thompson_, Nov 28 2021
%C A109657 a(43) > 5570000. - _Kevin P. Thompson_, Nov 28 2021
%H A109657 Kevin P. Thompson, <a href="/A109657/b109657.txt">Table of n, a(n) for n = 1..42</a>
%e A109657 6 is a member of the sequence since Sum_{k=1..6}(k!) = 1! + 2! + 3! + 4! + 5! + 6! = 1 + 2 + 6 + 24 + 120 + 720 = 873 which has a digit sum of 18 that is divisible by 6.
%t A109657 s = 0; Do[s += n!; k = Plus @@ IntegerDigits[s]; If[Mod[k, n] == 0, Print[n]], {n, 1, 10000}]
%t A109657 Module[{nn=2721000,sf},sf=Total[IntegerDigits[#]]&/@Accumulate[Range[nn]!];Select[ Thread[ {Range[nn],sf}],Mod[#[[2]],#[[1]]]==0&]][[;;,1]] (* _Harvey P. Dale_, Apr 16 2023 *)
%Y A109657 Cf. A000142, A007489, A349403.
%K A109657 nonn,base
%O A109657 1,2
%A A109657 _Ryan Propper_, Aug 06 2005
%E A109657 More terms from _Jon E. Schoenfield_, Jun 16 2010
%E A109657 a(26)-a(40) from _Kevin P. Thompson_, Nov 28 2021