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

1, 3, 6, 9, 12, 18, 54, 117, 272, 294, 296, 320, 783, 1125, 2088, 3375, 16164, 16407, 26286, 26777, 26784, 27516, 27568, 45945, 74970, 124236, 125589, 208116, 348705, 583746, 586218, 586353, 586368, 586536, 588567, 2712944, 2714655, 2714912, 2720288, 2720399

Offset: 1

Author

Ryan Propper, Aug 06 2005

Keywords

• nonn
• base

Comments

Most, but not all, of the terms in this sequence are divisible by 3; is this a coincidence?

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

Numbers n such that A349403(n) (mod n) == 0. - Kevin P. Thompson, Nov 28 2021

a(43) > 5570000. - Kevin P. Thompson, Nov 28 2021

Examples

			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.
		

Links

• Kevin P. Thompson, Table of n, a(n) for n = 1..42

Crossrefs

Cf. A000142, A007489, A349403.

Programs

Mathematica

s = 0; Do[s += n!; k = Plus @@ IntegerDigits[s]; If[Mod[k, n] == 0, Print[n]], {n, 1, 10000}]
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 *)

Extensions

More terms from Jon E. Schoenfield, Jun 16 2010

a(26)-a(40) from Kevin P. Thompson, Nov 28 2021