A109663 Numbers k such that the sum of the digits of (k^k + k!) is divisible by k.

1, 2, 3, 9, 15, 18, 27, 36, 51, 81, 93, 169, 181, 348, 444, 504, 528, 1881, 2031, 9843, 16479, 16685, 45435, 129056, 138510, 214008, 358326

Offset: 1

Author

Ryan Propper, Aug 06 2005

Keywords

• base
• more
• nonn

Comments

The quotients are 2, 3, 2, 6, 6, 5, 8, 7, 6, 9, 9, 10, 10, 12, 12, 12, 12, 15, 15, 18, 19, 19, 21, 23, 22, 24, 25.

No more terms < 500000. - Lars Blomberg, Jul 05 2011

Examples

			The digits of 1881^1881 + 1881! sum to 28215 and 28215 is divisible by 1881, so 1881 is in the sequence.
		

Crossrefs

Cf. A053042, A109662.

Programs

Mathematica

Do[s = n^n + n!; k = Plus @@ IntegerDigits[s]; If[Mod[k, n] == 0, Print[n]], {n, 1, 10000}]
Select[Range[360000],Divisible[Total[IntegerDigits[#^#+#!]],#]&] (* Harvey P. Dale, Dec 27 2018 *)

Extensions

a(21)-a(27) from Lars Blomberg, Jul 05 2011