login
A109663
Numbers k such that the sum of the digits of (k^k + k!) is divisible by k.
1
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
1,2
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
EXAMPLE
The digits of 1881^1881 + 1881! sum to 28215 and 28215 is divisible by 1881, so 1881 is in the sequence.
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 *)
CROSSREFS
Sequence in context: A245594 A078610 A108825 * A056702 A294126 A091361
KEYWORD
base,more,nonn
AUTHOR
Ryan Propper, Aug 06 2005
EXTENSIONS
a(21)-a(27) from Lars Blomberg, Jul 05 2011
STATUS
approved