login
A062927
Numbers k such that k divides the sum of digits of 9^k.
2
1, 3, 6, 9, 27, 90, 108, 144, 243
OFFSET
1,2
COMMENTS
Next term after 243, if it exists, is greater than 30000. - Ryan Propper, Jun 18 2005
No more terms < 1000000. - Lars Blomberg, May 05 2011
EXAMPLE
3 divides the sum of digits of 9^3 (i.e., 7 + 2 + 9 = 18), so 3 is a term.
MATHEMATICA
Do[If[Mod[Plus @@ IntegerDigits[9^n, 10], n] == 0, Print[n]], {n, 1, 30000}] (* Ryan Propper, Jun 18 2005 *)
Select[Range[300], Divisible[Total[IntegerDigits[9^#]], #]&] (* Harvey P. Dale, Jun 03 2015 *)
CROSSREFS
Sequence in context: A019461 A223556 A067862 * A349597 A178467 A193802
KEYWORD
hard,more,nonn,base
AUTHOR
STATUS
approved