A108827 Numbers k that divide the sum of the digits of k^k.

1, 2, 3, 9, 18, 27, 54, 90, 108, 163, 197, 254, 432, 1292, 2202, 9648, 10347, 16596, 17203, 46188, 46992, 77121, 130082, 167410, 216546, 596277

Offset: 1

Author

Ryan Propper, Jul 11 2005

Keywords

• more
• nonn
• base

Comments

Especially for larger terms k not divisible by 10, we can expect 4.5 times the number of digits in k^k to be close to some integer multiple (m) of k, so k should occur near 100^(m/9). E.g., for m = 10, 11, ..., 16, approximate (and corresponding actual) values would be 167 (163, 197), 278 (254), 464 (432), 774 (none), 1292 (1292), 2154 (2022) and 3594 (none). Larger terms k ending with exactly j zeros would be expected to occur near k = 10^j * 100^(m/9) for some integer m. - Jon E. Schoenfield, Jun 09 2007

The quotients are 1, 2, 3, 5, 6, 7, 7, 4, 9, 10, 10, 11, 12, 14, 15, 18, 18, 19, 19, 21, 21, 22, 23, 19, 24, 26.

Examples

			3^3 = 27; 2 + 7 = 9; 9 mod 3 = 0.
9^9 = 387420489; 3 + 8 + 7 + 4 + 2 + 4 + 8 + 9 = 45; 45 mod 9 = 0.
432 is a term because the sum of the digits of 432^432 = 5184 is divisible by 432.
		

Crossrefs

Cf. A125526, A125724, A108825.

Programs

Maple

a:=proc(n) local nn: nn:=convert(n^n,base,10): if type(add(nn[j],j=1..nops(nn))/n, integer)=true then n else fi end: seq(a(n),n=1..2000); # Emeric Deutsch

Mathematica

Do[If[Mod[Plus @@ IntegerDigits[n^n], n] == 0, Print[n]], {n, 1, 10000}]
Select[Range[600000],Divisible[Total[IntegerDigits[#^#]],#]&] (* Harvey P. Dale, Jan 28 2017 *)

Extensions

a(16)-a(19) from Simon Nickerson (simonn(AT)maths.bham.ac.uk) and Emeric Deutsch, Jul 15 2005

a(20)-a(22) from Ray Chandler, Jul 25 2005

Edited by N. J. A. Sloane, Apr 27 2008 at the suggestion of Stefan Steinerberger

a(23) from Robert G. Wilson v, May 17 2008

a(24) from Robert G. Wilson v, May 19 2008

a(25)-a(26) from Lars Blomberg, Jul 09 2011