%I #24 Sep 03 2023 11:08:47
%S 0,1,2,3,4,5,6,7,8,9,81,512,2401
%N Numbers k such that k = (sum of digits of k)^(number of digits of k).
%C To prove completeness, consider that k^m contains more than m digits for every k >= 10 and check 1 <= k <= 9 explicitly. - Ulrich Schimke (ulrschimke(AT)aol.com)
%C Subset of A023106. - _R. J. Mathar_, Oct 20 2008
%e 512 is in the sequence because (5 + 1 + 2)^3 = 512.
%t Select[Range[2500],#==Total[IntegerDigits[#]]^IntegerLength[#]&] (* _Harvey P. Dale_, Oct 26 2011 *)
%Y Cf. A007953, A055642, A023106.
%K nonn,nice,fini,full,base
%O 1,3
%A _Felice Russo_