A038178 Numbers k such that k = (sum of digits of k)^(number of digits of k).

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 81, 512, 2401

Offset: 1

Author

Felice Russo

Keywords

• nonn
• nice
• fini
• full
• base

Comments

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)

Subset of A023106. - R. J. Mathar, Oct 20 2008

Examples

			512 is in the sequence because (5 + 1 + 2)^3 = 512.
		

Crossrefs

Cf. A007953, A055642, A023106.

Programs

Mathematica

Select[Range[2500],#==Total[IntegerDigits[#]]^IntegerLength[#]&] (* Harvey P. Dale, Oct 26 2011 *)