A067864 Numbers k such that k divides the sum of digits of 6^k.

1, 3, 9, 18, 90

Offset: 1

Author

Amarnath Murthy and Shyam Sunder Gupta, Feb 16 2002

Keywords

• hard
• nonn
• base

Comments

Conjecture: the sequence is finite, since (sum of the digits of 6^k)/k -> log_10(6)*4.5 ~ 3.50168 as k->infinity (this is also a conjecture). - Robert Gerbicz, May 08 2008

Next term, if it exists, exceeds 100000. - Sean A. Irvine, Apr 05 2010

The keyword "hard" refers to the difficulty of finding, or disproving the existence of, the next term. - N. J. A. Sloane, Nov 09 2024

Examples

			a(1)=3, so 3 divides the sum of digits of 6^3 (i.e., 2 + 1 + 6 = 9).
		

Crossrefs

Cf. A000400, A066002, A175457, A175169, A067862, A067863.

Programs

Mathematica

Select[Range[100],Mod[Total[IntegerDigits[6^#]],#]==0&] (* Harvey P. Dale, Nov 09 2024 *)