A056888 a(n) = number of k such that sum of digits of 9^k is 9n.

2, 3, 2, 0, 4, 1, 3, 1, 1, 5, 2, 2, 3, 1, 0, 3, 6, 2, 3, 0, 0, 4, 1, 3, 1, 4, 1, 1, 0, 1, 3, 2, 3, 5, 1, 1, 3, 3, 2, 5, 0, 3, 3, 1, 1, 3, 2, 2, 0, 2, 1, 5, 2, 1, 1, 1, 1, 3, 4, 5, 1, 0, 1, 3, 2, 1, 2, 4, 5, 1, 1, 2, 1, 0, 1, 2, 4, 1, 2, 5, 0, 2, 4, 3, 2, 2, 1, 2, 2, 2, 0, 2, 3, 2, 1, 5, 1, 0, 4

Offset: 1

Author

N. J. A. Sloane, Sep 05 2000

Keywords

• nonn
• base

Comments

Proposed by Mark Sapir, Math. Dept., Vanderbilt University, who remarks (August 2000) that he can prove that a(n) is always finite and that a(1) = 2.

Values of a(n) for n>1 computed numerically by Michael Kleber, Sep 02 2000 and David W. Wilson, Sep 06 2000.

All terms except the first are only conjectures. For the theorem that a(n) is always finite, see Senge-Straus and Stewart. - N. J. A. Sloane, Jan 06 2011

Examples

			There are two powers of 9 with digit-sum 9, namely 9 and 81, so a(1) = 2.
		

References

• H. G. Senge and E. G. Straus, PV-numbers and sets of multiplicity, Periodica Math. Hungar., 3 (1971), 93-100.
• C. L. Stewart, On the representation of an integer in two different bases, J. Reine Angew. Math., 319 (1980), 63-72.

Crossrefs

Cf. A065999.