A046471 Number of numbers k>1 such that k equals the sum of digits in k^n.

8, 1, 5, 5, 4, 4, 8, 3, 3, 6, 3, 1, 11, 5, 7, 6, 4, 2, 9, 3, 3, 7, 3, 3, 13, 4, 2, 6, 5, 1, 10, 1, 7, 3, 5, 2, 8, 2, 2, 6, 1, 4, 9, 5, 3, 8, 8, 4, 11, 1, 3, 4, 4, 5, 2, 1, 6, 3, 4, 4, 5, 2, 3, 4, 4, 3, 8, 1, 5, 3, 2, 2, 5, 4, 5, 3, 3, 4, 8, 4, 2, 4, 4, 1, 5, 2, 6, 6, 3, 2, 7, 3, 3, 8, 5, 1, 7, 1, 4, 5, 2, 3, 9

Offset: 1

Author

Patrick De Geest, Aug 15 1998

Keywords

• nonn
• base

Comments

The number of digits in k^n is at most 1+n*log(k). Hence the maximum sum of digits of k^n is 9(1+n*log(k)). By solving k=9(1+n*log(k)), we can compute an upper bound on k for each n. Sequence A133509 lists the n for which a(n)=0.

Examples

			a(17)=4 -> sum-of-digits{x^17}=x for x=80,143,171 and 216 (x>1).
		

References

• Joe Roberts, "Lure of the Integers", The Mathematical Association of America, 1992, p. 172.

Links

• T. D. Noe, Table of n, a(n) for n=1..1000

Crossrefs

Cf. A046459, A046469, A046000.

a(n) = A046019(n) - 1.

Cf. A152147 (table of k such that the sum of digits of k^n equals k)

Extensions

Edited by T. D. Noe, Nov 25 2008