OFFSET
1,1
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.
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
EXAMPLE
a(17)=4 -> sum-of-digits{x^17}=x for x=80,143,171 and 216 (x>1).
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Patrick De Geest, Aug 15 1998
EXTENSIONS
Edited by T. D. Noe, Nov 25 2008
STATUS
approved