

A046471


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


5



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
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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 > sumofdigits{x^17}=x for x=80,143,171 and 216 (x>1).


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)
Sequence in context: A200120 A154861 A153495 * A126585 A157289 A194097
Adjacent sequences: A046468 A046469 A046470 * A046472 A046473 A046474


KEYWORD

nonn,base


AUTHOR

Patrick De Geest, Aug 15 1998


EXTENSIONS

Edited by T. D. Noe, Nov 25 2008


STATUS

approved



