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%I #12 Jul 07 2019 02:23:46
%S 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,
%T 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,
%U 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
%N Number of numbers k>1 such that k equals the sum of digits in k^n.
%C 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.
%D Joe Roberts, "Lure of the Integers", The Mathematical Association of America, 1992, p. 172.
%H T. D. Noe, <a href="/A046471/b046471.txt">Table of n, a(n) for n=1..1000</a>
%e a(17)=4 -> sum-of-digits{x^17}=x for x=80,143,171 and 216 (x>1).
%Y Cf. A046459, A046469, A046000.
%Y a(n) = A046019(n) - 1.
%Y Cf. A152147 (table of k such that the sum of digits of k^n equals k)
%K nonn,base
%O 1,1
%A _Patrick De Geest_, Aug 15 1998
%E Edited by _T. D. Noe_, Nov 25 2008
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