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A046471
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Number of numbers k>1 such that k equals the sum of digits in k^n.
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6
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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
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1,1
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COMMENTS
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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.
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REFERENCES
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Joe Roberts, "Lure of the Integers", The Mathematical Association of America, 1992, p. 172.
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LINKS
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EXAMPLE
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a(17)=4 -> sum-of-digits{x^17}=x for x=80,143,171 and 216 (x>1).
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CROSSREFS
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Cf. A152147 (table of k such that the sum of digits of k^n equals k)
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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