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A056888
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a(n) = number of k such that sum of digits of 9^k is 9n.
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1
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2, 3, 2, 0, 4, 1, 3, 1, 1, 5, 2, 2, 3, 1, 0, 3, 6, 2, 3, 0, 0, 4, 1, 3, 1, 4, 1, 1, 0, 1, 3, 2, 3, 5, 1, 1, 3, 3, 2, 5, 0, 3, 3, 1, 1, 3, 2, 2, 0, 2, 1, 5, 2, 1, 1, 1, 1, 3, 4, 5, 1, 0, 1, 3, 2, 1, 2, 4, 5, 1, 1, 2, 1, 0, 1, 2, 4, 1, 2, 5, 0, 2, 4, 3, 2, 2, 1, 2, 2, 2, 0, 2, 3, 2, 1, 5, 1, 0, 4
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OFFSET
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1,1
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COMMENTS
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Proposed by Mark Sapir, Math. Dept., Vanderbilt University, who remarks (August 2000) that he can prove that a(n) is always finite and that a(1) = 2.
All terms except the first are only conjectures. For the theorem that a(n) is always finite, see Senge-Straus and Stewart. - N. J. A. Sloane, Jan 06 2011
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REFERENCES
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H. G. Senge and E. G. Straus, PV-numbers and sets of multiplicity, Periodica Math. Hungar., 3 (1971), 93-100.
C. L. Stewart, On the representation of an integer in two different bases, J. Reine Angew. Math., 319 (1980), 63-72.
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LINKS
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EXAMPLE
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There are two powers of 9 with digit-sum 9, namely 9 and 81, so a(1) = 2.
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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