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A352218
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a(n) = least k such that A003592(n) | 20^k.
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2
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0, 1, 1, 1, 2, 1, 2, 1, 2, 3, 2, 2, 3, 2, 2, 3, 4, 3, 2, 3, 4, 3, 2, 3, 5, 4, 4, 3, 3, 5, 4, 4, 3, 3, 6, 4, 5, 5, 4, 3, 6, 4, 5, 5, 4, 3, 7, 4, 6, 5, 5, 6, 4, 7, 4, 6, 5, 5, 6, 4, 8, 4, 7, 5, 6, 6, 5, 8, 7, 4, 7, 5, 6, 6, 5, 9, 7, 4, 8, 5, 7, 6, 6, 9, 7, 5, 8
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OFFSET
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1,5
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COMMENTS
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Also, number of digits in the vigesimal (base 20) expansion of terminating unit fractions 1/A003592.
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REFERENCES
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G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Chapter IX: The Representation of Numbers by Decimals, Theorem 136. 8th ed., Oxford Univ. Press, 2008, 144-145.
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LINKS
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Eric Weisstein's World of Mathematics, Vigesimal
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EXAMPLE
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a(1) = 0 since A003592(1) = 1 | 20^0.
a(4) = 1 since A003592(4) = 5 | 20^1; 1/5 in base 20 = .4.
a(5) = 2 since A003592(5) = 8 | 20^2; 1/8 in base 20 = .2a, where "a" is digit 10), etc.
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MATHEMATICA
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With[{nn = 360000}, Sort[Join @@ Table[{2^a*5^b, Max[Ceiling[a/2], b]}, {a, 0, Log2[nn]}, {b, 0, Log[5, nn/(2^a)]}]][[All, -1]] ]
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CROSSREFS
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KEYWORD
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nonn,easy,base
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AUTHOR
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STATUS
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approved
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