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A055457
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5^a(n) exactly divides 5n. Or, 5-adic valuation of 5n.
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6
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1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 3, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 3, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 3, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 3, 1, 1, 1, 1, 2
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,5
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COMMENTS
| More generally, consider the sequence defined by p^a(n) exactly divides pn. For p = 3 we have A051064 and for p = 2 we have A001511.
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LINKS
| T. D. Noe, Table of n, a(n) for n=1..1000
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FORMULA
| G.f.: Sum(k>=0, x^(5^k)/(1-x^5^k)). - Ralf Stephan (ralf(AT)ark.in-berlin.de), Apr 12 2002
Multiplicative with a(p^e) = e+1 if p = 5, 1 otherwise
a(n)=(-1)*sum_{d divides n} mu(5d)*tau(n/d) - Benoit Cloitre (benoit7848c(AT)orange.fr), Jun 21 2007
Dirichlet g.f. zeta(s)/(1-1/5^s). - R. J. Mathar, Feb 09 2011
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EXAMPLE
| a(5) = 2 since 5^2 exactly divides 5 times 5, a(25) = 3 since 5^3 exactly divides 5 times 25 and a(125) = 4 since 5^4 exactly divides 5 times 125.
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PROG
| (PARI) a(n)=-sumdiv(n, d, moebius(5*d)*numdiv(n/d)) - Benoit Cloitre (benoit7848c(AT)orange.fr), Jun 21 2007
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CROSSREFS
| Cf. A001511, A007949, A051064.
Sequence in context: A139549 A130782 A177706 * A032542 A107038 A043278
Adjacent sequences: A055454 A055455 A055456 * A055458 A055459 A055460
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KEYWORD
| nonn,mult
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AUTHOR
| Alford Arnold (Alford1940(AT)aol.com), Jun 25 2000
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