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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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11
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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
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OFFSET
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1,5
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COMMENTS
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More generally, consider the sequence defined by p^a(n) exactly divides p*n. For p = 3 we have A051064 and for p = 2 we have A001511.
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LINKS
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Joseph Rosenbaum, Elementary Problem E319, American Mathematical Monthly, volume 45, number 10, December 1938, pages 694-696. (The A indices in P at equations 1' and 2' for p=5.)
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FORMULA
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G.f.: Sum_{k>=0} x^(5^k)/(1-x^5^k). - Ralf Stephan, Apr 12 2002
Multiplicative with a(p^e) = e+1 if p = 5, 1 otherwise.
Dirichlet g.f.: zeta(s)/(1-1/5^s). - R. J. Mathar, Feb 09 2011
a(5n) = 1 + a(n). a(5n+k) = 1 for k = 1..4. - Robert Israel, Dec 07 2015
G.f. satisfies A(x^5) = A(x) - x/(1-x). - Robert Israel, Dec 08 2015
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EXAMPLE
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a(5) = 2 since 5^2 exactly divides 5 times 5;
a(25) = 3 since 5^3 exactly divides 5 times 25;
a(125) = 4 since 5^4 exactly divides 5 times 125.
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MAPLE
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seq(padic:-ordp(5*n, 5), n=1..1000); # Robert Israel, Dec 07 2015
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MATHEMATICA
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max = 1000; s = (1/x)*Sum[x^(5^k)/(1-x^5^k), {k, 0, Log[5, max] // Ceiling }] + O[x]^max; CoefficientList[s, x] (* Jean-François Alcover, Dec 04 2015 *)
Table[IntegerExponent[n, 5] + 1, {n, 1, 100}] (* Amiram Eldar, Sep 21 2020 *)
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PROG
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(PARI) a(n)=-sumdiv(n, d, moebius(5*d)*numdiv(n/d)) \\ Benoit Cloitre, Jun 21 2007
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CROSSREFS
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KEYWORD
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nonn,mult,easy
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AUTHOR
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STATUS
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approved
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