login
A055457
5^a(n) exactly divides 5n. Or, 5-adic valuation of 5n.
17
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
OFFSET
1,5
COMMENTS
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.
LINKS
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.)
FORMULA
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.
a(n) = -Sum_{d|n} mu(5d)*tau(n/d). - Benoit Cloitre, Jun 21 2007
Dirichlet g.f.: zeta(s)/(1-1/5^s). - R. J. Mathar, Feb 09 2011
a(n) = A112765(5n). - R. J. Mathar, Jul 17 2012
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
a(n) = A112765(n) + 1. - Amiram Eldar, Sep 21 2020
Sum_{k=1..n} a(k) ~ 5*n/4. - Vaclav Kotesovec, Sep 21 2020
EXAMPLE
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.
MAPLE
seq(padic:-ordp(5*n, 5), n=1..1000); # Robert Israel, Dec 07 2015
MATHEMATICA
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 *)
PROG
(PARI) a(n)=-sumdiv(n, d, moebius(5*d)*numdiv(n/d)) \\ Benoit Cloitre, Jun 21 2007
(PARI) a(n)=valuation(5*n, 5) \\ Anders Hellström, Dec 04 2015
CROSSREFS
Cf. A001511, A007949, A051064, A112765, A191610 (partial sums).
Sequence in context: A177706 A130782 A364358 * A277873 A032542 A107038
KEYWORD
nonn,mult,easy
AUTHOR
Alford Arnold, Jun 25 2000
STATUS
approved