A055457 5^a(n) exactly divides 5n. Or, 5-adic valuation of 5n.

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

Author

Alford Arnold, Jun 25 2000

Keywords

• nonn
• mult
• easy

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.

The number of powers of 5 that divide n. - Amiram Eldar, Mar 29 2025

Examples

			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.
		

Links

• T. D. Noe, Table of n, a(n) for n=1..1000
• 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.)

Crossrefs

Cf. A007949, A112765, A191610 (partial sums).

Cf. A001511, A051064, A115362, A373216, A373217.

Programs

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 *)

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

Python

def A055457(n):
    c = 1
    while not (a:=divmod(n,5))[1]:
        c += 1
        n = a[0]
    return c # Chai Wah Wu, Feb 28 2025

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

G.f.: Sum_{i>=1, j>=0} x^(i*5^j). - Seiichi Manyama, Mar 23 2025