A112765 Exponent of highest power of 5 dividing n. Or, 5-adic valuation of n.

0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 2, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 2, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 2, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 2, 0, 0, 0, 0, 1

Offset: 1

Author

Reinhard Zumkeller, Sep 18 2005

Keywords

• nonn
• easy

Comments

A027868 gives partial sums.

This is also the 5-adic valuation of Fibonacci(n). See Lengyel link. - Michel Marcus, May 06 2017

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
• Dario T. de Castro, P-adic Order of Positive Integers via Binomial Coefficients, INTEGERS, Electronic J. of Combinatorial Number Theory, Vol. 22, Paper A61, 2022.
• T. Lengyel, The order of the Fibonacci and Lucas numbers, Fibonacci Quart. 33 (1995), no. 3, 234-239. See Lemma 1 p. 235.

Crossrefs

Cf. A007814, A007949, A112762, A022337, A122840, A027868, A054899, A122841, A160093, A160094, A196563, A196564.

Cf. A343251, A000351 (positions of records, greedy inverse).

Programs

Haskell

a112765 n = fives n 0 where
   fives n e | r > 0     = e
             | otherwise = fives n' (e + 1) where (n',r) = divMod n 5
-- Reinhard Zumkeller, Apr 08 2011

Maple

A112765 := proc(n)
    padic[ordp](n,5) ;
end proc: # R. J. Mathar, Jul 12 2016

Mathematica

a[n_] := IntegerExponent[n, 5]; Array[a, 105] (* Jean-François Alcover, Jan 25 2018 *)

PARI

A112765(n)=valuation(n,5); /* Joerg Arndt, Apr 08 2011 */

Python

def a(n):
    k = 0
    while n > 0 and n%5 == 0: n //= 5; k += 1
    return k
print([a(n) for n in range(1, 106)]) # Michael S. Branicky, Aug 06 2021

Formula

Totally additive with a(p) = 1 if p = 5, 0 otherwise.

From Hieronymus Fischer, Jun 08 2012: (Start)

With m = floor(log_5(n)), frac(x) = x-floor(x):

a(n) = Sum_{j=1..m} (1 - ceiling(frac(n/5^j))).

a(n) = m + Sum_{j=1..m} (floor(-frac(n/5^j))).

a(n) = A027868(n) - A027868(n-1).

G.f.: Sum_{j>0} x^5^j/(1-x^5^j). (End)

a(5n) = A055457(n). - R. J. Mathar, Jul 17 2012

Asymptotic mean: lim_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1/4. - Amiram Eldar, Feb 14 2021

a(n) = 5*Sum_{j=1..floor(log(n)/log(5))} frac(binomial(n, 5^j)*5^(j-1)/n). - Dario T. de Castro, Jul 10 2022