%I #84 Jul 23 2022 14:53:53

%S 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,

%T 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,

%U 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

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

%C A027868 gives partial sums.

%C This is also the 5-adic valuation of Fibonacci(n). See Lengyel link. - _Michel Marcus_, May 06 2017

%H Reinhard Zumkeller, <a href="/A112765/b112765.txt">Table of n, a(n) for n = 1..10000</a>

%H Dario T. de Castro, <a href="http://math.colgate.edu/~integers/w61/w61.pdf">P-adic Order of Positive Integers via Binomial Coefficients</a>, INTEGERS, Electronic J. of Combinatorial Number Theory, Vol. 22, Paper A61, 2022.

%H T. Lengyel, <a href="http://www.fq.math.ca/Scanned/33-3/lengyel.pdf">The order of the Fibonacci and Lucas numbers</a>, Fibonacci Quart. 33 (1995), no. 3, 234-239. See Lemma 1 p. 235.

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

%F From _Hieronymus Fischer_, Jun 08 2012: (Start)

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

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

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

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

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

%F a(5n) = A055457(n). - _R. J. Mathar_, Jul 17 2012

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

%F 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

%p A112765 := proc(n)

%p padic[ordp](n,5) ;

%p end proc: # _R. J. Mathar_, Jul 12 2016

%t a[n_] := IntegerExponent[n, 5]; Array[a, 105] (* _Jean-François Alcover_, Jan 25 2018 *)

%o (Haskell)

%o a112765 n = fives n 0 where

%o fives n e | r > 0 = e

%o | otherwise = fives n' (e + 1) where (n',r) = divMod n 5

%o -- _Reinhard Zumkeller_, Apr 08 2011

%o (PARI) A112765(n)=valuation(n,5); /* _Joerg Arndt_, Apr 08 2011 */

%o (Python)

%o def a(n):

%o k = 0

%o while n > 0 and n%5 == 0: n //= 5; k += 1

%o return k

%o print([a(n) for n in range(1, 106)]) # _Michael S. Branicky_, Aug 06 2021

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

%Y Cf. A343251.

%K nonn,easy

%O 1,25

%A _Reinhard Zumkeller_, Sep 18 2005