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A112765
Exponent of highest power of 5 dividing n. Or, 5-adic valuation of n.
60
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
1,25
COMMENTS
A027868 gives partial sums.
This is also the 5-adic valuation of Fibonacci(n). See Lengyel link. - Michel Marcus, May 06 2017
LINKS
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.
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
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 *)
PROG
(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
(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
KEYWORD
nonn,easy
AUTHOR
Reinhard Zumkeller, Sep 18 2005
STATUS
approved