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A191610 Possible number of trailing zeros in n!. 6
0, 1, 2, 3, 4, 6, 7, 8, 9, 10, 12, 13, 14, 15, 16, 18, 19, 20, 21, 22, 24, 25, 26, 27, 28, 31, 32, 33, 34, 35, 37, 38, 39, 40, 41, 43, 44, 45, 46, 47, 49, 50, 51, 52, 53, 55, 56, 57, 58, 59, 62, 63, 64, 65, 66, 68, 69, 70, 71, 72, 74, 75, 76, 77, 78, 80, 81, 82, 83, 84, 86, 87, 88, 89, 90, 93, 94, 95, 96, 97, 99, 100, 101, 102, 103, 105, 106, 107, 108, 109, 111, 112, 113, 114, 115, 117, 118, 119, 120, 121, 124, 125, 126, 127, 128, 130, 131, 132, 133, 134, 136 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,3
COMMENTS
Equivalently, possible values of 10-adic valuation of n!. - Joerg Arndt, Sep 21 2020
LINKS
FORMULA
a(n) ~ 5*n/4. - Vaclav Kotesovec, Sep 21 2020
G.f.: 1/(1-x) * Sum_{k>=0} x^(5^k)/(1-x^5^k). - Joerg Arndt, Sep 21 2020
MATHEMATICA
zOF[n_Integer?Positive]:=Module[{maxpow=0}, While[5^maxpow<=n, maxpow++]; Plus@@Table[ Quotient[n, 5^i], {i, maxpow-1}]]; Attributes[zOF]={Listable}; zOF[Range[1000]]//Union (* Harvey P. Dale, Dec 06 2023 *)
PROG
(Haskell)
a191610 1 = 0
a191610 n = sum $ takeWhile (> 0) $ map ((n - 1) `div`) a000351_list
-- Reinhard Zumkeller, Oct 31 2012
(Python)
# requires Python 3.2 and higher
from itertools import accumulate
from sympy import multiplicity
A191610 = [0]+list(accumulate(multiplicity(5, n) for n in range(5, 10**3, 5)))
# Chai Wah Wu, Sep 05 2014
CROSSREFS
Cf. A027868, A000351, A055457 (first differences).
Complement of A000966.
Sequence in context: A047226 A059537 A039259 * A037465 A157846 A285126
KEYWORD
nonn
AUTHOR
STATUS
approved

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Last modified March 19 01:57 EDT 2024. Contains 370952 sequences. (Running on oeis4.)