A381239 -id:A381239 - cp's OEIS frontend

A191610 Possible number of trailing zeros in k!.

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

Offset: 1

José María Grau Ribas, Jun 09 2011

nonn

Equivalently, possible values of 10-adic valuation of k!. - Joerg Arndt, Sep 21 2020

			3 is a term because 15! = 1307674368000 has 3 trailing 0's.
5 is not a term because 24! has 4 trailing 0's, but 25! has 6 trailing 0's.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A027868, A000351, A055457 (first differences).
Complement of A000966.
Cf. A380662, A381239.

a191610 1 = 0
a191610 n = sum $ takeWhile (> 0) $ map ((n - 1) `div`) a000351_list
-- Reinhard Zumkeller, Oct 31 2012

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 *)
Table[Sum[Floor[(n - 1)/5^k], {k, 0, Floor[Log[5, n]]}], {n, 1, 200}] (* Clark Kimberling, Feb 17 2025 *)

# 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

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
a(n) = Sum_{k>=0} floor((n-1)/5^k). - Clark Kimberling, Feb 17 2025

A380662 Numbers m such that Sum_{k>=0} floor(m/5^k) is prime.

Original entry on oeis.org

2, 3, 6, 11, 16, 25, 30, 34, 35, 39, 44, 49, 58, 68, 73, 79, 82, 84, 87, 89, 92, 103, 106, 111, 113, 121, 123, 126, 131, 146, 154, 155, 159, 160, 170, 183, 188, 193, 202, 207, 212, 217, 219, 224, 226, 228, 236, 248, 251, 266, 271, 279, 280, 284, 289, 295

Offset: 1

Clark Kimberling, Feb 19 2025

nonn

			floor(16/1) + floor(16/5) = 19, so 16 is in the sequence.

Cf. A191610, A381239.

f[n_] := Sum[Floor[(n-1)/5^k], {k, 0, Floor[Log[5, n]]}]  (* A191610 *)
Select[Range[400], PrimeQ[f[#]] &]

A381897 a(n) = least integer m >= 2 such that prime(n) is a sum of the form Sum_{k>=0} floor(h/m^k) for some integer h >= 1.

Original entry on oeis.org

3, 2, 3, 2, 2, 3, 3, 2, 2, 4, 2, 4, 2, 4, 2, 2, 3, 3, 2, 2, 2, 2, 5, 2, 2, 2, 3, 3, 2, 2, 2, 2, 2, 3, 2, 3, 3, 3, 3, 2, 3, 2, 2, 2, 2, 4, 3, 3, 3, 2, 3, 2, 4, 3, 3, 2, 3, 2, 2, 2, 3, 2, 4, 2, 3, 3, 3, 2, 4, 2, 2, 2, 2, 3, 3, 2, 2, 2, 2, 3, 3, 2, 2, 4, 2, 4

Offset: 1

Clark Kimberling, Mar 09 2025

nonn

			a(10) = 4 because 4 is the least m such that prime(10) is a sum of the form Sum_{k>=0} [h/m^k] for some h >= 1; that sum is 29 = [23/1] + [23/4] + [23/16], where [ ] = floor.

Cf. A000040, A381239, A382278.

f[h_, m_] := Sum[Floor[h/m^k], {k, 0, Floor[Log[m, h]]}]
{rng, n} = {1000, 6};
Table[u[m] = Select[Range[rng], PrimeQ[f[#, m]] &], {m, 2, n}];
tmp = SortBy[Map[#[[1]] &, GatherBy[Flatten[Table[
       Transpose[{ConstantArray[m, Length[u[m]]],
         Map[PrimePi[f[#, m]] &, u[m]]}], {m, 2, n}],1], #[[2]] &]], #[[2]] &];
tmp = Map[#[[1]] &, Take[tmp, Position[Differences[Map[#[[2]] &, tmp]], x_ /; x != 1, 1, 1][[1]][[1]]]]
(* Peter J. C. Moses, Feb 19 2025 *)

a(n) = A382278(prime(n)). - Pontus von Brömssen, Mar 22 2025

cp's OEIS Frontend

A191610 Possible number of trailing zeros in k!.

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A380662 Numbers m such that Sum_{k>=0} floor(m/5^k) is prime.

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A381897 a(n) = least integer m >= 2 such that prime(n) is a sum of the form Sum_{k>=0} floor(h/m^k) for some integer h >= 1.

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