A132029 -id:A132029 - cp's OEIS frontend

A027868 Number of trailing zeros in n!; highest power of 5 dividing n!.

0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 18, 18, 18, 18, 18, 19

Offset: 0

Warut Roonguthai

Also the highest power of 10 dividing n! (different from A054899). - Hieronymus Fischer, Jun 18 2007
Alternatively, a(n) equals the expansion of the base-5 representation A007091(n) of n (i.e., where successive positions from right to left stand for 5^n or A000351(n)) under a scale of notation whose successive positions from right to left stand for (5^n - 1)/4 or A003463(n); for instance, n = 7392 has base-5 expression 2*5^5 + 1*5^4 + 4*5^3 + 0*5^2 + 3*5^1 + 2*5^0, so that a(7392) = 2*781 + 1*156 + 4*31 + 0*6 + 3*1 + 2*0 = 1845. - Lekraj Beedassy, Nov 03 2010
Partial sums of A112765. - Hieronymus Fischer, Jun 06 2012
Also the number of trailing zeros in A000165(n) = (2*n)!!. - Stefano Spezia, Aug 18 2024

			a(100)  = 24.
a(10^3) = 249.
a(10^4) = 2499.
a(10^5) = 24999.
a(10^6) = 249998.
a(10^7) = 2499999.
a(10^8) = 24999999.
a(10^9) = 249999998.
a(10^n) = 10^n/4 - 3 for 10 <= n <= 15 except for a(10^14) = 10^14/4 - 2. - _M. F. Hasler_, Dec 27 2019

M. Gardner, "Factorial Oddities." Ch. 4 in Mathematical Magic Show: More Puzzles, Games, Diversions, Illusions and Other Mathematical Sleight-of-Mind from Scientific American. New York: Vintage, 1978, pp. 50-65.

Hieronymus Fischer, Table of n, a(n) for n = 0..10000 (first 1000 terms from T. D. Noe)
David S. Hart, James E. Marengo, Darren A. Narayan, and David S. Ross, On the number of trailing zeros in n!, College Math. J., 39(2):139-145, 2008.
Enrique Pérez Herrero, Trailing Zeros in n!, Psychedelic Geometry Blogspot.
Soichi Ikeda and Kaneaki Matsuoka, On transcendental numbers generated by certain integer sequences, Siauliai Math. Semin., 8 (16) 2013, 63-69.
Shyam Sunder Gupta, Fascinating Factorials, Exploring the Beauty of Fascinating Numbers, Springer (2025) Ch. 16, 411-442.
Shu-Chung Liu and Jean C.-C. Yeh, Catalan numbers modulo 2^k, J. Int. Seq. 13 (2010), 10.5.4, eq (5).
Antonio M. Oller-Marcén, A new look at the trailing zeros of n!, arXiv:0906.4868v1 [math.NT], 2009.
Antonio M. Oller-Marcén and J. Maria Grau, On the Base-b Expansion of the Number of Trailing Zeros of b^k!, J. Int. Seq. 14 (2011) 11.6.8
Eric Weisstein's World of Mathematics, Factorial.
Index entries for sequences related to factorial numbers

See A000966 for the missing numbers. See A011371 and A054861 for analogs involving powers of 2 and 3.
Cf. A054899, A007953, A112765, A067080, A098844, A132027, A067080, A098844, A132029, A054999, A112765, A191610, A000351.
Cf. also A000142, A004154.
Cf. A000165, A008904.

a027868 n = sum $ takeWhile (> 0) $ map (n `div`) $ tail a000351_list
-- Reinhard Zumkeller, Oct 31 2012

[Valuation(Factorial(n), 5): n in [0..80]]; // Bruno Berselli, Oct 11 2021

0, seq(add(floor(n/5^i),i=1..floor(log[5](n))), n=1..100); # Robert Israel, Nov 13 2014

Table[t = 0; p = 5; While[s = Floor[n/p]; t = t + s; s > 0, p *= 5]; t, {n, 0, 100} ]
Table[ IntegerExponent[n!], {n, 0, 80}] (* Robert G. Wilson v *)
zOF[n_Integer?Positive]:=Module[{maxpow=0},While[5^maxpow<=n,maxpow++];Plus@@Table[Quotient[n,5^i],{i,maxpow-1}]]; Attributes[zOF]={Listable}; Join[{0},zOF[ Range[100]]] (* Harvey P. Dale, Apr 11 2022 *)

a(n)={my(s);while(n\=5,s+=n);s} \\ Charles R Greathouse IV, Nov 08 2012, edited by M. F. Hasler, Dec 27 2019

a(n)=valuation(n!,5) \\ Charles R Greathouse IV, Nov 08 2012

apply( A027868(n)=(n-sumdigits(n,5))\4, [0..99]) \\ M. F. Hasler, Dec 27 2019

from sympy import multiplicity
A027868, p5 = [0,0,0,0,0], 0
for n in range(5,10**3,5):
    p5 += multiplicity(5,n)
    A027868.extend([p5]*5) # Chai Wah Wu, Sep 05 2014

def A027868(n): return 0 if n<5 else n//5 + A027868(n//5) # David Radcliffe, Jun 26 2016

from sympy.ntheory.factor_ import digits
def A027868(n): return n-sum(digits(n,5)[1:])>>2 # Chai Wah Wu, Oct 18 2024

a(n) = Sum_{i>=1} floor(n/5^i).
a(n) = (n - A053824(n))/4.
From Hieronymus Fischer, Jun 25 2007 and Aug 13 2007, edited by M. F. Hasler, Dec 27 2019: (Start)
G.f.: g(x) = Sum_{k>0} x^(5^k)/(1-x^(5^k))/(1-x).
a(n) = Sum_{k=5..n} Sum_{j|k, j>=5} (floor(log_5(j)) - floor(log_5(j-1))).
G.f.: g(x) = L[b(k)](x)/(1-x) where L[b(k)](x) = Sum_{k>=0} b(k)*x^k/(1-x^k) is a Lambert series with b(k) = 1, if k>1 is a power of 5, else b(k) = 0.
G.f.: g(x) = Sum_{k>0} c(k)*x^k/(1-x), where c(k) = Sum_{j>1, j|k} floor(log_5(j)) - floor(log_5(j - 1)).
Recurrence:
a(n) = floor(n/5) + a(floor(n/5));
a(5*n) = n + a(n);
a(n*5^m) = n*(5^m-1)/4 + a(n).
a(k*5^m) = k*(5^m-1)/4, for 0 <= k < 5, m >= 0.
Asymptotic behavior:
a(n) = n/4 + O(log(n)),
a(n+1) - a(n) = O(log(n)), which follows from the inequalities below.
a(n) <= (n-1)/4; equality holds for powers of 5.
a(n) >= n/4 - 1 - floor(log_5(n)); equality holds for n = 5^m-1, m > 0.
lim inf (n/4 - a(n)) = 1/4, for n -> oo.
lim sup (n/4 - log_5(n) - a(n)) = 0, for n -> oo.
lim sup (a(n+1) - a(n) - log_5(n)) = 0, for n -> oo.
(End)
a(n) <= A027869(n). - Reinhard Zumkeller, Jan 27 2008
10^a(n) = A000142(n) / A004154(n). - Reinhard Zumkeller, Nov 24 2012
a(n) = Sum_{k=1..floor(n/2)} floor(log_5(n/k)). - Ammar Khatab, Feb 01 2025

Examples added by Hieronymus Fischer, Jun 06 2012

A098844 a(1)=1, a(n) = n*a(floor(n/2)).

Original entry on oeis.org

1, 2, 3, 8, 10, 18, 21, 64, 72, 100, 110, 216, 234, 294, 315, 1024, 1088, 1296, 1368, 2000, 2100, 2420, 2530, 5184, 5400, 6084, 6318, 8232, 8526, 9450, 9765, 32768, 33792, 36992, 38080, 46656, 47952, 51984, 53352, 80000, 82000, 88200, 90300

Offset: 1

Benoit Cloitre, Nov 03 2004

nonn

			a(10) = floor(10/2^0)*floor(10/2^1)*floor(10/2^2)*floor(10/2^3) = 10*5*2*1 = 100;
a(17) = 1088 since 17 = 10001(base 2) and so a(17) = 10001*1000*100*10*1(base 2) = 17*8*4*2*1 = 1088.

Hieronymus Fischer, Table of n, a(n) for n = 1..1000

Cf. A048651, A067080, A132027, A132028, A132029, A132030, A132019, A132026, A132038.
For formulas regarding a general parameter p (i.e., terms floor(n/p^k)) see A132264.
For the product of terms floor(n/p^k) for p=3 to p=12 see A132027(p=3)-A132033(p=9), A067080(p=10), A132263(p=11), A132264(p=12).
For the products of terms 1+floor(n/p^k) see A132269-A132272, A132327, A132328.

lst={};Do[p=n;s=1;While[p>1,p=IntegerPart[p/2];s*=p;];AppendTo[lst,s],{n,1,6!,2}];lst (* Vladimir Joseph Stephan Orlovsky, Jul 28 2009 *)

PARI
```
a(n)=if(n<2,1,n*a(floor(n/2)))
```

from math import prod
def A098844(n): return n*prod(n//2**k for k in range(1,n.bit_length()-1)) # Chai Wah Wu, Jun 07 2022

a(2^n) = 2^(n*(n+1)/2) = A006125(n+1).
From Hieronymus Fischer, Aug 13 2007: (Start)
a(n) = Product_{k=0..floor(log_2(n))} floor(n/2^k), n>=1.
Recurrence:
a(n*2^m) = n^m*2^(m(m+1)/2)*a(n).
a(n) <= n^((1+log_2(n))/2) = 2^A000217(log_2(n)); equality iff n is a power of 2.
a(n) >= c(n)*(n+1)^((1 + log_2(n+1))/2) for n != 2,
where c(n) = Product_{k=1..floor(log_2(n))} (1 - 1/2^k); equality holds iff n+1 is a power of 2.
a(n) > c*(n+1)^((1 + log_2(n+1))/2)
where c = 0.288788095086602421... (see constant A048651).
lim inf a(n)/n^((1+log_2(n))/2)=0.288788095086602421... for n-->oo.
lim sup a(n)/n^((1+log_2(n))/2) = 1 for n-->oo.
lim inf a(n)/a(n+1) = 0.288788095086602421... for n-->oo (see constant A048651).
a(n) = O(n^((1+log_2(n))/2)). (End)

Formula section edited by Hieronymus Fischer, Jun 13 2012

A132021 Decimal expansion of Product_{k>=0} 1-1/(2*5^k).

Original entry on oeis.org

4, 3, 8, 7, 9, 6, 8, 3, 7, 2, 0, 3, 6, 3, 8, 5, 3, 1, 2, 6, 6, 7, 2, 9, 9, 9, 7, 1, 7, 7, 2, 5, 8, 3, 5, 9, 6, 0, 4, 5, 7, 4, 6, 3, 1, 2, 3, 9, 3, 5, 1, 1, 6, 5, 4, 1, 7, 7, 3, 6, 7, 5, 6, 4, 3, 6, 7, 9, 1, 0, 6, 6, 5, 6, 9, 8, 6, 6, 5, 0, 0, 6, 9, 2, 8, 9, 6, 6, 7, 2, 3, 8, 9, 8, 5, 4, 4, 0, 0, 6, 0, 2, 8

Offset: 0

Hieronymus Fischer, Aug 14 2007

			0.438796837203638531...

Cf. A048651, A098844, A067080, A132019, A132026, A132029, A100222, A000217, A020729.

digits = 103; NProduct[1-1/(2*5^k), {k, 0, Infinity}, NProductFactors -> 100, WorkingPrecision -> digits+5] // N[#, digits+5]& // RealDigits[#, 10, digits]& // First (* Jean-François Alcover, Feb 18 2014 *)
RealDigits[QPochhammer[1/2, 1/5], 10, 120][[1]] (* Amiram Eldar, May 08 2023 *)

Equals lim inf_{n->oo} Product_{k=0..floor(log_5(n))} floor(n/5^k)*5^k/n.
Equals lim inf_{n->oo} A132029(n)/n^(1+floor(log_5(n)))*5^(1/2*(1+floor(log_5(n)))*floor(log_5(n))).
Equals lim inf_{n->oo} A132029(n)/n^(1+floor(log_5(n)))*5^A000217(floor(log_5(n))).
Equals (1/2)*exp(-Sum_{n>0} 5^(-n)*Sum_{k|n} 1/(k*2^k)).
Equals lim inf_{n->oo} A132029(n)/A132029(n+1).
Equals Product_{n>=0} (1 - 1/A020729(n)). - Amiram Eldar, May 08 2023

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A027868 Number of trailing zeros in n!; highest power of 5 dividing n!.

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A098844 a(1)=1, a(n) = n*a(floor(n/2)).

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A132021 Decimal expansion of Product_{k>=0} 1-1/(2*5^k).

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