A027868 -id:A027868 - cp's OEIS frontend

A137582 Numbers having no inner zeros in decimal representation of their factorial.

0, 1, 2, 3, 4, 5, 6, 9, 10, 11, 14, 31, 40

Offset: 1

Reinhard Zumkeller, Jan 27 2008

Conjecture: the sequence is finite.

			A000142(a(10)) = 11! = 399168*10^2;
A000142(a(11)) = 14! = 871782912*10^2;
A000142(a(12)) = 31! = 822283865417792281772556288*10^7.

Index entries for sequences related to factorial numbers.

Cf. A000142, A027869, A027868, A137581.

isok(k) = my(f=k!); while(!(f % 10), f \= 10); #select(x->(x == 0), digits(f)) == 0; \\ Michel Marcus, Jun 28 2023

A137581(a(n)) = 0.

A165753 Number of trailing zeros in sequence of factorials of Fibonacci numbers.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 2, 4, 7, 13, 20, 34, 56, 93, 150, 244, 396, 643, 1043, 1689, 2733, 4425, 7161, 11587, 18754, 30344, 49100, 79449, 128552, 208007, 336563, 544573, 881140, 1425715, 2306860, 3732583, 6039449, 9772038, 15811490, 25583533, 41395029

Offset: 1

Carmine Suriano, Sep 26 2009

nonn

a(n) is the number of zeros at rightmost place of F(n)!

			F(7)!=13!=6227020800 which ends with 2 zeros; therefore a(7)=2.

C. Suriano, Miniature Matematiche, Il rosone, 2009, pp. 149-152.

Harvey P. Dale, Table of n, a(n) for n = 1..1000

zOF[n_Integer?Positive]:=Module[{maxpow=0},While[5^maxpow<=n,maxpow++];Plus@@Table[Quotient[n,5^i],{i,maxpow-1}]];zOF[#]&/@ Fibonacci[ Range[ 50]] (* Harvey P. Dale, Jan 10 2020 *)

a(n) = A027868(A000045(n))

a(n) = k * phi^n + O(n), for k = 1/sqrt(80).

Extended and edited by Charles R Greathouse IV, Mar 23 2010

A175417 Exponent of 2 minus sum of all other exponents, in the prime power factorization of n!

Original entry on oeis.org

0, 0, 1, 0, 2, 1, 1, 0, 3, 1, 1, 0, 1, 0, 0, -2, 2, 1, 0, -1, 0, -2, -2, -3, -1, -3, -3, -6, -5, -6, -7, -8, -3, -5, -5, -7, -7, -8, -8, -10, -8, -9, -10, -11, -10, -13, -13, -14, -11, -13, -14, -16, -15, -16, -18, -20, -18, -20, -20, -21, -21, -22, -22, -25, -19, -21, -22

Offset: 0

Zak Seidov, May 08 2010

sign

a(n)=0 for n={0,1,3,7,11,13,14,18,20}.

			a(20) = 0 because 20! = 2432902008176640000 = ((2^18)*(3^8)*(5^4)*(7^2)*(11^1)*(13^1)*(17^1)*(19^1)) and 18-(8+4+2+1+1+1+1) = 0.

Robert Israel, Table of n, a(n) for n = 0..10000

Cf. A000142, A011371, A027868, A054861, A054896.

f:= proc(n) local t,p,k;
      p:= 2: t:= add(floor(n/2^k),k=1..ilog2(n)):
      do
        p:= nextprime(p);
        if n < p then return t fi;
        t:= t - add(floor(n/p^k),k=1..ilog[p](n))
      od
end proc:
map(f, [$0..100]); # Robert Israel, Nov 10 2024

Table[2*IntegerExponent[m!,2]-Total[Last/@FactorInteger[m! ]],{m,0,130}]

a(n)=2*A011371(n)-A022559(n).

A181578 The number k such that each of the five factorials (5k+j)!, j=0..4, has exactly n trailing zeros in its base-10 representation; 0 if no such k exists.

Original entry on oeis.org

0, 1, 2, 3, 4, 0, 5, 6, 7, 8, 9, 0, 10, 11, 12, 13, 14, 0, 15, 16, 17, 18, 19, 0, 20, 21, 22, 23, 24, 0, 0, 25, 26, 27, 28, 29, 0, 30, 31, 32, 33, 34, 0, 35, 36, 37, 38, 39, 0, 40, 41, 42, 43, 44, 0, 45, 46, 47, 48, 49, 0, 0, 50, 51, 52, 53, 54, 0, 55, 56, 57, 58, 59, 0, 60, 61, 62, 63

Offset: 0

Lekraj Beedassy, Nov 02 2010

Entries are zero if n is an element of A000966.

This traces 5 consecutive occurrences of n in A027868.

Cf. A181579, A000142.

A181581 Smallest k such that each of the five factorials (5k)!, (5k+1)!, (5k+2)!, (5k+3)! and (5k+4)! has exactly 10^n trailing 0's. Zero, if no such k exists.

Original entry on oeis.org

1, 9, 81, 801, 8002, 80001, 800001, 8000002, 80000003, 800000003, 8000000003, 80000000003, 800000000002, 8000000000003, 80000000000004, 0, 0, 0, 800000000000000004, 8000000000000000003, 0, 800000000000000000005, 8000000000000000000006, 0

Offset: 0

Lekraj Beedassy, Nov 02 2010

nonn

a(n) = A173558(n)/5 or 0.

Hiroaki Yamanouchi, Table of n, a(n) for n = 0..100
Index to sequences related to factorial numbers.

Cf. A027868, A173558.

a(15)-a(17) corrected and a(18)-a(23) added by Hiroaki Yamanouchi, Oct 06 2014

A218976 a(n) is the smallest positive integer such that 10^(2 + floor(k/a(1)) + floor(k/a(2)) + ... + floor(k/a(n))) divides (k+9)! for all k > 0.

Original entry on oeis.org

6, 16, 116, 241, 242, 491, 991, 2491, 3331, 14966, 15556, 62491, 78116, 83331, 249991, 264866, 546841, 1109366, 2265491, 4999861, 4999991, 5837041, 12499996, 25249861, 26011861, 36249091, 80070866, 190999991, 242090611, 365038621, 976562241, 1210466866, 1830622801

Offset: 1

R. J. Cano and Charles R Greathouse IV, Nov 08 2012

nonn

Every factorial of the form (k+9)! for every integer k > 0 ends at least in two zeros. This sequence gives a lower bound on the number of zeros.
This sequence is infinite and increasing, with 1/a(1) + 1/a(2) + ... = 1/4.
Conjecture: All terms except a(5) are 1 mod 5. - R. J. Cano, Nov 11 2012

			10^(2 + floor(5/5)) does not divide 14!, so a(1) > 5. But 10^(2 + floor(k/6)) divides (k+9)! for all k > 0, so a(1) = 6.

Table of n, a(n) for n = 1..33
Dickson, Leonard Eugene (2005) [1919], History of the theory of numbers. Vol. I: Divisibility and primality., New York: Dover Publications, ISBN 978-0-486-44232-7, [MR], page 263. De Polignac's formula
Wikipedia, De Polignac's formula
Wikipedia, Trailing zero

Cf. A027868, A217626.

searchLimit(s1)={
    my(e,s2,f=(e,s)->(e+2-9*s)/(s-s1));
    while(s2<=s1, s2 += 1/5^e++);
    min(f(e,s2), f(e++,s2+=1/5^e))\1
};
v5(n)=my(s);while(n\=5,s+=n);s;
nxt(v=[6])={
    my(S=sum(i=1,#v,1/v[i]), candidate=max(v[#v],1\(1/4-S))+1, k=candidate, lm=searchLimit(S+1/candidate));
    while(k<=lm,
        if(v5(k+9)<2+sum(i=1,#v,k\v[i])+k\candidate,
            candidate++;
            lm=searchLimit(S+1/candidate)
        ,
            k++
        )
    );
    candidate
};
steps(n)={
    my(v=[6],t);
    print1(6);
    for(i=2,n,
        t=nxt(v);
        print1(", "t);
        v=concat(v,t)
    );
    v
};
steps(20)

Let Psi(k) = 2 + sum_{n >= 1} floor(k/a(n)). Then 10^Psi(k) divides (k+9)!.

a(32)-a(33) from Charles R Greathouse IV, Nov 19 2012

A228311 Numbers k such that the sum of digits of k! is itself a factorial.

Original entry on oeis.org

0, 1, 2, 3, 4, 21966, 176755, 182624820

Offset: 1

Charles R Greathouse IV, Aug 27 2013

The sum of digits of k! is approximately (9/2)*(d-z), where d=A034886(k) is the number of digits of k!, which is about (log(k/E)*k + log(2*k*Pi)/2)/log(10), and z=A027868(k) is the number of trailing zeros of k!, which is Sum_{j>=1} floor(k/5^j). - Giovanni Resta, Aug 28 2013
a(9) > 2.235*10^9. - Hans Havermann, May 16 2014
k! has ~ k log_10(k) digits, so its digit sum is typically close to C*k*log_10(k) for some constant C. A random number around j has probability something like log(j)/(j log log(j)) of being a factorial, so the probability that the digit sum of k! is a factorial should be something like const/(k log log k). The sum of this diverges, so we should expect infinitely many terms in the sequence. - Robert Israel, Aug 08 2014

			The sum of the digits of 21966! is 362880 = 9!.
The sum of the digits of 176755! is 3628800 = 10!.
The sum of the digits of 182624820! is 6227020800 = 13!.

Shyam Sunder Gupta, Fascinating Factorials, Exploring the Beauty of Fascinating Numbers, Springer (2025) Ch. 16, 411-442.
"Mouhaha" Digit sums and factorials

Cf. A229024, A004152.

lst = {0}; k = p = 1; fctl = Range@ 15!; While[k < 180000, p = p*k; While[ Mod[p, 10] == 0, p /= 10]; If[ MemberQ[ fctl, Plus @@ IntegerDigits@ p], Print[k]; AppendTo[lst, k]]; k++]; lst (* Robert G. Wilson v, Feb 18 2014 *)
With[{fcts=Range[20]!},Select[Range[0,22000],MemberQ[fcts,Total[IntegerDigits[#!]]]&]] (* Harvey P. Dale, Jan 06 2024 *)

lpf(n)=my(f=factor(n)[,1]); f[1]
factorial_lval(n, p)={
    my(s);
    while(n\=p, s+=n);
    s
};
isfactorial(n)={
    if(n<3, return(n>0));
    my(v2=valuation(n,2),mn=v2+1,mx=mn+log(v2+1.5)\log(2),t,c);
    while (mx - mn > 1,
        t = mn + (mx - mn)\2;
        c = factorial_lval(t, 2);
        if (c < v2,
            mn = t+1
        ,
            if (c > v2,
                mx = t-1
            ,
                mx = bitor(t,1);
                mn = max(mn, mx-1)
            )
        )
    );
    if (mn < mx,
        my(p=lpf(mx));
        t = valuation(n, p);
        c = factorial_lval(mx, p);
        if (t > c,return(0));
        if (t == c,
            mn = mx
        )
    );
    n==mn!
};
is(n)=isfactorial(sumdigits(n!))

a(8) from Hans Havermann, Mar 24 2014

A255400 a(n) is the smallest nonnegative integer such that a(n)! contains a string of exactly n consecutive 0's.

Original entry on oeis.org

0, 5, 10, 15, 20, 264, 25, 30, 35, 40, 45, 101805, 50, 55, 60, 65, 70

Offset: 0

Martin Y. Champel, Feb 22 2015

Most multiples of 5 belong to the sequence (if not all).
All terms whose indices are included in A000966 are far bigger than their neighboring terms whose indices are multiples of 5.
a(11) is a multiple of 5, we can verify a(11) = a(25448).

			a(0) = 0 as 0! = 1 does not contain '0'.
a(1) = 5 as 5! = 120 contains '0'.
a(2) = 10 as 10! = 3628800 contains '00' and 10 is the smallest integer for which the condition is met.

Cf. A027868, A000966.
Cf. A254042, A254447, A254448, A254449, A254500, A254501, A254502, A254716, A254717.
Cf. A252652.

\\ uses is() from A000966
f(k, special, sz, sz1) = my(f=k!); if (special, s=Str(f/10^valuation(f, 10)), s=Str(k!)); #strsplit(s, sz) - #strsplit(s, sz1);
a(n) = if (n==0, return(0)); my(sz= concat(vector(n, k, "0")), sz1=concat(sz, "0"), k=1,special=is(n)); while (f(k, special, sz, sz1) != 1, k++); k; \\ Michel Marcus, Oct 25 2023

# Python version 2.7
from math import factorial as fct
def trailing_zero(n):
    k=0
    while n!=0:
        n/=5
        k+=n
    return k
def A255400():
    index = 1
    f = 1
    while True:
        if trailing_zero(f) == index:
            print("A255400("+str(index)+") = " +str(f))
            index += 1
        elif trailing_zero(f) > index:
            while True:
                clnzer = str(fct(f))[:-trailing_zero(f)]
                if index*'0' in clnzer and (index+1)*'0' not in clnzer:
                    print("A255400("+str(index)+") = " +str(f))
                    index += 1
                    f = 0
                    break
                f +=1
        f +=1
    return

import re
def A255400(n):
    f, i, s = 1, 0, re.compile('[0-9]*[1-9]0{'+str(n)+'}[1-9][0-9]*')
    while s.match(str(f)+'1') is None:
        i += 1
        f *= i
    return i # Chai Wah Wu, Apr 02 2015

A354463 a(n) is the number of trailing zeros in (2^n)!.

Original entry on oeis.org

0, 0, 0, 1, 3, 7, 14, 31, 63, 126, 253, 509, 1021, 2045, 4094, 8189, 16380, 32763, 65531, 131067, 262140, 524285, 1048571, 2097146, 4194297, 8388603, 16777208, 33554424, 67108858, 134217720, 268435446, 536870902, 1073741816, 2147483642, 4294967289, 8589934584, 17179869176, 34359738358, 68719476729

Offset: 0

William Boyles, May 31 2022

			For n = 4, (2^4)! = 20922789888000, which has a(4) = 3 trailing zeros.

William Boyles, Table of n, a(n) for n = 0..650

Cf. A000079, A027868.

seq n = aux (2 ^ n) 0
  where
    aux x acc
      | x < 5 = acc
      | otherwise = aux y (acc + y)
      where
        y = x `div` 5

a[n_]:=IntegerExponent[(2^n)!]; Array[a,38,0] (* Stefano Spezia, Jun 01 2022 *)

a(n) = val(1<David A. Corneth, Jun 01 2022

from sympy import factorial, multiplicity
def a(n): return multiplicity(5, factorial(2**n, evaluate=False))
print([a(n) for n in range(39)]) # Michael S. Branicky, Jun 01 2022

def A354463(n):
    c, m = 0, 2**n
    while m >= 5:
        m //= 5
        c += m
    return c # Chai Wah Wu, Jun 02 2022

a(n) = A027868(A000079(n)). - Michel Marcus, Jun 01 2022

a(n) = 2^(n-2) - A055223(n) for n >= 2. - John Keith, Jun 06 2022

A381886 Triangle read by rows: T(n, k) = Sum_{j=1..floor(log[k](n))} floor(n / k^j) if k >= 2, T(n, 1) = n, T(n, 0) = 0^n.

Original entry on oeis.org

1, 0, 1, 0, 2, 1, 0, 3, 1, 1, 0, 4, 3, 1, 1, 0, 5, 3, 1, 1, 1, 0, 6, 4, 2, 1, 1, 1, 0, 7, 4, 2, 1, 1, 1, 1, 0, 8, 7, 2, 2, 1, 1, 1, 1, 0, 9, 7, 4, 2, 1, 1, 1, 1, 1, 0, 10, 8, 4, 2, 2, 1, 1, 1, 1, 1, 0, 11, 8, 4, 2, 2, 1, 1, 1, 1, 1, 1, 0, 12, 10, 5, 3, 2, 2, 1, 1, 1, 1, 1, 1

Offset: 0

Peter Luschny, Apr 03 2025

			Triangle starts:
  [ 0] 1;
  [ 1] 0,  1;
  [ 2] 0,  2,  1;
  [ 3] 0,  3,  1, 1;
  [ 4] 0,  4,  3, 1, 1;
  [ 5] 0,  5,  3, 1, 1, 1;
  [ 6] 0,  6,  4, 2, 1, 1, 1;
  [ 7] 0,  7,  4, 2, 1, 1, 1, 1;
  [ 8] 0,  8,  7, 2, 2, 1, 1, 1, 1;
  [ 9] 0,  9,  7, 4, 2, 1, 1, 1, 1, 1;
  [10] 0, 10,  8, 4, 2, 2, 1, 1, 1, 1, 1;
  [11] 0, 11,  8, 4, 2, 2, 1, 1, 1, 1, 1, 1;
  [12] 0, 12, 10, 5, 3, 2, 2, 1, 1, 1, 1, 1, 1;

Jeffrey C. Lagarias and Wijit Yangjit, The factorial function and generalizations, extended, arXiv:2310.12949 [math.NT], 2023.
A. M. Oller-Marcen 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.

Cf. A011371 (column 2), A054861 (column 3), A054893 (column 4), A027868 (column 5), A054895 (column 6), A054896 (column 7), A054897 (column 8), A054898 (column 9), A078651 (row sums).

Cf. A078632, A078567, A153216, A366471.

T := (n, b) -> local i; ifelse(b = 0, b^n, ifelse(b = 1, n, add(iquo(n, b^i), i = 1..floor(log(n, b))))): seq(seq(T(n, b), b = 0..n), n = 0..12);
# Alternative:
T := (n, k) -> local j; ifelse(k = 0, k^n, ifelse(k = 1, n, add(padic:-ordp(j, k), j = 1..n))): for n from 0 to 12 do seq(T(n, k), k = 0..n) od;

T[n_, 0] := If[n == 0, 1, 0]; T[n_, 1] := n;
T[n_, k_] := Last@Accumulate[IntegerExponent[Range[n], k]];
Table[T[n, k], {n, 0, 12}, {k, 0, n}] // MatrixForm
(* Alternative: *)
T[n_, k_] := Sum[Floor[n/k^j], {j, Floor[Log[k, n]]}]; T[n_, 1] := n; T[n_, 0] := 0^n; T[0, 0] = 1; Flatten@ Table[T[n, k], {n, 0, 12}, {k, 0, n}] (* Michael De Vlieger, Apr 03 2025 *)

T(n,k) = if (n==0, 1, if (n==1, k, if (k==0, 0, if (k==1, n, sum(j=1, n, valuation(j, k))))));
row(n) = vector(n+1, k, T(n,k-1)); \\ Michel Marcus, Apr 04 2025

from math import log
def T(n: int, b: int) -> int:
    return (b**n if b == 0 else n if b == 1 else
        sum(n // (b**i) for i in range(1, 1 + int(log(n, b)))))
print([[T(n, b) for b in range(n+1)] for n in range(12)])

def T(n, b): return (b^n if b == 0 else n if b == 1 else sum(valuation(j, b) for j in (1..n)))
print(flatten([[T(n, b) for b in range(n+1)] for n in srange(13)]))

T(n, k) = Sum_{j=1..n} valuation(j, k) for n >= 2.

cp's OEIS Frontend

A137582 Numbers having no inner zeros in decimal representation of their factorial.

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A165753 Number of trailing zeros in sequence of factorials of Fibonacci numbers.

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A175417 Exponent of 2 minus sum of all other exponents, in the prime power factorization of n!

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A181578 The number k such that each of the five factorials (5k+j)!, j=0..4, has exactly n trailing zeros in its base-10 representation; 0 if no such k exists.

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A181581 Smallest k such that each of the five factorials (5k)!, (5k+1)!, (5k+2)!, (5k+3)! and (5k+4)! has exactly 10^n trailing 0's. Zero, if no such k exists.

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A218976 a(n) is the smallest positive integer such that 10^(2 + floor(k/a(1)) + floor(k/a(2)) + ... + floor(k/a(n))) divides (k+9)! for all k > 0.

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A228311 Numbers k such that the sum of digits of k! is itself a factorial.

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Mathematica

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A255400 a(n) is the smallest nonnegative integer such that a(n)! contains a string of exactly n consecutive 0's.

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