A056851 -id:A056851 - cp's OEIS frontend

A034886 Number of digits in n!.

1, 1, 1, 1, 2, 3, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16, 18, 19, 20, 22, 23, 24, 26, 27, 29, 30, 31, 33, 34, 36, 37, 39, 41, 42, 44, 45, 47, 48, 50, 52, 53, 55, 57, 58, 60, 62, 63, 65, 67, 68, 70, 72, 74, 75, 77, 79, 81, 82, 84, 86, 88, 90, 91, 93, 95, 97, 99, 101, 102

Offset: 0

Erich Friedman

Most counterexamples to the Kamenetsky formula (see below) must belong to A177901.
Noam D. Elkies reported on MathOverflow (see link):
"A counterexample [to Kamenetsky's formula] is n_1 := 6561101970383, with log_10((n_1/e)^n_1*sqrt(2*Pi*n_1)) = 81244041273652.999999999999995102482, but log_10(n_1!) = 81244041273653.000000000000000618508. [...] n_1 is the first counterexample, and the only one up to 10^14."
From Bernard Schott, Dec 07 2019: (Start)
a(n) < n iff 2 <= n <= 21;
a(n) = n iff n = 1, 22, 23, 24;
a(n) > n iff n = 0 or n >= 25. (End)

Martin 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, pp. 50-65, 1978

T. D. Noe, Table of n, a(n) for n = 0..10000
Noam D. Elkies, A counterexample to Kamenetsky's formula for the number of digits in n-factorial.
Wikipedia, Stirling's Formula.
Index entries for sequences related to factorial numbers.

Cf. A000142, A055642.
Cf. A006488 (a(n) is a square), A056851 (a(n) is a cube), A035065 (a(n) is a prime), A333431 (a(n) is a factorial), A333598 (a(n) is a palindrome), A067367 (p and a(p) are primes), A058814 (n divides a(n)).
Cf. A137580 (number of distinct digits in n!), A027868 (number of trailing zeros in n!).

a034886 = a055642 . a000142  -- Reinhard Zumkeller, Apr 08 2012

[Floor(Log(Factorial(n))/Log(10)) + 1: n in [0..30]]; // G. C. Greubel, Feb 26 2018

A034886 := n -> `if`(n<2,1,`if`(n<6561101970383, ceil((ln(2*Pi)-2*n+ln(n)*(1+2*n))/(2*ln(10))),length(n!))); # Peter Luschny, Aug 26 2011

Join[{1, 1}, Table[Ceiling[Log[10, 2 Pi n]/2 + n*Log[10, n/E]], {n, 2, 71}]]
f[n_] := Floor[(Log[2Pi] - 2n + Log[n]*(1 + 2n))/(2Log[10])] + 1; f[0] = f[1] = 1; Array[f, 72, 0] (* Robert G. Wilson v, Jan 09 2013 *)
IntegerLength/@(Range[0,80]!) (* Harvey P. Dale, Aug 07 2022 *)

for(n=0,30, print1(floor(log(n!)/log(10)) + 1, ", ")) \\ G. C. Greubel, Feb 26 2018

a(n) = floor(log(n!)/log(10)) + 1.
a(n) = A027869(n) + A079680(n) + A079714(n) + A079684(n) + A079688(n) + A079690(n) + A079691(n) + A079692(n) + A079693(n) + A079694(n); a(n) = A055642(A000142(n)). - Reinhard Zumkeller, Jan 27 2008
Using Stirling's formula we can derive an approximation, which is very fast to compute in practice: ceiling(log_10(2*Pi*n)/2 + n*(log_10(n/e))). This approximation gives the exact answer for 2 <= n <= 5*10^7. - Dmitry Kamenetsky, Jul 07 2008
a(n) = ceiling(log_10(1) + log_10(2) + ... + log_10(n)). - Dmitry Kamenetsky, Nov 05 2010

Explained that the formula is an approximation. Made the formula easier to read. - Dmitry Kamenetsky, Dec 15 2010

A006488 Numbers n such that n! has a square number of digits.

Original entry on oeis.org

0, 1, 2, 3, 7, 12, 18, 32, 59, 81, 105, 132, 228, 265, 284, 304, 367, 389, 435, 483, 508, 697, 726, 944, 1011, 1045, 1080, 1115, 1187, 1454, 1494, 1617, 1659, 1788, 1921, 2012, 2105, 2200, 2248, 2395, 2445, 2861, 2915, 3192, 3480, 3539, 3902, 3964, 4476

Offset: 1

N. J. A. Sloane and Robert G. Wilson v

Numbers whose square is represented by the number of digits of n!: 1, 2, 3, 4, 6, 9, 11, 13, 15, 21, 23, 24, 25, 28, 29, ..., . - Robert G. Wilson v, May 14 2014
From Bernard Schott, Jan 04 2020: (Start)
In M. Gardner's book, see reference, there is a tree printout of 105! with 169 digits, where the bottom row consists of the 25 trailing zeros of 105!. M. Gardner does not explain if this is the only factorial that can be displayed in a similar tree form.
Proof: If m! has q^2 digits, hence the number of trailing zeros in m! must be equal to 2*q-1 to satisfy this triangular look; m = 105 satisfies these two conditions with q = 13 because 105! has 13^2 = 169 digits and 2*13-1 = 25 trailing zeros.
When m < 105 and m! has q^2 digits (m <= 81), then q <= 11 and the number of trailing zeros is <= 2*q - 3.
When m > 105 and m! has q^2 digits (m >= 132), then q >= 15 and the number of trailing zeros is >= 2*q + 2.
Hence, only 105! presents such a tree printout.
              1
             081
            39675
           8240290
          900504101
         30580032964
        9720646107774
       902579144176636
      57322653190990515
     3326984536526808240
    339776398934872029657
   99387290781343681609728
  0000000000000000000000000
(End)

M. Gardner, Mathematical Magic Show. Random House, NY, 1978, p. 55.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Robert G. Wilson v, Table of n, a(n) for n = 1..1311
D. S. Kluk and N. J. A. Sloane, Correspondence, 1979
Eric Weisstein's World of Mathematics, Stirling's Approximation and Stirling's Series
Index entries for sequences related to factorial numbers

Cf. A000142, A027868 (trailing zeros), A034886 (number of digits), A056851.

[k:k in [0..5000]| IsSquare(#Intseq(Factorial(k)))]; // Marius A. Burtea, Jan 04 2020

LogBase10Stirling[n_] := Floor[Log[10, 2 Pi n]/2 + n*Log[10, n/E] + Log[10, 1 + 1/(12n) + 1/(288n^2) - 139/(51840n^3) - 571/(2488320n^4) + 163879/(209018880n^5)]]; Select[ Range[ 4500], IntegerQ[ Sqrt[ (LogBase10Stirling[ # ] + 1)]] & ] (* The Mathematica coding comes from J. Stirling's expansion for the Gamma function; see the links. For more terms inside the last Log_10 function, use A001163 & A001164. Robert G. Wilson v, Apr 27 2014 *)
Select[Range[0,4500],IntegerQ[Sqrt[IntegerLength[#!]]]&] (* Harvey P. Dale, Sep 27 2018 *)

isok(n) = issquare(#Str(n!)); \\ Michel Marcus, Sep 05 2015

A333598 Numbers m such that m! has a palindromic number of decimal digits.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 22, 30, 37, 44, 57, 63, 69, 70, 81, 86, 91, 106, 111, 116, 126, 131, 140, 145, 154, 163, 168, 177, 186, 199, 221, 225, 238, 242, 255, 259, 288, 292, 368, 372, 384, 388, 407, 411, 419, 423, 438, 450, 532

Offset: 1

Bernard Schott, Mar 28 2020

The corresponding palindromic numbers are 1, 1, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44, 55, 77, 88, 99, 101, ...

Nice result: 22 is a palindrome and 22! has 22 digits, and also, 44! has 55 digits.

			14! = 87178291200 that has 11 digits, 11 is a palindrome, hence 14 is a term.

Cf. A000142, A002113, A034886.

Cf. A006488 (similar, with square), A035065 (similar, with prime), A056851 (similar, with cube), A333431 (similar, with factorial).

Select[Range[0, 532], PalindromeQ @ Length @ IntegerDigits[#!] &] (* Amiram Eldar, Mar 28 2020 *)
Select[Range[0,550],PalindromeQ[IntegerLength[#!]]&] (* Harvey P. Dale, Oct 30 2023 *)

isok(m) = my(d=digits(#Str(m!))); d == Vecrev(d); \\ Michel Marcus, Mar 28 2020

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A034886 Number of digits in n!.

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A006488 Numbers n such that n! has a square number of digits.

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A333598 Numbers m such that m! has a palindromic number of decimal digits.

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