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

Author

Erich Friedman

Keywords

• nonn
• base
• easy

Comments

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)

References

• 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

Links

• 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.

Crossrefs

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!).

Programs

Haskell

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

Magma

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

Maple

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

Mathematica

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 *)

PARI

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

Formula

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

Extensions

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