A079584 Number of ones in the binary expansion of n!.

1, 1, 1, 2, 2, 4, 4, 6, 6, 6, 11, 7, 12, 12, 12, 18, 18, 22, 23, 17, 22, 25, 28, 31, 29, 30, 35, 38, 42, 40, 48, 42, 42, 46, 51, 56, 51, 58, 59, 64, 63, 66, 64, 71, 74, 70, 77, 81, 89, 87, 89, 90, 88, 94, 87, 99, 103, 98, 101, 109, 113, 103, 113, 120, 120, 109, 123, 121, 130, 121

Offset: 0

Author

Jose R. Brox (tautocrona(AT)terra.es), Jan 26 2003

Keywords

• nonn
• base

Examples

			a(5) = 4 because 5! = 120 and 120_10 = 1111000_2, with 4 ones.
		

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 0..10000
• Florian Luca, The number of non-zero digits of n!, Canad. Math. Bull. 45 (2002), pp. 115-118.
• Carlo Sanna, On the sum of digits of the factorial, arXiv:1409.4912 [math.NT], 2014.
• Carlo Sanna, On the sum of digits of the factorial, Journal of Number Theory 147 (2015), pp. 836-841.
• Index entries for sequences related to factorial numbers

Crossrefs

Cf. A000120 (binary weight), A000142 (factorial), A004152 (sum of decimal digits).

Programs

Maple

seq(convert(convert(n!,base,2),`+`), n=0..1000); # Robert Israel, Sep 18 2014

Mathematica

Table[DigitCount[n!,2,1],{n,70}] (* Harvey P. Dale, Jul 10 2012 *)

PARI

for(n=1,300,b=binary(n!); print1(sum(k=1,length(b),b[k])","))

PARI

a(n)=hammingweight(n!) \\ Charles R Greathouse IV, Mar 27 2013

Python

import math
def a(n):
    return bin(math.factorial(n))[2:].count("1") # Indranil Ghosh, Dec 23 2016

Formula

a(n) << n log n. - Charles R Greathouse IV, Mar 27 2013

a(n) = A000120(A000142(n)). - Michel Marcus, Sep 18 2014

Extensions

a(0)=1 prepended by Alois P. Heinz, Mar 07 2023