A046820 Number of 1's in binary expansion of 5n.

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

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• base

Comments

a(n) is also the largest integer such that 2^a(n) divides binomial(10n, 5n). - Benoit Cloitre, Mar 27 2002

Examples

			For n = 10, 5*n = 50 = 110010_2, having 3 1's. So, a(10) = 3. - _Indranil Ghosh_, Jan 18 2017
		

Links

• Indranil Ghosh, Table of n, a(n) for n = 0..10000
• Michael Gilleland, Some Self-Similar Integer Sequences.

Crossrefs

Cf. A000120.

Programs

Mathematica

a[n_] := DigitCount[5*n, 2, 1]; Array[a, 100, 0] (* Amiram Eldar, Jul 18 2023 *)

PARI

a(n) = hammingweight(5*n); \\ Michel Marcus, Aug 19 2018

Python

def A046820(n):
    return bin(5*n)[2:].count("1") # Indranil Ghosh, Jan 18 2017

Formula

a(n) = floor(log(gcd(binomial(10*n, 5*n), 2^floor(log(binomial(10*n, 5*n))/log(2))))/log(2)). - Benoit Cloitre, Mar 27 2002

a(n) = A000120(5*n). - Indranil Ghosh, Jan 18 2017