A118738 Number of ones in binary expansion of 5^n.

1, 2, 3, 6, 5, 6, 7, 8, 12, 13, 11, 15, 13, 14, 17, 20, 20, 20, 24, 19, 26, 29, 25, 27, 30, 19, 31, 33, 29, 36, 37, 33, 39, 34, 42, 40, 44, 42, 38, 46, 53, 54, 49, 52, 52, 53, 50, 49, 54, 60, 58, 60, 54, 64, 58, 74, 61, 67, 74, 65, 61, 77, 74, 81, 86, 78, 87, 85, 82, 89, 83, 79

Offset: 0

Author

Zak Seidov, May 22 2006

Keywords

• base
• nonn
• easy

Comments

Also binary weight of 10^n, which is verified easily enough: 10^n = 2^n * 5^n; it is obvious that 2^n in binary is a single 1 followed by n 0's, therefore, in the binary expansion of 2^n * 5^n, the 2^n contributes only the trailing zeros. - Alonso del Arte, Oct 28 2012

Conjecture: a(n)/n -> log_4(5) = 1.160964... as n -> oo. - M. F. Hasler, Apr 17 2024

Examples

			a(2) = 3 because 5^2 = 25 is 11001, which has 3 on bits.
		

Links

• Robert Israel, Table of n, a(n) for n = 0..10000
• Hugo Pfoertner, Plot of a(n) - 1.160964*n, +-4*sqrt(n), n up to 10^6.

Crossrefs

Cf. A000120 (Hamming weight), A000351 (5^n), A061785 (floor(log_2(5^n))), A118737 (number of bits 0 in 5^n).

Cf. A011754 (analog for 3^n).

Programs

Magma

[&+Intseq(5^n, 2): n in [0..100]]; // Vincenzo Librandi, Nov 13 2024

Maple

seq(convert(convert(5^n,base,2),`+`),n=0..100); # Robert Israel, Dec 24 2017

Mathematica

Table[DigitCount[5^n, 2, 1], {n, 0, 71}] (* Ray Chandler, Sep 29 2006 *)

PARI

a(n) = hammingweight(5^n) \\ Iain Fox, Dec 24 2017

Python

A118738 = lambda n: (5**n).bit_count() # For Python 3.10 and later. - M. F. Hasler, Apr 17 2024

Formula

a(n) + A118737(n) = A061785(n) + 1 for n >= 1. - Robert Israel, Dec 24 2017 [corrected by Amiram Eldar, Jul 27 2023]

a(n) = A000120(A000351(n)) = Hammingweight(5^n). - M. F. Hasler, Apr 17 2024