A118737 -id:A118737 - cp's OEIS frontend

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

Zak Seidov, May 22 2006

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

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

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.

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).

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

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

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

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

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

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

cp's OEIS Frontend

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

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