A245788 n times the number of 1's in the binary expansion of n.

0, 1, 2, 6, 4, 10, 12, 21, 8, 18, 20, 33, 24, 39, 42, 60, 16, 34, 36, 57, 40, 63, 66, 92, 48, 75, 78, 108, 84, 116, 120, 155, 32, 66, 68, 105, 72, 111, 114, 156, 80, 123, 126, 172, 132, 180, 184, 235, 96, 147, 150, 204, 156, 212, 216, 275, 168, 228, 232, 295, 240

Offset

0,3

Links

Jens Kruse Andersen, Table of n, a(n) for n = 0..1000

Project Euler, Problem 759, sequence f(n).

Formula

a(2*n) = 2*a(n).

a(2*n+1) = 2*n + 1 + (2+1/n)*a(n). - Robert Israel, Aug 01 2014

G.f.: x * (d/dx) (1/(1 - x))*Sum_{k>=0} x^(2^k)/(1 + x^(2^k)). - Ilya Gutkovskiy, Mar 27 2018

Example

G.f. = x + 2*x^2 + 6*x^3 + 4*x^4 + 10*x^5 + 12*x6 + 21*x^7 + 8*x^8 + 18*x^9 + ...

Maple

a:= n -> n * convert(convert(n, base, 2), `+`):
seq(a(n), n=0..100); # Robert Israel, Aug 01 2014

Mathematica

Table[n*DigitCount[n, 2, 1], {n, 0, 100}] (* Harvey P. Dale, Dec 16 2014 *)

Prog

(PARI) sumbit(n) = my(r); while(n>0, r+=n%2; n\=2); r
a(n) = n*sumbit(n)
(PARI) {a(n) = if( n<0, 0, n * sumdigits(n, 2))}; /* Michael Somos, Aug 05 2014 */ /* since version 2.6.0 */
(Python) [n*bin(n)[2:].count('1') for n in range(1000)] # Chai Wah Wu, Aug 03 2014

Crossrefs

Cf. A000120 (number of 1's), A057147 (decimal version).

Sequence in context: A264647 A094748 A245579 * A065879 A065880 A335063

Adjacent sequences: A245785 A245786 A245787 * A245789 A245790 A245791

Keyword

nonn,base

Author

Franklin T. Adams-Watters, Aug 01 2014

Status

approved