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

%I A245788 #28 Mar 20 2023 14:36:44
%S A245788 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,
%T A245788 75,78,108,84,116,120,155,32,66,68,105,72,111,114,156,80,123,126,172,
%U A245788 132,180,184,235,96,147,150,204,156,212,216,275,168,228,232,295,240
%N A245788 n times the number of 1's in the binary expansion of n.
%H A245788 Jens Kruse Andersen, <a href="/A245788/b245788.txt">Table of n, a(n) for n = 0..1000</a>
%H A245788 Project Euler, <a href="https://projecteuler.net/problem=759">Problem 759</a>, sequence f(n).
%F A245788 a(2*n) = 2*a(n).
%F A245788 a(2*n+1) = 2*n + 1 + (2+1/n)*a(n). - _Robert Israel_, Aug 01 2014
%F A245788 G.f.: x * (d/dx) (1/(1 - x))*Sum_{k>=0} x^(2^k)/(1 + x^(2^k)). - _Ilya Gutkovskiy_, Mar 27 2018
%e A245788 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 + ...
%p A245788 a:= n -> n * convert(convert(n,base,2),`+`):
%p A245788 seq(a(n),n=0..100); # _Robert Israel_, Aug 01 2014
%t A245788 Table[n*DigitCount[n,2,1],{n,0,100}] (* _Harvey P. Dale_, Dec 16 2014 *)
%o A245788 (PARI) sumbit(n) = my(r);while(n>0,r+=n%2;n\=2);r
%o A245788 a(n) = n*sumbit(n)
%o A245788 (PARI) {a(n) = if( n<0, 0, n * sumdigits(n, 2))}; /* _Michael Somos_, Aug 05 2014 */ /* since version 2.6.0 */
%o A245788 (Python) [n*bin(n)[2:].count('1') for n in range(1000)] # _Chai Wah Wu_, Aug 03 2014
%Y A245788 Cf. A000120 (number of 1's), A057147 (decimal version).
%K A245788 nonn,base
%O A245788 0,3
%A A245788 _Franklin T. Adams-Watters_, Aug 01 2014