%I #32 Feb 18 2024 01:25:03
%S 1,4,5,9,10,13,14,16,17,20,21,25,26,29,30,25,26,29,30,34,35,38,39,41,
%T 42,45,46,50,51,54,55,36,37,40,41,45,46,49,50,52,53,56,57,61,62,65,66,
%U 61,62,65,66,70,71,74,75,77,78,81,82,86,87,90,91,49,50,53,54,58,59,62
%N If 2n = Sum 2^e(k) then a(n) = Sum e(k)^2.
%H T. D. Noe, <a href="/A008935/b008935.txt">Table of n, a(n) for n=1..1023</a>
%F G.f.: 1/(1-x) * Sum_{k>=0} (k+1)^2*x^2^k/(1+x^2^k). - _Ralf Stephan_, Jun 23 2003
%e To get a(5), we write 10 = 2 + 8 = 2^1 + 2^3 so a(5) = 1^2 + 3^2 = 10.
%p a:= n-> (l-> add(l[i]*i^2, i=1..nops(l)))(convert(n, base, 2)):
%p seq(a(n), n=1..80); # _Alois P. Heinz_, Nov 20 2019
%t a[n_] := Total[Flatten[Position[Reverse[IntegerDigits[n, 2]], 1]]^2]; Table[a[n],{n,1,70}] (* Jean-François Alcover Mar 21 2011 *)
%o (C)
%o #include <stdio.h>
%o #include <stdlib.h>
%o #define USAGE "Usage: 'A008935 num'\n where num is the index of the desired ending value in the sequence.\n"
%o #define MAX 1023
%o #define SHIFT_MAX 9
%o int main(int argc, char *argv[]) { unsigned short mask, i, j, end; unsigned long sum; if (argc < 2) { fprintf(stderr, USAGE); return EXIT_FAILURE; } end = atoi(argv[1]); end = (end >= MAX) ? MAX : end;
%o fprintf(stdout, "Values: "); for (i = 1; i <= end; i++) { sum = 0; mask = 1; for (j = 0; j < SHIFT_MAX; j++, mask *= 2) if (i & mask) sum += (j+1) * (j+1); fprintf(stdout, "%ld", sum); if (i < end) fprintf(stdout, ","); } fprintf(stdout, "\n"); return EXIT_SUCCESS; }
%o (Haskell)
%o a008935 = f 1 where
%o f k x | x == 0 = 0
%o | r == 0 = f (k+1) x'
%o | otherwise = k^2 + f (k+1) x' where (x',r) = divMod x 2
%o -- _Reinhard Zumkeller_, Jul 05 2011
%Y Gives A003995 if sorted and duplicates removed.
%K nonn,nice,easy
%O 1,2
%A _N. J. A. Sloane_
%E Corrected and extended by Larry Reeves (larryr(AT)acm.org), Mar 22 2000