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If 2n = Sum 2^e(k) then a(n) = Sum e(k)^2.
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%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