A359495 Sum of positions of 1's in binary expansion minus sum of positions of 1's in reversed binary expansion, where positions in a sequence are read starting with 1 from the left.

0, 0, -1, 0, -2, 0, -2, 0, -3, 0, -2, 1, -4, -1, -3, 0, -4, 0, -2, 2, -4, 0, -2, 2, -6, -2, -4, 0, -6, -2, -4, 0, -5, 0, -2, 3, -4, 1, -1, 4, -6, -1, -3, 2, -5, 0, -2, 3, -8, -3, -5, 0, -7, -2, -4, 1, -9, -4, -6, -1, -8, -3, -5, 0, -6, 0, -2, 4, -4, 2, 0, 6

Offset: 0

Author

Gus Wiseman, Jan 05 2023

Keywords

• sign
• look
• base

Comments

Also the sum of partial sums of reversed binary expansion minus sum of partial sums of binary expansion.

Examples

			The binary expansion of 158 is (1,0,0,1,1,1,1,0), with positions of 1's {1,4,5,6,7} with sum 23, reversed {2,3,4,5,8} with sum 22, so a(158) = 1.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..16383

Crossrefs

Indices of positive terms are A359401.

Indices of 0's are A359402.

A030190 gives binary expansion, reverse A030308.

A070939 counts binary digits.

A230877 adds up positions of 1's in binary expansion, reverse A029931.

Cf. A000120, A048793, A053632, A222955, A231204, A291166, A326669, A326672, A326673, A359042, A359043.

Programs

Maple

a:= n-> (l-> add(i*(l[-i]-l[i]), i=1..nops(l)))(Bits[Split](n)):
seq(a(n), n=0..127);  # Alois P. Heinz, Jan 09 2023

Mathematica

sap[q_]:=Sum[q[[i]]*(2i-Length[q]-1),{i,Length[q]}];
Table[sap[IntegerDigits[n,2]],{n,0,100}]

Python

def A359495(n):
    k = n.bit_length()-1
    return sum((i<<1)-k for i, j in enumerate(bin(n)[2:]) if j=='1') # Chai Wah Wu, Jan 09 2023

Formula

a(n) = A029931(n) - A230877(n).

If n = Sum_{i=1..k} q_i * 2^(i-1), then a(n) = Sum_{i=1..k} q_i * (2i-k-1).