A359400 Sum of positions of zeros in the reversed binary expansion of n, where positions in a sequence are read starting with 1 from the left.

1, 0, 1, 0, 3, 2, 1, 0, 6, 5, 4, 3, 3, 2, 1, 0, 10, 9, 8, 7, 7, 6, 5, 4, 6, 5, 4, 3, 3, 2, 1, 0, 15, 14, 13, 12, 12, 11, 10, 9, 11, 10, 9, 8, 8, 7, 6, 5, 10, 9, 8, 7, 7, 6, 5, 4, 6, 5, 4, 3, 3, 2, 1, 0, 21, 20, 19, 18, 18, 17, 16, 15, 17, 16, 15, 14, 14, 13

Offset: 0

Author

Gus Wiseman, Jan 05 2023

Keywords

• nonn
• base

Examples

			The reversed binary expansion of 100 is (0,0,1,0,0,1,1), with zeros at positions {1,2,4,5}, so a(100) = 12.
		

Crossrefs

The number of zeros is A023416, partial sums A059015.

Row sums of A368494.

For positions of 1's we have A029931, non-reversed A230877.

The non-reversed version is A359359.

A003714 lists numbers with no successive binary indices.

A030190 gives binary expansion, reverse A030308.

A039004 lists the positions of zeros in A345927.

Cf. A000120, A048793, A069010, A070939, A073642, A328594, A328595, A344618, A359402, A359495.

Programs

C

long A359400(long n) {
  long result = 0, counter = 1;
  do {
    if (n % 2 == 0)
      result += counter;
    counter++;
    n /= 2;
  } while (n > 0);
  return result; } // Frank Hollstein, Jan 06 2023

Mathematica

Table[Total[Join@@Position[Reverse[IntegerDigits[n,2]],0]],{n,0,100}]

Python

def a(n): return sum(i for i, bi in enumerate(bin(n)[:1:-1], 1) if bi=='0')
print([a(n) for n in range(78)]) # Michael S. Branicky, Jan 09 2023

Formula

a(n) = binomial(A029837(n)+1, 2) - A029931(n), for n>0.