A334592 Number of zeros in XOR-triangle with first row generated from the binary expansion of n.

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

Offset: 1

Author

Peter Kagey, May 07 2020

Keywords

• nonn
• base

Comments

An XOR-triangle is an inverted 0-1 triangle formed by choosing a top row and having each entry in the subsequent rows be the XOR of the two values above it.

Conjecture: Records occur at powers of two.

Examples

			For n = 53, a(53) = 9 because 53 = 110101_2 in binary, and the corresponding XOR-triangle has 9 zeros:
  1 1 0 1 0 1
   0 1 1 1 1
    1 0 0 0
     1 0 0
      1 0
       1
		

Links

• Peter Kagey, Table of n, a(n) for n = 1..8191
• MathOverflow user DSM, Number triangle
• Index entries for sequences related to binary expansion of n

Crossrefs

Cf. A000217, A070939.

Cf. also A334556, A334591, A334593, A334594, A334595, A334596.

Programs

Mathematica

Array[Count[Flatten@ NestWhileList[Map[BitXor @@ # &, Partition[#, 2, 1]] &, IntegerDigits[#, 2], Length@ # > 1 &], 0] &, 77] (* Michael De Vlieger, May 08 2020 *)

PARI

a(n) = {my(b=binary(n), nb=#b-hammingweight(n)); for (n=1, #b-1, b = vector(#b-1, k, bitxor(b[k], b[k+1])); nb += #b-vecsum(b);); nb;} \\ Michel Marcus, May 08 2020

Formula

a(n) = A000217(A070939(n)) - A334593(n).