A101080 Table of Hamming distances between binary vectors representing i and j, for i >= 0, j >= 0, read by antidiagonals.

0, 1, 1, 1, 0, 1, 2, 2, 2, 2, 1, 1, 0, 1, 1, 2, 2, 1, 1, 2, 2, 2, 1, 2, 0, 2, 1, 2, 3, 3, 3, 3, 3, 3, 3, 3, 1, 2, 1, 2, 0, 2, 1, 2, 1, 2, 2, 2, 2, 1, 1, 2, 2, 2, 2, 2, 1, 2, 1, 1, 0, 1, 1, 2, 1, 2, 3, 3, 3, 3, 2, 2, 2, 2, 3, 3, 3, 3, 2, 2, 1, 2, 2, 1, 0, 1, 2, 2, 1, 2, 2, 3, 3, 2, 2, 3, 3, 1, 1, 3, 3, 2, 2, 3, 3

Offset: 0

Author

Marc LeBrun, Nov 29 2004

Keywords

• easy
• nonn
• tabl

Comments

a(n,0) = a(0,n) = A000120(n).

Examples

			a(3,5) = 2 because the binary Hamming distance (number of differing bits) between ...0011 and ...0101 is 2.
From _Indranil Ghosh_, Mar 31 2017: (Start)
Array begins:
  0, 1, 1, 2, 1, 2, 2, 3, ...
  1, 0, 2, 1, 2, 1, 3, 2, ...
  1, 2, 0, 1, 2, 3, 1, 2, ...
  2, 1, 1, 0, 3, 2, 2, 1, ...
  1, 2, 2, 3, 0, 1, 1, 2, ...
  2, 1, 3, 2, 1, 0, 2, 1, ...
  2, 3, 1, 2, 1, 2, 0, 1, ...
  3, 2, 2, 1, 2, 1, 1, 0, ...
  ...
Triangle formed when the array is read by antidiagonals:
  0;
  1, 1;
  1, 0, 1;
  2, 2, 2, 2;
  1, 1, 0, 1, 1;
  2, 2, 1, 1, 2, 2;
  2, 1, 2, 0, 2, 1, 2;
  3, 3, 3, 3, 3, 3, 3, 3;
  ...
(End)
		

Links

• Alois P. Heinz, Antidiagonals n = 0..200, flattened

Crossrefs

Cf. A000120, A003987.

Programs

Maple

H:= (i, j)-> add(v, v=convert(Bits[Xor](i, j), base, 2)):
seq(seq(H(n, d-n), n=0..d), d=0..20);  # Alois P. Heinz, Nov 18 2015

Mathematica

a[i_, j_] := Total[IntegerDigits[BitXor[i, j], 2]]; Table[a[i-j, j], {i, 0, 20}, {j, 0, i}] // Flatten (* Jean-François Alcover, Apr 07 2016 *)

PARI

b(n) = if(n<1, 0, b(n\2) + n%2);
tabl(nn) = {for(n=0, nn, for(k=0, n, print1(b(bitxor(k, n - k)),", ");); print(););};
tabl(20) \\ Indranil Ghosh, Mar 31 2017

Python

for n in range(20):
    print([bin(k^(n - k))[2:].count("1") for k in range(n + 1)]) # Indranil Ghosh, Mar 31 2017

Formula

a(i,j) = A000120(A003987(i,j)).