A261283 a(n) = bitwise XOR of all the bit numbers for the bits that are set in n, using number 1 for the LSB.

0, 1, 2, 3, 3, 2, 1, 0, 4, 5, 6, 7, 7, 6, 5, 4, 5, 4, 7, 6, 6, 7, 4, 5, 1, 0, 3, 2, 2, 3, 0, 1, 6, 7, 4, 5, 5, 4, 7, 6, 2, 3, 0, 1, 1, 0, 3, 2, 3, 2, 1, 0, 0, 1, 2, 3, 7, 6, 5, 4, 4, 5, 6, 7, 7, 6, 5, 4, 4, 5, 6, 7, 3, 2, 1, 0, 0, 1, 2, 3, 2, 3, 0, 1, 1, 0

Offset: 0

Author

M. F. Hasler, Aug 14 2015, following the original version A253315 by Philippe Beaudoin, Dec 30 2014

Keywords

• nonn
• base
• easy

Comments

If the least significant bit is numbered 0, then a(2n) = a(2n+1) if one uses the "natural" definition reading "...set in n": see A253315 for that version. To avoid the duplication, we chose here to start numbering the bits with 1 for the LSB; equivalently, we can start numbering the bits with 0 but use the definition "...bits set in 2n". In any case, a(n) = A253315(2n) = A253315(2n+1).

Since the XOR operation is associative, one can define XOR of an arbitrary number of terms in a recursive way, there is no ambiguity about the order in which the operations are performed.

Examples

			a(7) = a(4+2+1) = a(2^2+2^1+2^0) = (2+1) XOR (1+1) XOR (0+1) = 3 XOR 3 = 0.
a(12) = a(8+4) = a(2^3+2^2) = (3+1) XOR (2+1) = 4+3 = 7.
		

Links

• Philippe Beaudoin, Table of n, a(n) for n = 0..8190
• Oliver Nash, Yet another prisoner puzzle, coins on a chessboard problem.
• Tilman Piesk, Illustration of the first 128 terms

Crossrefs

Cf. A075926 (indices of 0's), A253315, A327041 (OR equivalent).

Programs

Mathematica

A261283[n_] := If[n == 0, 0, BitXor @@ PositionIndex[Reverse[IntegerDigits[n, 2]]][1]]; Array[A261283, 100, 0] (* Paolo Xausa, May 29 2024 *)

PARI

A261283(n,b=bittest(n,0))={for(i=1,#binary(n),bittest(n,i)&&b=bitxor(b,i+1));b}