

A261283


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


5



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
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OFFSET

0,3


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.


LINKS

Philippe Beaudoin, Table of n, a(n) for n = 0..8190
O. Nash, Yet another prisoner puzzle, coins on a chessboard problem.


EXAMPLE

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.


PROG

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


CROSSREFS

Cf. A075926 (indices of 0's).
Sequence in context: A279316 A074989 A307429 * A123548 A131838 A274885
Adjacent sequences: A261280 A261281 A261282 * A261284 A261285 A261286


KEYWORD

nonn,base,easy


AUTHOR

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


STATUS

approved



