A144078 a(n) = the number of digits in the binary representation of n that differ from the corresponding digit in the binary reversal of n. (I.e., a(n) = number of 1's in n XOR A030101(n).)

0, 2, 0, 2, 0, 2, 0, 2, 0, 4, 2, 4, 2, 2, 0, 2, 0, 4, 2, 2, 0, 4, 2, 4, 2, 2, 0, 4, 2, 2, 0, 2, 0, 4, 2, 4, 2, 6, 4, 4, 2, 6, 4, 2, 0, 4, 2, 4, 2, 2, 0, 6, 4, 4, 2, 6, 4, 4, 2, 4, 2, 2, 0, 2, 0, 4, 2, 4, 2, 6, 4, 2, 0, 4, 2, 4, 2, 6, 4, 4, 2, 6, 4, 2, 0, 4, 2, 4, 2, 6, 4, 2, 0, 4, 2, 4, 2, 2, 0, 6, 4, 4, 2, 4, 2

Offset

1,2

Comments

a(n) + A144079(n) = A070939(n), the number of binary digits in n.

Links

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Formula

From Rémy Sigrist, Oct 07 2018: (Start)

a(n) = 0 iff n is a binary palindrome (A006995).

a(A143960(n)) = 2*n (in fact A143960(n) is the least k such that a(k) = 2*n).

(End)

Example

20 in binary is 10100. Compare this with its digit reversal, 00101. XOR each pair of corresponding digits: 1 XOR 0 = 1, 0 XOR 0 = 0, 1 XOR 1 = 0, 0 XOR 0 = 0, 0 XOR 1 = 1. There are two bit pairs that differ, so a(20) = 2.

Maple

A144078 := proc(n) local a, dgs, i; a := 0 ; dgs := convert(n, base, 2) ; for i from 1 to nops(dgs) do if op(i, dgs)+op(-i, dgs) = 1 then a := a+1 ; fi; od; RETURN(a) ; end: for n from 1 to 240 do printf("%d, ", A144078(n)) ; od: # R. J. Mathar, Sep 14 2008

Mathematica

brd[n_]:=Module[{idn2=IntegerDigits[n, 2]}, Count[Transpose[{idn2, Reverse[ idn2]}], _?(#[[1]]!=#[[2]]&)]]; Array[brd, 110] (* Harvey P. Dale, May 09 2016 *)

Prog

(PARI) a(n) = hammingweight(bitxor(n, fromdigits(Vecrev(binary(n)), 2))) \\ Rémy Sigrist, Oct 07 2018

Crossrefs

Cf. A006995, A030101, A143960, A144079.

Sequence in context: A096158 A264782 A053471 * A277158 A318282 A359287

Adjacent sequences: A144075 A144076 A144077 * A144079 A144080 A144081

Keyword

base,nonn

Author

Leroy Quet, Sep 09 2008

Extensions

More terms from R. J. Mathar, Sep 14 2008

Status

approved