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

Author

Leroy Quet, Sep 09 2008

Keywords

• base
• nonn

Comments

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

Examples

			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.
		

Links

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

Crossrefs

Cf. A006995, A030101, A143960, A144079.

Programs

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 *)

PARI

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

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)

Extensions

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