A144079 a(n) = the number of digits in the binary representation of n that equal the corresponding digit in the binary reversal of n. (I.e., a(n) = number of 0's in n XOR A030101(n).)

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

Offset: 1

Author

Leroy Quet, Sep 09 2008

Keywords

• base
• nonn

Comments

A144078(n) + a(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 three bit pairs that contain the same values, so a(20) = 3.
		

Crossrefs

Cf. A030101, A144078.

Programs

Maple

A144079 := 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,",A144079(n)) ; od: # R. J. Mathar, Sep 14 2008

Mathematica

Table[With[{c=IntegerDigits[n,2]},Count[BitXor[c,Reverse[c]],0]],{n,110}] (* Harvey P. Dale, Sep 03 2015 *)

Extensions

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