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%I #20 May 16 2023 12:58:59
%S 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,
%T 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,
%U 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
%N 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).)
%C A144078(n) + a(n) = A070939(n), the number of binary digits in n.
%e 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.
%p 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
%t Table[With[{c=IntegerDigits[n,2]},Count[BitXor[c,Reverse[c]],0]],{n,110}] (* _Harvey P. Dale_, Sep 03 2015 *)
%Y Cf. A030101, A144078.
%K base,nonn
%O 1,3
%A _Leroy Quet_, Sep 09 2008
%E More terms from _R. J. Mathar_, Sep 14 2008
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