A277238 Folding numbers (see comments for the definition).

1, 2, 6, 10, 12, 22, 28, 38, 42, 52, 56, 78, 90, 108, 120, 142, 150, 170, 178, 204, 212, 232, 240, 286, 310, 346, 370, 412, 436, 472, 496, 542, 558, 598, 614, 666, 682, 722, 738, 796, 812, 852, 868, 920, 936, 976, 992, 1086, 1134, 1206, 1254, 1338, 1386, 1458, 1506, 1596, 1644, 1716, 1764, 1848, 1896, 1968

Offset: 1

Author

Taylor J. Smith, Oct 06 2016

Keywords

• nonn
• base

Comments

Folding numbers: Numbers with an even number of bits in their binary expansion such that the XOR of the left half and the reverse of the right half is the all-1's string. Numbers with an odd number of bits in their binary expansion such that the central bit is 1, and the XOR of the left (n-1)/2 bits and the reverse of the right (n-1)/2 bits is the all-1's string.

Folding numbers with an even (resp. odd) number of bits form A035928 (resp. A276795). - N. J. A. Sloane, Nov 03 2016

Examples

			178 in base 2 is 10110010. Taking the XOR of 1011 and 0100 (which is 0010 reversed) gives the result 1111, so 178 is in the sequence.
		

Links

• Lars Blomberg, Table of n, a(n) for n = 1..10000
• Programming Puzzles and Code Golf Stack Exchange, Folding Numbers.

Crossrefs

Cf. A035928, A276795.

Programs

Maple

N:= 16: # to get all terms < 2^N
M[1]:= [[1]]: M[2]:= [[1,0]]:
for d from 3 to N by 2 do
  M[d]:= map(L -> [op(L[1..(d-1)/2]),1,op(L[(d+1)/2..-1])], M[d-1]);
  if d < N then
    M[d+1]:= map(L -> ([op(L[1..(d-1)/2]),0,1,op(L[(d+1)/2..-1])],[op(L[1..(d-1)/2]),1,0,op(L[(d+1)/2..-1])]), M[d-1])
  fi
od:
seq(seq(add(L[-i]*2^(i-1),i=1..d),L=M[d]),d=1..N); # Robert Israel, Nov 09 2016

Mathematica

{1}~Join~Select[Range@ 2000, If[OddQ@ Length@ # && Take[#, {Ceiling[ Length[#]/2]}] == {0}, False, Union[Take[#, Floor[Length[#]/2]] + Reverse@ Take[#, -Floor[Length[#]/2]]] == {1}] &@ IntegerDigits[#, 2] &] (* Michael De Vlieger, Oct 07 2016 *)

PARI

isok(n) = {if (n==1, return(1)); b = binary(n); if ((#b % 2) && (b[#b\2+1] == 0), return (0)); vecmin(vector(#b1, k, bitxor(b[k], b[#b-k+1]))) == 1;} \\ Michel Marcus, Oct 07 2016