%I
%S 1,2,6,10,12,22,28,38,42,52,56,78,90,108,120,142,150,170,178,204,212,
%T 232,240,286,310,346,370,412,436,472,496,542,558,598,614,666,682,722,
%U 738,796,812,852,868,920,936,976,992,1086,1134,1206,1254,1338,1386,1458,1506,1596,1644,1716,1764,1848,1896,1968
%N Folding numbers (see comments for the definition).
%C 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 all1'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 (n1)/2 bits and the reverse of the right (n1)/2 bits is the all1's string.
%C Folding numbers with an even (resp. odd) number of bits form A035928 (resp. A276795).  _N. J. A. Sloane_, Nov 03 2016
%H Lars Blomberg, <a href="/A277238/b277238.txt">Table of n, a(n) for n = 1..10000</a>
%H Stack Exchange, <a href="http://codegolf.stackexchange.com/questions/95458/foldingnumbers">Programming Puzzles and Code Golf: Folding Numbers</a>
%e 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.
%p N:= 16: # to get all terms < 2^N
%p M[1]:= [[1]]: M[2]:= [[1,0]]:
%p for d from 3 to N by 2 do
%p M[d]:= map(L > [op(L[1..(d1)/2]),1,op(L[(d+1)/2..1])], M[d1]);
%p if d < N then
%p M[d+1]:= map(L > ([op(L[1..(d1)/2]),0,1,op(L[(d+1)/2..1])],[op(L[1..(d1)/2]),1,0,op(L[(d+1)/2..1])]), M[d1])
%p fi
%p od:
%p seq(seq(add(L[i]*2^(i1),i=1..d),L=M[d]),d=1..N); # _Robert Israel_, Nov 09 2016
%t {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 *)
%o (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[#bk+1]))) == 1;} \\ _Michel Marcus_, Oct 07 2016
%Y Cf. A035928, A276795.
%K nonn,base
%O 1,2
%A _Taylor J. Smith_, Oct 06 2016
