%I
%S 1111,110110,111011,1011110,1101111,10011001,10100110,11001010,
%T 11010011,110011111,111010011,111011101,1001101111,11001101111,
%U 11010010100,100101110100,101000111011,101011001010,101111110111,110011001100,110111001101,111011110011
%N Numbers n such that n occurs within its base 2 representation regarded as a fixed necklace, but n is not a substring of the base 2 representation regarded as a string.
%C No power of 10 can occur in this sequence.
%H Eric W. Weisstein, <a href="http://mathworld.wolfram.com/Necklace.html">MathWorld: Necklace</a>
%e 111011 (in base 10) = 11011000110100011 (in base 2). Regarding this base 2 representation as a fixed necklace, we can list characters in the order 11110110001101000 by starting with the characters "11" at the end of the base 2 representation. In this listing 111011 occurs (1{111011}0001101000). 111011 however does not occur in the original base 2 representation 11011000110100011. Thus 111011 is in the sequence.
%o (PARI) {inseq(w)=local(bw,mm,texp,btod,bigb,lbb,swsq,ii, hwf);
%o bw=binary(w); mm=length(bw); texp=0; btod=0;
%o forstep(i=mm, 1, 1, btod=btod+bw[i]*10^texp; texp++);
%o bigb=binary(btod); lbb=length(bigb);
%o for(k=0, lbb  1 , swsq=1;
%o for(j=1, mm, ii=(j+k)%lbb; if(ii==0, ii=lbb);
%o if(bw[j]!=bigb[ii], swsq=1)); if(swsq==1, hwf=k; break));
%o if(swsq==1,if(hwf>lbbmm, swsq=btod, swsq=1)); return(swsq)}
%o {ptd=0;for(w=0, 10^9, jj=inseq(w); if(jj>=0, ptd++; print1(jj,", "); if(ptd>23,break)))}
%Y The union of A038102 and this sequence is A225237.
%K nonn,base
%O 1,1
%A _Douglas Latimer_, May 04 2013
