%I
%S 0,1,2,3,4,6,8,16,23,32,64,91,128,256,512,1024,2048,4096,5793,8192,
%T 16384,32768,46341,65536,92682,131072,185364,262144,370728,524288,
%U 1048576,2097152,2965821,4194304,5931642,8388608,16777216,33554432,47453133,67108864,94906266
%N Numbers n with property that binary expansion of n^3 begins with the binary expansion of n.
%C 2^k is always a term in this sequence.
%C It appears that all solutions are either a power of 2 or approximately sqrt(2) * a power of 2.  _Andrew Howroyd_, Dec 24 2019
%H Andrew Howroyd, <a href="/A261751/b261751.txt">Table of n, a(n) for n = 1..1000</a>
%e 23 is a term of this sequence because its cube written in base 2 (10111110000111) starts with its representation in base 2 (10111).
%t SetBeginSet[set1_, set2_] :=
%t Do[For[i = 1, i <= Length[set1], i++,If[! set1[[i]] == set2[[i]], Return[False]]];Return[True], {1}];
%t For[k = 0; set = {}, k <= 100000, k++,If[SetBeginSet[IntegerDigits[k, 2], IntegerDigits[k^3, 2]],Print[k]]]
%o (PARI) ok(n)={my(t=n^3); t == 0  t>>(logint(t,2)logint(n,2))==n} \\ _Andrew Howroyd_, Dec 23 2019
%o (PARI) \\ for larger values
%o viable(b,k)={my(p=b^3, q=(b+2^k1)^3, s=logint(q,2), t=slogint(b,2)+k); (p>>s)==0  ((p>>t)<=(b>>k) && (b>>k)<=(q>>t))}
%o upto(n)={
%o local(L=List([0]));
%o my(recurse(b,k)=; if(b <= n && viable(b,k), k; if(k<0, listput(L, b), self()(b,k); self()(b+2^k,k))));
%o for(k=0, logint(n,2), recurse(2^k, k));
%o Vec(L);
%o } \\ _Andrew Howroyd_, Dec 24 2019
%Y Base 2 version of A052210.
%Y Cf. A004539.
%K nonn,base,easy
%O 1,3
%A _Dhilan Lahoti_, Aug 30 2015
%E Terms a(31) and beyond from _Andrew Howroyd_, Dec 23 2019
