A261751 Numbers n with property that binary expansion of n^3 begins with the binary expansion of n.

0, 1, 2, 3, 4, 6, 8, 16, 23, 32, 64, 91, 128, 256, 512, 1024, 2048, 4096, 5793, 8192, 16384, 32768, 46341, 65536, 92682, 131072, 185364, 262144, 370728, 524288, 1048576, 2097152, 2965821, 4194304, 5931642, 8388608, 16777216, 33554432, 47453133, 67108864, 94906266

Offset: 1

Author

Dhilan Lahoti, Aug 30 2015

Keywords

• nonn
• base
• easy

Comments

2^k is always a term in this sequence.

It appears that all solutions are either a power of 2 or approximately sqrt(2) * a power of 2. - Andrew Howroyd, Dec 24 2019

Examples

			23 is a term of this sequence because its cube written in base 2 (10111110000111) starts with its representation in base 2 (10111).
		

Links

• Andrew Howroyd, Table of n, a(n) for n = 1..1000

Crossrefs

Base 2 version of A052210.

Cf. A004539.

Programs

Mathematica

SetBeginSet[set1_, set2_] :=
  Do[For[i = 1, i <= Length[set1], i++,If[! set1[[i]] == set2[[i]], Return[False]]];Return[True], {1}];
For[k = 0; set = {}, k <= 100000, k++,If[SetBeginSet[IntegerDigits[k, 2], IntegerDigits[k^3, 2]],Print[k]]]

PARI

ok(n)={my(t=n^3); t == 0 || t>>(logint(t,2)-logint(n,2))==n} \\ Andrew Howroyd, Dec 23 2019

PARI

\\ for larger values
viable(b,k)={my(p=b^3, q=(b+2^k-1)^3, s=logint(q,2), t=s-logint(b,2)+k); (p>>s)==0 || ((p>>t)<=(b>>k) && (b>>k)<=(q>>t))}
upto(n)={
  local(L=List([0]));
  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))));
  for(k=0, logint(n,2), recurse(2^k, k));
  Vec(L);
} \\ Andrew Howroyd, Dec 24 2019

Extensions

Terms a(31) and beyond from Andrew Howroyd, Dec 23 2019