A090049 Length of longest contiguous block of 0's in binary expansion of n^3.

1, 0, 3, 1, 6, 1, 3, 1, 9, 2, 3, 2, 6, 3, 3, 2, 12, 3, 3, 2, 6, 4, 3, 4, 9, 4, 3, 3, 6, 3, 3, 3, 15, 4, 3, 2, 6, 3, 3, 2, 9, 4, 4, 2, 6, 3, 4, 3, 12, 3, 4, 6, 6, 3, 3, 3, 9, 2, 3, 4, 6, 2, 3, 4, 18, 5, 4, 2, 6, 6, 3, 4, 9, 2, 3, 2, 6, 1, 3, 4, 12, 6, 4, 3, 6, 2, 3, 5, 9, 5, 3, 3, 6, 6, 3, 3, 15, 4, 3

Offset: 0

Author

Reinhard Zumkeller, Nov 20 2003

Keywords

• nonn
• base
• easy

Links

• Robert Israel, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A000578, A007814, A023416, A070939, A087117, A090003, A090047, A090048.

Programs

Maple

f:= proc(n) local R;
  R:= convert(2*n^3+1,base,2);
  R:= select(t -> R[t]=1, [$1..nops(R)]);
  max(R[2..-1]-R[1..-2])-1
end proc:
f(0):= 1:
map(f, [$0..100]); # Robert Israel, Mar 10 2020

Mathematica

a[n_] := Module[{d = IntegerDigits[n^3, 2]}, If[! MemberQ[d, 0], 0, Max[Length /@ Split[d][[2 ;; -1 ;; 2]]]]]; a[0] = 1; Array[a, 102, 0] (* Amiram Eldar, Jul 29 2025 *)

PARI

a(n)=my(r,k,t=n^3); for(i=0,exponent(t), if(bittest(t,i), k=0, k++>r, r=k)); if(n, r, 1) \\ Charles R Greathouse IV, Mar 10 2020

Formula

a(n) = A087117(A000578(n)).

a(n) <= 3*A070939(n) - 3 for n > 0. - Charles R Greathouse IV, Mar 10 2020

a(n) >= 3*A007814(n). Conjecture: if n < 2^k then a(n) < 3*k. - Robert Israel, Mar 10 2020