A335901 a(n) = 2*a(floor((n-1)/a(n-1))) with a(1) = 1.

1, 2, 2, 2, 4, 2, 4, 2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 8, 4, 8, 4, 4, 4, 4, 4, 8, 4, 8, 4, 4, 4, 4, 4, 8, 4, 8, 4, 8, 8, 8, 8, 8, 8, 8, 8, 4, 8, 4, 8, 4, 8, 4, 8, 8, 8, 8, 8, 8, 8, 8, 8, 4, 8, 4, 8, 4, 8, 4, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8

Offset: 1

Author

Altug Alkan, following a suggestion from Andrew R. Booker, Jun 29 2020

Keywords

• nonn
• easy

Comments

Least k such that a(k) = 2^n are 1, 2, 5, 21, 169, 2705, ... (Conjecture: This sequence is A117261).

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A130147, A132424, A130535, A283207, A288914, A335898.

Programs

Maple

f:= proc(n) option remember;
  2*procname(floor((n-1)/procname(n-1))) end proc:
f(1):= 1:
map(f, [$1..105]); # Robert Israel, Jul 08 2020

Mathematica

a[1] = 1; a[n_] := a[n] = 2 * a[Floor[(n-1)/a[n-1]]]; Array[a, 100] (* Amiram Eldar, Jun 29 2020 *)

PARI

a=vector(10^3); a[1]=1; for(n=2, #a, a[n]=2*a[(n-1)\a[n-1]]); a