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a(n) = 2*a(floor((n-1)/a(n-1))) with a(1) = 1.
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%I #16 Jul 08 2020 13:05:57

%S 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,

%T 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,

%U 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

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

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

%H Robert Israel, <a href="/A335901/b335901.txt">Table of n, a(n) for n = 1..10000</a>

%p f:= proc(n) option remember;

%p 2*procname(floor((n-1)/procname(n-1))) end proc:

%p f(1):= 1:

%p map(f, [$1..105]); # _Robert Israel_, Jul 08 2020

%t a[1] = 1; a[n_] := a[n] = 2 * a[Floor[(n-1)/a[n-1]]]; Array[a, 100] (* _Amiram Eldar_, Jun 29 2020 *)

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

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

%K nonn,easy

%O 1,2

%A _Altug Alkan_, following a suggestion from _Andrew R. Booker_, Jun 29 2020