%I #17 Sep 27 2024 09:27:09

%S 1,6,40,896,294912,23622320128,119903836479112085504,

%T 2552117751907038475975309555738261585920,

%U 984232758517187661100353372573847216752794869657944794335389464067261601939456

%N Solution to f(2) = 1, f(n) = sqrt(n) f(sqrt(n)) + n at values n = 2^2^i.

%H Alois P. Heinz, <a href="/A001367/b001367.txt">Table of n, a(n) for n = 0..11</a>

%F a(n) = 2^(2^n-1)*(2*n+1). - _Christian Krause_, Sep 27 2024

%p f:= proc(n) f(n):= `if`(n=2, 1, sqrt(n) *f(sqrt(n)) +n) end:

%p a:= n-> f(2^(2^n)):

%p seq(a(n), n=0..10); # _Alois P. Heinz_, Jun 27 2012

%t f[ 2 ] := 1; f[ n_ ] := Sqrt[ n ]*f[ Sqrt[ n ] ] + n; Table[ f[ 2^2^i ], {i, 0, 7} ]

%K nonn

%O 0,2

%A Alexander Sorg (sorg(AT)bu.edu)