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%I #12 Jun 27 2012 17:58:01
%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>
%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)
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