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Powernacci numbers: a(n) = 2^(a(n-1) + a(n-2)) with a(0) = 0 and a(1) = 1.
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%I #44 Dec 12 2018 08:44:57

%S 0,1,2,8,1024

%N Powernacci numbers: a(n) = 2^(a(n-1) + a(n-2)) with a(0) = 0 and a(1) = 1.

%C a(5) = 4.602094... * 10^310, a(6) = 1.734184... * 10^(1.385368... * 10^310). - _Amiram Eldar_, Dec 12 2018

%H Francois Alcover, <a href="/A322142/b322142.txt">Table of n, a(n) for n = 0..5</a>

%e a(0) = 0.

%e a(1) = 1.

%e a(2) = 2^(0+1) = 2.

%e a(3) = 2^(1+2) = 8.

%t a[0] = 0; a[1] = 1; a[n_] := a[n] = 2^(a[n-2] + a[n-1]); Array[a, 6, 0] (* _Amiram Eldar_, Dec 12 2018 *)

%Y Cf. A000045.

%K nonn

%O 0,3

%A _Francois Alcover_, Dec 11 2018