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a(n) = 2^(a(n-2) + log_{2}(a(n-1))), a(1)=1, a(2)=2.
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%I #17 Jun 28 2021 13:02:01

%S 1,2,4,16,256,16777216,

%T 1942668892225729070919461906823518906642406839052139521251812409738904285205208498176

%N a(n) = 2^(a(n-2) + log_{2}(a(n-1))), a(1)=1, a(2)=2.

%C The subsequent terms are too large to include.

%C Apparently a(n) = A000643(n+2) - A000643(n+1). - _R. J. Mathar_, Apr 22 2007

%F a(n) = a(n-1) * 2^a(n-2) for n >= 3. - _Alois P. Heinz_, Jun 28 2021

%t RecurrenceTable[{a[1]==1,a[2]==2,a[n]==2^(a[n-2]+Log[2,a[n-1]])},a,{n,6}] (* _Harvey P. Dale_, Aug 23 2013 *)

%Y Cf. A112237, A112866.

%K nonn

%O 1,2

%A _Yasutoshi Kohmoto_, Feb 18 2006

%E Definition adapted to offset by _Georg Fischer_, Jun 18 2021