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a(n) = a(n-1)*a(n-3) + a(n-2), with a(0)=a(1)=0 and a(2)=1.
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%I #43 Sep 29 2021 15:41:23

%S 0,0,1,0,1,1,1,2,3,5,13,44,233,3073,135445,31561758,96989417779,

%T 13136731722638413,414618347540933702027833,

%U 40213592128486236142855326045681320,528275171395527518169753769210241662354568290572993

%N a(n) = a(n-1)*a(n-3) + a(n-2), with a(0)=a(1)=0 and a(2)=1.

%F a(n) ~ c^(A092526^n), where c = A344388 = 1.0574735961... (very close to A201506). - _Vaclav Kotesovec_, Aug 16 2021

%p A048634 := proc(n) option remember; if n<=1 then 0 elif n=2 then 1 else A048634(n-1)*A048634(n-3)+A048634(n-2); fi; end;

%t RecurrenceTable[{a[n] == a[n-1]*a[n-3] + a[n-2], a[0] == 0, a[1] == 0, a[2] == 1}, a, {n, 0, 20}] (* _Vaclav Kotesovec_, Aug 16 2021 *)

%K nonn,easy

%O 0,8

%A _N. J. A. Sloane_, David(AT)interface.co.uk

%E Name clarified by _Michel Marcus_, Aug 16 2021