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a(0) = a(1) = 1, a(2) = 2; a(n) = 2*a(n-2) + a(n-1)*a(n-3).
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%I #10 Oct 01 2015 17:45:31

%S 1,1,2,4,8,24,112,944,22880,2564448,2420884672,55389846424256,

%T 142044380887832220032,343873064435082883562892998016,

%U 19047076228497528742755382412205052966716160

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

%C This is a recurrence relation which has 1,1 and 2 as the base cases and the n-th term is obtained by multiplying the (n-2)th term by 2 and adding it with the product of (n-1)th and (n-3)rd term.

%F a(n) ~ c^(d^n), where d = 1.465571231876768026... is the root of the equation d^3 = d^2 + 1 and c = 1.604048928929157460568... . - _Vaclav Kotesovec_, Oct 01 2015

%p f:=proc(n) option remember;

%p if n <= 1 then 1 elif n=2 then 2 else

%p f(n-1)*f(n-3)+2*f(n-2); fi; end;

%p [seq(f(n),n=0..15)]; # _N. J. A. Sloane_, Oct 01 2015

%t RecurrenceTable[{a[0]==a[1]==1,a[2]==2,a[n]==2a[n-2]+a[n-1]a[n-3]},a,{n,20}] (* _Harvey P. Dale_, Oct 01 2015 *)

%K easy,nonn

%O 0,3

%A Mohit Maheshwari (mohitmahe1989(AT)gmail.com), Jan 19 2008

%E Corrected by _Harvey P. Dale_, Oct 01 2015