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a(1) = a(2) = a(3) = 1; for n>3, a(n) = (a(n-3) + a(n-1))*(a(n-2) + a(n-3)).
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%I #22 Sep 08 2022 08:46:14

%S 1,1,1,4,10,55,826,54340,47921995,2643710343286,126835535679488180710,

%T 335322495784116748418182251685105,

%U 42530809264656915340847577048392358554130713446770436

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

%F a(n) = (a(n-3) + a(n-1))*(a(n-2) + a(n-3)) for n>=4; a(1)=a(2)=a(3)=1.

%t RecurrenceTable[{a[1] == a[2] == a[3] == 1, a[n] == (a[n - 3] + a[n - 1]) (a[n - 2] + a[n - 3])}, a, {n, 15}] (* _Bruno Berselli_, Oct 09 2015 *)

%o (PARI) a(n) = if(n<4, 1, (a(n-3)+a(n-1))*(a(n-2)+a(n-3)));

%o vector(15, n, a(n)) \\ _Altug Alkan_, Oct 08 2015

%o (Magma) [n le 3 select 1 else (Self(n-3)+ Self(n-1))*(Self(n-2)+Self(n-3)): n in [1..15]]; // _Vincenzo Librandi_, Oct 09 2015

%K nonn

%O 1,4

%A _Gianmarco Giordano_, Oct 08 2015

%E More terms from _Altug Alkan_, Oct 08 2015