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a(n) = Fibonacci(binomial(n+3, 3)).
0

%I #13 Sep 08 2022 08:46:20

%S 1,3,55,6765,9227465,225851433717,160500643816367088,

%T 5358359254990966640871840,13598018856492162040239554477268290,

%U 4244200115309993198876969489421897548446236915,263621064469290555679241849789653324393054271110084140201023

%N a(n) = Fibonacci(binomial(n+3, 3)).

%C Fourth diagonal of A045995.

%F a(n) = A000045(A000292(n+1)) = A045995(n+3, 3).

%t Table[Fibonacci[Binomial[n + 3, 3]], {n, 0, 20}]

%o (Magma) [Fibonacci(Binomial(n+3,3)): n in [0..20]];

%o (PARI) a(n) = fibonacci(binomial(n+3, 3)) \\ _Felix Fröhlich_, Jan 12 2018

%Y Cf. A000045, A000292, A045995, A081667.

%K nonn,easy

%O 0,2

%A _Vincenzo Librandi_, Jan 10 2018