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a(n) = (1/3)*Product_{k=0..n} (F(k)+3), where F=A000045 (Fibonacci numbers).
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%I #15 Aug 04 2024 20:46:41

%S 1,4,16,80,480,3840,42240,675840,16220160,600145920,34808463360,

%T 3202378629120,470749658480640,111096919401431040,

%U 42216829372543795200,25878916405369346457600,25620127241315652993024000,40992203586105044788838400000

%N a(n) = (1/3)*Product_{k=0..n} (F(k)+3), where F=A000045 (Fibonacci numbers).

%C Trivially, a(n+1)/a(n) is an integer for n>=0, so (a(n)) is a divisibility sequence.

%t q[n_] := Fibonacci[n]

%t p[n_] := Product[q[k] + 3, {k, 0, n}]

%t Table[(1/3)*Simplify[p[n]], {n, 0, 20}]

%o (PARI) a(n) = prod(k=0, n, fibonacci(k)+3)/3; \\ _Michel Marcus_, Aug 04 2024

%Y Cf. A000045, A082480.

%K nonn

%O 0,2

%A _Clark Kimberling_, Aug 03 2024