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a(n) = Product_{k=1..n} (Fibonacci(k+2) - 1).
1

%I #12 Apr 25 2018 11:56:58

%S 1,1,2,8,56,672,13440,443520,23950080,2107607040,301387806720,

%T 69921971159040,26290661155799040,16011012643881615360,

%U 15786858466867272744960,25195826113120167300956160,65080818850189392138369761280,272037822793791659138385602150400

%N a(n) = Product_{k=1..n} (Fibonacci(k+2) - 1).

%C a(n) = determinant of (n + 1) X (n + 1) matrix whose main diagonal consists of the consecutive Fibonacci numbers starting with Fibonacci(2) (1, 2, 3, 5, 8, 13, ...) and all other elements are 1's (see example).

%H Alois P. Heinz, <a href="/A302999/b302999.txt">Table of n, a(n) for n = 0..97</a>

%F a(n) = Product_{k=1..n} A000071(k+2).

%F a(n) = Product_{k=1..n} Sum_{j=1..k} A000045(j).

%F a(n) ~ c * ((1 + sqrt(5))/2)^(n*(n+5)/2) / 5^(n/2), where c = 0.1972502311584232476952451740107000852343536766534965116633336539193... - _Vaclav Kotesovec_, Apr 17 2018

%F a(n) = A190535(n-3) for n > 3. - _Alois P. Heinz_, Apr 25 2018

%e The matrix begins:

%e 1 1 1 1 1 1 1 1 ...

%e 1 2 1 1 1 1 1 1 ...

%e 1 1 3 1 1 1 1 1 ...

%e 1 1 1 5 1 1 1 1 ...

%e 1 1 1 1 8 1 1 1 ...

%e 1 1 1 1 1 13 1 1 ...

%e 1 1 1 1 1 1 21 1 ...

%e 1 1 1 1 1 1 1 34 ...

%p b:= proc(n) b(n):= `if`(n<1, [1$2][], (f->

%p [f, b(n-1)[2]*(f-1)][])(b(n-1)+b(n-2)))

%p end:

%p a:= n-> b(n)[2]:

%p seq(a(n), n=0..20); # _Alois P. Heinz_, Apr 24 2018

%t Table[Product[Fibonacci[k + 2] - 1, {k, 1, n}], {n, 0, 17}]

%t Table[Product[Sum[Fibonacci[j], {j, 1, k}], {k, 1, n}], {n, 0, 17}]

%t Table[Det[Table[If[i == j, Fibonacci[i + 1], 1], {i, 1, n + 1}, {j, 1, n + 1}]], {n, 0, 17}]

%Y Cf. A000045, A000071, A003266, A067962, A190535, A194157, A194158, A196870.

%K nonn

%O 0,3

%A _Ilya Gutkovskiy_, Apr 17 2018