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A349272 a(n) = Product_{k = 1..2*n+1} Fibonacci(2*k) / Sum_{k = 1..2*n+1} Fibonacci(2*k).

%I A349272 #13 Aug 31 2023 06:38:21
%S A349272 1,2,315,2471040,918185538816,16047302734562299200,
%T A349272 13178031727820369629763174400,
%U A349272 508406658175888466343652105865846784000,921456090985190879093613420564815806955580862464000,78458394721620642094151397745899367347021362840662985785265356800
%N A349272 a(n) = Product_{k = 1..2*n+1} Fibonacci(2*k) / Sum_{k = 1..2*n+1} Fibonacci(2*k).
%C A349272 Let m be an even positive integer. We conjecture that the sequence defined by b_m(n) = Product_{k = 1..2*n+1} Fibonacci(m*k) / Sum_{k = 1..2*n+1} Fibonacci(m*k) is integral. The formula given below proves the conjecture in the present case m = 2. The cases m = 4 and m = 6 of the conjecture can be proved in a similar manner.
%C A349272 More generally, if F(n,x) denotes the n-th Fibonacci polynomial we conjecture that, for each n, the rational function Product_{k = 1..2*n+1} F(m*k,x) / Sum_{k = 1..2*n+1} F(m*k,x) is an integral polynomial.
%H A349272 Wikipedia, <a href="https://en.wikipedia.org/wiki/Fibonacci_polynomials">Fibonacci polynomials</a>
%F A349272 a(n) = F(2*n+1)/F(2*n+2) * Product_{k = 1..2*n} Fibonacci(2*k), shows a(n) to be integral. Cf. A159951.
%F A349272 a(n) ~ A194159 * phi^(4*n^2 + 2*n - 1) / 5^n, where phi = A001622 is the golden ratio. - _Vaclav Kotesovec_, Aug 31 2023
%p A349272 with(combinat):
%p A349272 seq(mul(fibonacci(2*k), k = 1..2*n+1)/add(fibonacci(2*k), k = 1..2*n+1), n = 0..10);
%t A349272 Table[Product[ Fibonacci[2k],{k,2n+1}]/Sum[Fibonacci[2k],{k,2n+1}],{n,0,9}] (* _Stefano Spezia_, Nov 13 2021 *)
%o A349272 (PARI) a(n) = prod(k = 1, 2*n+1, fibonacci(2*k)) / sum(k = 1, 2*n+1, fibonacci(2*k)); \\ _Michel Marcus_, Nov 12 2021
%Y A349272 Cf. A000045, A159951, A175553.
%Y A349272 Cf. A003266, A194157.
%K A349272 nonn,easy
%O A349272 0,2
%A A349272 _Peter Bala_, Nov 12 2021