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

1, 2, 315, 2471040, 918185538816, 16047302734562299200, 13178031727820369629763174400, 508406658175888466343652105865846784000, 921456090985190879093613420564815806955580862464000, 78458394721620642094151397745899367347021362840662985785265356800

Offset

0,2

Comments

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.

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.

Links

Table of n, a(n) for n=0..9.

Wikipedia, Fibonacci polynomials

Formula

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.

a(n) ~ A194159 * phi^(4*n^2 + 2*n - 1) / 5^n, where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Aug 31 2023

Maple

with(combinat):
seq(mul(fibonacci(2*k), k = 1..2*n+1)/add(fibonacci(2*k), k = 1..2*n+1), n = 0..10);

Mathematica

Table[Product[ Fibonacci[2k], {k, 2n+1}]/Sum[Fibonacci[2k], {k, 2n+1}], {n, 0, 9}] (* Stefano Spezia, Nov 13 2021 *)

Prog

(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

Crossrefs

Cf. A000045, A159951, A175553.

Cf. A003266, A194157.

Sequence in context: A106718 A119779 A118579 * A221190 A367927 A192725

Adjacent sequences: A349269 A349270 A349271 * A349273 A349274 A349275

Keyword

nonn,easy

Author

Peter Bala, Nov 12 2021

Status

approved