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
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
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Nov 12 2021
STATUS
approved