A350902 a(n) = (5*F(n)*F(n-1)*F(2*n-1)*a(n-1) + F(n-1)*L(n)*a(n-2))/(L(n-1)*F(n)), with a(0) = 1, a(1) = 0, where F(n) = A000045(n) and L(n) = A000032(n).

1, 0, 3, 25, 816, 59475, 12031005, 6229446000, 8517168411895, 30387269735449725, 284188952072106783648, 6954889250543118311091775, 445684855849546942072130113089, 74767094861864103592878982016253600, 32838249084789127737424410920015676309123

Offset: 0

Author

Amiram Eldar, Jan 21 2022

Keywords

• nonn
• easy

Comments

Although the recurrence relation involves fractions, all the terms are integers.

The sequence of fractions b(n) = A350903(n)/A350904(n) is defined by the same recurrence relation, but with the initial terms 0 and 1 instead of 1 and 0.

André-Jeannin (1991) used this sequence and the sequence b(n) to prove that s = Sum_{n>=1} 1/F(n) (A079586) is an irrational number.

The sequence of ratios r(n) = b(n)/a(n) rapidly converges to s. For example, abs(r(16)-s) < 10^(-100) and abs(r(49)-s) < 10^(-1000).

Links

• Amiram Eldar, Table of n, a(n) for n = 0..69
• Richard André-Jeannin, Irrationalité de la somme des inverses de certaines suites récurrentes, Comptes Rendus de l'Académie des Sciences - Series I - Mathematics 308:19 (1989), pp. 539-541.
• Richard André-Jeannin, Sequences of Integers Satisfying Recurrence Relations, The Fibonacci Quarterly, Vol. 29, No. 3 (1991), pp. 205-208.

Crossrefs

Cf. A000032, A000045, A079586, A350903, A350904.

Programs

Mathematica

With[{F = Fibonacci, L = LucasL}, a[0] = 1; a[1] = 0; a[n_] := a[n] = (5*F[n]*F[n - 1]*F[2*n - 1]*a[n - 1] + F[n - 1]*L[n]*a[n - 2])/(L[n - 1]*F[n]); Array[a, 15, 0]]

Formula

Limit_{n->oo} A350903(n)/(A350904(n)*a(n)) = A079586 (André-Jeannin, 1991).