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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).
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%I #13 Oct 16 2023 01:49:50

%S 1,0,3,25,816,59475,12031005,6229446000,8517168411895,

%T 30387269735449725,284188952072106783648,6954889250543118311091775,

%U 445684855849546942072130113089,74767094861864103592878982016253600,32838249084789127737424410920015676309123

%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)), with a(0) = 1, a(1) = 0, where F(n) = A000045(n) and L(n) = A000032(n).

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

%C 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.

%C 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.

%C 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).

%H Amiram Eldar, <a href="/A350902/b350902.txt">Table of n, a(n) for n = 0..69</a>

%H Richard André-Jeannin, <a href="http://gallica.bnf.fr/ark:/12148/bpt6k5686125p/f9.image">Irrationalité de la somme des inverses de certaines suites récurrentes</a>, Comptes Rendus de l'Académie des Sciences - Series I - Mathematics 308:19 (1989), pp. 539-541.

%H Richard André-Jeannin, <a href="https://www.fq.math.ca/Scanned/29-3/andre-jeannin2.pdf">Sequences of Integers Satisfying Recurrence Relations</a>, The Fibonacci Quarterly, Vol. 29, No. 3 (1991), pp. 205-208.

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

%t 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]]

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

%K nonn,easy

%O 0,3

%A _Amiram Eldar_, Jan 21 2022