A350901 a(n) = F(n) * (2*F(n-1)^2 + (-1)^(n-1)) * (2*F(n)^2 + (-1)^n), where F(n) is the n-th Fibonacci number.

0, 1, 3, 42, 399, 4655, 50568, 565149, 6248991, 69380842, 769112355, 8530996299, 94604226192, 1049202243593, 11635724020011, 129042610760010, 1431102560300007, 15871178746661911, 176014035001069464, 1952025706821035013, 21648296204009443815, 240083286518079466826

Offset: 0

Author

Amiram Eldar, Jan 21 2022

Keywords

• nonn
• easy

Comments

This sequence is notable for having sum of reciprocals that is smaller by 2 than the sum of the reciprocals of the Fibonacci numbers (see the Formula section).

The series with the reciprocals of the positive terms of this sequence is converging much more rapidly. For example, to calculate the sum with an error that is smaller than 10^(-100) there is a need to sum up the reciprocals of the first 482 Fibonacci numbers, while with the reciprocals of the terms of this sequence the first 97 terms are enough.

Links

• Amiram Eldar, Table of n, a(n) for n = 0..300
• Richard André-Jeannin, Problem H-450, Advanced Problems and Solutions, The Fibonacci Quarterly, Vol. 29, No. 1 (1991), p. 89; Comparable, Solution to Problem H-450 by Paul S. Bruckman, ibid., Vol. 30, No. 2 (1992), pp. 191-192.
• Index entries for linear recurrences with constant coefficients, signature (8,40,-60,-40,8,1).

Crossrefs

Cf. A000045, A061646, A079586.

Programs

Mathematica

a[n_] := Fibonacci[n]*(2*Fibonacci[n - 1]^2 + (-1)^(n - 1))*(2*Fibonacci[n]^2 + (-1)^n); Array[a, 25, 0]
(* or *)
LinearRecurrence[{8, 40, -60, -40, 8, 1}, {0, 1, 3, 42, 399, 4655}, 25]

Formula

a(n) = A000045(n) * A061646(n-1) * A061646(n).

a(n) = 8*a(n-1) + 40*a(n-2) - 60*a(n-3) - 40*a(n-4) + 8*a(n-5) + a(n-6), for n > 5.

Sum_{n>=1} 1/a(n) = -2 + A079586 (André-Jeannin, 1991).