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

0,3

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).

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).

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]

Crossrefs

Cf. A000045, A061646, A079586.

Sequence in context: A114943 A119577 A051273 * A160873 A084512 A084522

Adjacent sequences: A350898 A350899 A350900 * A350902 A350903 A350904

Keyword

nonn,easy

Author

Amiram Eldar, Jan 21 2022

Status

approved