A333209 a(n) is the denominator of Sum_{i >= 0} 1/(Lucas(i)*Lucas(i+2n)), with Lucas(i) as defined in A000032.

2, 36, 7392, 1688148, 197412831, 21085413226416, 101768454084335346, 60343478516053297339236, 73240105330540144095414793632, 1956470757376233684880813258936380492, 32802418997525523144166495047229414174839, 202042966989952174292936124782341088713724476716231

Offset

1,1

Comments

The numerators are given in A333208.

See A333088 and A333089 for similar fractions for infinite sums of Fibonacci numbers.

Sum_{i >= 0} 1/(Lucas(i)*Lucas(i+2n)) is a fraction for n > 0.

Sum_{i >= 0} 1/Lucas(i)^2 = 1/4 + A105394, i.e., the n = 0 case, is believed to be transcendental.

Links

A.H.M. Smeets, Table of n, a(n) for n = 1..54

Formula

a(n) = denominator of (1/Fibonacci(2n)) * Sum_{i = 1..n} 1/(Lucas(2i-2)*Lucas(2i-1)).

Lim_{n -> inf} (A333208(n)/a(n)) / (A333208(n-1)/a(n-1)) = 1 - 1/phi = 1/phi^2 = A132318.

The following generalization holds: (Start)

Let H_(a,b) (n) be defined by H_(a,b) (0) = a, H_(a,b) (1) = b and H_(a,b) (n) = H_(a,b) (n-1) + H_(a,b) (n-2) for n > 1, then

Sum_{i >= 0} 1/(H_(a,b) (i)*H_(a,b) (i+2n)) = (1/Fibonacci(2n)) * Sum_{i=1..n} 1/(H_(a,b) (2i-2)*H_(a,b) (2i-1)) for n > 0, and are thus all fractions. Specially, H_(0,1) are the Fibonacci numbers A000045, H_(2,1) as here, are the Lucas numbers A000032, and H_(3,1) are the Pibonacci numbers A104449. (End)

Example

These infinite sums begin: 1/2, 7/36, 551/7392, ...

Mathematica

a[n_] := Denominator[Sum[1/(LucasL[2 i - 2]*LucasL[2 i - 1]), {i, 1, n}]/Fibonacci[2 n]]; Array[a, 12] (* Amiram Eldar, Mar 11 2020 *)

Prog

(Python)
from math import gcd
f0, f1, g0, g1, snum, sden, n = 1, 1, 1, 2, 0, 1, 0
while n < 12:
    n = n+1
    snum, sden = g0*g1*snum+sden, sden*g0*g1
    d = gcd(snum, sden*f0)
    print(n, sden*f0//d)
f0, f1, g0, g1 = 2*f0+f1, f0+f1, 2*g0+g1, g0+g1 # A.H.M. Smeets, Nov 30 2020
(Python)
from math import gcd
f0, f1, g0, g1, snum, sden, n = 1, 1, 1, 2, 0, 1, 0
while n < 12:
    n = n+1
    snum, sden = g0*g1*snum+sden, sden*g0*g1
    d = gcd(snum, sden*f0)
    print(n, sden*f0//d)
    f0, f1, g0, g1 = 2*f0+f1, f0+f1, 2*g0+g1, g0+g1 # A.H.M. Smeets, Nov 30 2020

Crossrefs

Cf. A000032, A105394, A132318, A333088, A333089, A333208.

Sequence in context: A004003 A369676 A060739 * A224733 A264953 A308942

Adjacent sequences: A333206 A333207 A333208 * A333210 A333211 A333212

Keyword

nonn,frac

Author

A.H.M. Smeets, Mar 11 2020

Status

approved