A333208 - OEIS

%I #29 Apr 11 2020 15:43:47

%S 1,7,551,48091,2148268,87644575267,161577754532123,

%T 36595152483523582367,16965509829762630129638831,

%U 173107561150078104051618631740949,1108595900580419409151086339986148307,2608169750203411467722731179728125652086612772

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

%C The denominators are given in A333209.

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

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

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

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

%F Lim_{n -> inf} (a(n)/A333209(n)) / (a(n-1)/A333209(n-1)) = 1 - 1/phi = 1/phi^2 = A132318.

%F The following generalization holds: (Start)

%F 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

%F 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)

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

%t a[n_] := Numerator[Sum[1/(LucasL[2 i - 2]*LucasL[2 i - 1]), {i, 1, n}]/Fibonacci[2 n]]; Array[a, 12] (* _Amiram Eldar_, Mar 11 2020 *)

%Y Cf. A000032, A105394, A132318, A333088, A333089, A333209.

%K nonn

%O 1,2

%A _A.H.M. Smeets_, Mar 11 2020