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%I #25 May 20 2024 16:50:31
%S 1,6,24,105,440,1872,7917,33558,142120,602085,2550384,10803744,
%T 45765161,193864710,821223480,3478759473,14736260008,62423801712,
%U 264431463285,1120149660630,4745030096456,20100270061581,85146110318304,360684711374400,1527884955751825
%N a(n) = Fibonacci(n+1)*Fibonacci(2n).
%C Consider the sum of the 4*n Lucas numbers from index 1 through 4*n. It is divisible by the (n+1)st Lucas number and the ratio is 5*a(n).
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (3,6,-3,-1).
%F G.f.: x*(1 + 3*x)/((1 + x - x^2)*(1 - 4*x - x^2)). - _Stefano Spezia_, May 20 2024
%e a(3) = Fibonacci(4)*Fibonacci(6) = 3*8 = 24. The sum of the first 12 Lucas numbers is 840, which is the 4th Lucas number (7) times 5*a(3).
%t Table[Fibonacci[k + 1] Fibonacci[2 k], {k, 30}]
%Y Cf. A000045, A001906.
%K nonn,easy
%O 1,2
%A _Tanya Khovanova_ and the MIT PRIMES STEP senior group, May 18 2024