|
%I #25 Sep 08 2022 08:46:03
%S 0,2,276,34561,4261992,524393210,64499742738,7933009283134,
%T 975696814205904,120002796170968643,14759368609635548580,
%U 1815282342961539780022,223264968937188026209956,27459775899111901985784506
%N a(n) = F(12*n)/(24*L(2*n)), n >= 0, with F = A000045 (Fibonacci) and L = A000032 (Lucas).
%C 24*a(n) is the third example for the Riordan transition matrix introduced in a comment on A078812 (with offset [0,0]). Take there l -> n, n -> 2. See the second formula below.
%H Vincenzo Librandi, <a href="/A215043/b215043.txt">Table of n, a(n) for n = 0..100</a>
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (144,-2640,6930,-2640,144,-1).
%F a(n) = F(12*n)/(24*L(2*n)), n >= 0, with F = A000045 (Fibonacci) and L = A000032 (Lucas).
%F a(n) = 3*F(2*n) + 20*F(2*n)^3 + 25*F(2*n)^5, n >= 0 (see the comment above).
%F O.g.f.: x*(2 - 12*x + 97*x^2 - 12*x^3 + 2*x^4)/((1 - 3*x + x^2)*(1 - 18*x + x^2)*(1 - 123*x + x^2)). From the o.g.f.s for the sequences appearing in the preceding formula, see A001906, A215039 and A215044.
%F a(n) = (L(8*n) + 1)*F(2*n)/24. - _Ehren Metcalfe_, Jun 04 2019
%t Table[Fibonacci[12*n]/(24*LucasL[2*n]), {n,0,15}] (* _G. C. Greubel_, Jun 30 2019 *)
%o (Magma) [Fibonacci(12*n)/(24*Lucas(2*n)): n in [0..15]] // _Vincenzo Librandi_, Sep 02 2012
%o (PARI) lucas(n) = fibonacci(n+1) + fibonacci(n-1);
%o vector(15, n, n--; fibonacci(12*n)/(24*lucas(2*n))) \\ _G. C. Greubel_, Jun 30 2019
%o (Sage) [fibonacci(12*n)/(24*lucas_number2(2*n,1,-1)) for n in (0..15)] # _G. C. Greubel_, Jun 30 2019
%Y Cf. A215042 (for F(8*n)/L(2*n)).
%K nonn,easy
%O 0,2
%A _Wolfdieter Lang_, Aug 31 2012
| 