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%I #19 Sep 08 2022 08:46:11
%S 1,13,130,1106,8575,62475,435576,2939208,19342285,124800361,792586270,
%T 4969028750,30822650251,189500937303,1156406300340,7012380492516,
%U 42294614785465,253926386816725,1518506730836026,9050029200532298,53778595325886295,318762380704793571,1885254096749834160
%N 2nd diagonal of triangle in A255494.
%H G. C. Greubel, <a href="/A255495/b255495.txt">Table of n, a(n) for n = 0..1000</a>
%H S. Falcon, <a href="http://saspublisher.com/wp-content/uploads/2014/06/SJET24C669-675.pdf">On The Generating Functions of the Powers of the K-Fibonacci Numbers</a>, Scholars Journal of Engineering and Technology (SJET), 2014; 2 (4C):669-675.
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (14,-56,14,189,84,-20).
%F G.f.: (1 -x +4*x^2)/((1+x)*(1-5*x)*(1-6*x+x^2)*(1-4*x-4*x^2)). - _R. J. Mathar_, Jun 14 2015
%F From _G. C. Greubel_, Sep 20 2021: (Start)
%F a(n) = (1/2)*(P(n+3)*P(n+4) + 2^(n+4)*P(n+4) - 2*5^(n+3)), where P(n) = A000129(n).
%F a(n) = 5*a(n-1) + P(n+1)*(P(n+3) - 2^(n+2)) = 5*a(n) + P(n+1)*A094706(n+1). (End)
%t a[n_]:= (1/2)*(Fibonacci[n+3, 2]*Fibonacci[n+4, 2] + 2^(n+4)*Fibonacci[n+4, 2] - 2*5^(n+3));
%t Table[a[n], {n,0,30}] (* _G. C. Greubel_, Sep 20 2021 *)
%o (Magma) I:=[1,13,130,1106,8575,62475]; [n le 6 select I[n] else 14*Self(n-1) - 56*Self(n-2) +14*Self(n-3) +189*Self(n-4) + 84*Self(n-5) -20*Self(n-6): n in [1..31]]; // _G. C. Greubel_, Sep 20 2021
%o (Sage)
%o def P(n): return lucas_number1(n, 2, -1)
%o def A255495(n): return (1/2)*(P(n+3)*P(n+4) + 2^(n+4)*P(n+4) - 2*5^(n+3))
%o [A255495(n) for n in (0..30)] # _G. C. Greubel_, Sep 20 2021
%Y Cf. A000129, A094706, A255494.
%K nonn
%O 0,2
%A _N. J. A. Sloane_, Mar 06 2015
%E Terms a(13) onward from _G. C. Greubel_, Sep 20 2021
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