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%I #30 Oct 01 2024 15:36:00
%S 0,1,243,32768,4084101,503284375,61917364224,7615646045657,
%T 936668172433707,115202670521319424,14168993617568728125,
%U 1742671044798615789551,214334370099947863277568,26361384861716322814590193
%N a(n) = F(2*n)^5 with F=A000045 (Fibonacci numbers).
%H Vincenzo Librandi, <a href="/A215044/b215044.txt">Table of n, a(n) for n = 0..200</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 O.g.f.: x*(1 + 99*x + 416*x^2 + 99*x^3 + x^4)/((1-3*x+x^2)*(1-18*x+x^2)*(1-123*x+x^2)), (from the even part of the bisection of A056572).
%F a(n) = (5*F(4*n) - 4*F(8*n) + F(12*n))/(5^2*L(2*n)), with L=A000032 (Lucas). See the third row in the signed triangle A039598, called in a general comment S.
%F a(n) = (10*F(2*n) - 5*F(6*n) + F(10*n))/5^2, from the partial fraction decomposition of the o.g.f. - _Wolfdieter Lang_, Oct 11 2012
%t Table[Fibonacci[2*n]^5, {n,0,15}] (* _Vincenzo Librandi_, Sep 02 2012 *)
%o (Magma) [Fibonacci(2*n)^5: n in [0..15]]; // _Vincenzo Librandi_, Sep 02 2012
%o (PARI) a(n)=fibonacci(2*n)^5 \\ _Charles R Greathouse IV_, Oct 16 2015
%Y Cf. A000045, A056572, A215045 (odd part).
%K nonn,easy
%O 0,3
%A _Wolfdieter Lang_, Aug 31 2012