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Let F(n) denote the Fibonacci numbers, A000045: a(n) = Sum_{k=0..n} C(n,k)^2*(n-k)!*F(k).
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%I #13 Jan 05 2021 21:55:43

%S 0,1,5,29,203,1680,16058,173865,2099957,27952999,406125305,6389713034,

%T 108157272720,1958821525361,37779732341077,772829270394685,

%U 16708083353842267,380563529091632760,9106983116342966818,228393730451588322201,5989333028770423686565

%N Let F(n) denote the Fibonacci numbers, A000045: a(n) = Sum_{k=0..n} C(n,k)^2*(n-k)!*F(k).

%C If the e.g.f. of F(n) is E(x) and a(n) = Sum_{k=0..n} C(n,k)^2*(n-k)!*F(k), then the e.g.f. of a(n) is E(x/(1-x))/(1-x).

%F E.g.f.: (2/sqrt(5))*exp(x/2/(1-x))*sinh(sqrt(5)*(x/2)/(1-x))/(1-x).

%F a(n) = (4*n - 3)*a(n-1) - 2*(n-1)*(3*n - 5)*a(n-2) + (n-2)^2*(4*n - 7)*a(n-3) - (n-3)^2*(n-2)^2*a(n-4). - _Vaclav Kotesovec_, Nov 13 2017

%F a(n) ~ n^(n + 1/4) / (sqrt(10) * phi^(1/4) * exp(n - 2*sqrt(phi*n) + phi/2)), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - _Vaclav Kotesovec_, Nov 13 2017

%e F(n) = 0,1,1,2,3,5,8,13,21,34,55,...

%e a(3) = C(3,0)^2*3!*F(0) + C(3,1)^2*2!*F(1) + C(3,2)^2*1!*F(2) + C(3,3)^2*0!*F(3) = 1*6*0 + 9*2*1 + 9*1*1 + 1*1*2 = 0 + 18 + 9 + 2 = 29.

%p b[0]:=0:b[1]:=1:for n from 2 to 30 do b[n]:=b[n-1]+b[n-2] od:

%p seq(sum('binomial(n,k)^2*(n-k)!*b[k]', 'k'=0..n),n=0..30);

%t Table[Sum[Binomial[n,k]^2 * (n-k)! * Fibonacci[k], {k,0,n}], {n,0,20}] (* _Vaclav Kotesovec_, Nov 13 2017 *)

%Y Cf. A000045.

%K easy,nonn

%O 0,3

%A _Miklos Kristof_, Apr 25 2005