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a(n) = Sum_{k=0..n} k^3 * binomial(n-k, k).
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%I #20 Mar 21 2023 09:22:21

%S 0,0,1,2,11,28,80,194,461,1036,2263,4800,9960,20268,40577,80086,

%T 156115,301004,574744,1087918,2043229,3810320,7060079,13004832,

%U 23826480,43437240,78827425,142446698,256400411,459826588,821834624,1464149690,2600662157,4606368148

%N a(n) = Sum_{k=0..n} k^3 * binomial(n-k, k).

%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (4,-2,-8,5,8,-2,-4,-1).

%F a(n) = (n*(3*n^2 + 3*n - 2)*A000032(n) - (5*n^3 + 3*n^2 - 2*n - 2)*A000045(n))/50.

%F a(n) ~ (phi^(n-2)*n^3)/25, phi = (1+sqrt(5))/2.

%F G.f.: x^2*(1 + x*(x-2)*(x-1)^2)/(x^2 + x - 1)^4.

%F D-finite with recurrence: (n^5 + 3*n^4 - 5*n^3 - 5*n^2 - n + 2)*a(n) + (n+1)^2*(n^3 + 4*n^2 + n - 1)*a(n-1) = n*(n^4 - 5*n^2 + 5*n - 1)*a(n+1).

%F E.g.f.: exp(x/2)*(5*x*(x^2 + 9*x - 1)*cosh(sqrt(5)*x/2) + sqrt(5)*(5*x^3 + 3*x^2 + 7*x + 2)*sinh(sqrt(5)*x/2))/125. - _Stefano Spezia_, Mar 20 2023

%t Table[(n (3 n^2 + 3 n - 2) LucasL[n] - (5 n^3 + 3 n^2 - 2 n - 2) Fibonacci[n])/50, {n, 0, 30}] (* or *)

%t Table[Sum[k^3 Binomial[n - k, k], {k, 0, n}], {n, 0, 30}]

%o (PARI) a(n) = sum(k=0, n, k^3*binomial(n-k, k)); \\ _Michel Marcus_, Oct 11 2016

%o (PARI) a(n) = my(f1=fibonacci(n-1),f2=fibonacci(n)); (n*(3*n^2 + 3*n - 2)*(2*f1+f2) - (5*n^3 + 3*n^2 - 2*n - 2)*f2)/50 \\ _Charles R Greathouse IV_, Oct 13 2016

%Y Cf. A000032, A000045, A224227.

%K nonn,easy

%O 0,4

%A _Vladimir Reshetnikov_, Oct 10 2016