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%I #25 Jul 02 2023 09:08:49
%S 4,26,81,184,350,594,931,1376,1944,2650,3509,4536,5746,7154,8775,
%T 10624,12716,15066,17689,20600,23814,27346,31211,35424,40000,44954,
%U 50301,56056,62234,68850,75919,83456,91476,99994,109025,118584,128686,139346,150579
%N Scaled convolution of (n^3)*A000984(n) with A000984(n).
%C S(3,n) := Sum_{j=0..n} j^3*binomial(2*j,j)*binomial(2*(n-j),n-j).
%C a(n) = 2^3*S(3,n)/4^n, n >= 1.
%C O.g.f. for S(3,n) is G(k=3,x). See triangle A142963 for the general G(k,x) formula.
%C The author was led to compute such sums by a question asked by M. Greiter, Jun 27 2008.
%H Vincenzo Librandi, <a href="/A142962/b142962.txt">Table of n, a(n) for n = 1..1000</a>
%F a(n) = n^2*(3+5*n)/2.
%F a(n) = (2^3)*S(3,n)/4^n with the convolution S(3,n) defined above.
%F G.f.: x*(4+10*x+x^2)/(1-x)^4. - _Joerg Arndt_, Jul 02 2023
%t Rest@ CoefficientList[Series[x (4 + 10 x + x^2)/(1 - x)^4, {x, 0, 39}], x] (* _Michael De Vlieger_, Jul 02 2023 *)
%Y Cf. A142961 triangle: row k=3: [3, 5], with the row polynomial 3+5*n.
%Y Cf. A049451 (scaled k=2 case).
%K nonn,easy
%O 1,1
%A _Wolfdieter Lang_, Sep 15 2008
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