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%I #28 Jun 04 2023 15:15:58
%S 0,0,3,22,88,258,623,1316,2520,4476,7491,11946,18304,27118,39039,
%T 54824,75344,101592,134691,175902,226632,288442,363055,452364,558440,
%U 683540,830115,1000818,1198512,1426278,1687423,1985488,2324256,2707760,3140291,3626406
%N Convolution of the even-indexed triangular numbers (A014105) and the squares (A000290).
%C This sequence was originally proposed in a comment on A071245 by _J. M. Bergot_ as giving the first differences. Therefore, a(n) gives the partial sums of A071245.
%H Colin Barker, <a href="/A277228/b277228.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (6,-15,20,-15,6,-1).
%F O.g.f.: x^2*(1 + x)*(3 + x)/(1 - x)^6 = (x*(3 + x)/(1 - x)^3)*(x*(1 + x)/(1 - x)^3).
%F a(n) = Sum_{k=0..n} A014105(n-k)*A000290(k).
%F a(n) = binomial(n+1, 3)*(4*n^2 + 5*n + 4)/10 = (n - 1)*n*(n + 1)*(4*n^2 + 5*n + 4)/60.
%F a(n) = Sum_{k=0..n} A071245(k).
%F a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6) for n>5. - _Colin Barker_, Oct 21 2016
%t Table[(n - 1) n (n + 1) (4 n^2 + 5 n + 4)/60, {n, 0, 40}] (* _Bruno Berselli_, Oct 21 2016 *)
%t LinearRecurrence[{6,-15,20,-15,6,-1},{0,0,3,22,88,258},40] (* _Harvey P. Dale_, Jun 04 2023 *)
%o (PARI) concat(vector(2), Vec(x^2*(1+x)*(3+x)/(1-x)^6 + O(x^50))) \\ _Colin Barker_, Oct 21 2016
%o (Magma) [Binomial(n+1, 3)*(4*n^2 +5*n +4)/10: n in [0..40]]; // _G. C. Greubel_, Oct 22 2018
%Y Cf. A000217, A000290, A014105, A071245.
%K nonn,easy
%O 0,3
%A _Wolfdieter Lang_, Oct 20 2016
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