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%I #19 Dec 11 2019 06:49:27
%S 0,1,10,48,158,413,924,1848,3396,5841,9526,14872,22386,32669,46424,
%T 64464,87720,117249,154242,200032,256102,324093,405812,503240,618540,
%U 754065,912366,1096200,1308538,1552573,1831728,2149664,2510288,2917761,3376506,3891216
%N Convolution of the odd-indexed triangular numbers (A000384(n+1)) and the squares (A000290).
%C This sequence was originally proposed in a comment on A071238 by J. M. Bergot as giving the first differences. Therefore, a(n) gives the partial sums of A071238.
%H Colin Barker, <a href="/A277229/b277229.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*(1 + x)*(1 + 3*x)/(1 - x)^6 = ((1 + 3*x)/(1 - x)^3)*(x*(1 + x)/(1 - x)^3).
%F a(n) = Sum_{k=0..n} A000384(n+1-k)*A000290(k).
%F a(n) = binomial(n+2, 3)*(4*n^2 + 3*n + 3)/10 = n*(n + 1)*(n + 2)*(4*n^2 + 3*n + 3)/60.
%F a(n) = Sum_{k=0..n} A071238(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 (n + 1) (n + 2) (4 n^2 + 3 n + 3)/60, {n, 0, 40}] (* _Bruno Berselli_, Oct 21 2016 *)
%o (PARI) concat(0, Vec(x*((1+x)*(1+3*x))/(1-x)^6 + O(x^50))) \\ _Colin Barker_, Oct 21 2016
%Y Cf. A000217, A000290, A000384, A071238.
%K nonn,easy
%O 0,3
%A _Wolfdieter Lang_, Oct 20 2016
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