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%I #16 Sep 08 2022 08:45:29
%S 3,23,72,162,305,513,798,1172,1647,2235,2948,3798,4797,5957,7290,8808,
%T 10523,12447,14592,16970,19593,22473,25622,29052,32775,36803,41148,
%U 45822,50837,56205,61938,68048,74547,81447,88760,96498,104673,113297
%N a(n) = n*(4*n^2+5*n-3)/2.
%C Inner product of two arithmetic series (A016777, A005408): (1,4,7,...,3n-2)*(3,5,7,...,2n+1) = sum((3i-2)*(2i+1),i=1...n) = 1*3+4*3+7*7+...+(3n-2)*(2n+1) = (1/2)*n*(4*n^2+5*n-3).
%H Vincenzo Librandi, <a href="/A126335/b126335.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F a(1)=3, a(2)=23, a(3)=72, a(4)=162; for n>4, a(n)=4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4).
%F G.f.: x*(3 + 11*x - 2*x^2)/(1 - x)^4.
%t CoefficientList[Series[(3 + 11 x - 2 x^2)/(1 - x)^4, {x, 0, 50}], x] (* _Vincenzo Librandi_, Oct 12 2013 *)
%o (PARI) a(n) = n*(4*n^2 + 5*n - 3)/2; \\ _Michel Marcus_, Oct 11 2013
%o (Magma) [n*(4*n^2+5*n-3)/2: n in [1..40]]; // _Vincenzo Librandi_, Oct 12 2013
%K nonn,easy
%O 1,1
%A _Zak Seidov_, Mar 11 2007
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