%I #23 Sep 08 2022 08:45:59
%S 0,15,120,435,1128,2415,4560,7875,12720,19503,28680,40755,56280,75855,
%T 100128,129795,165600,208335,258840,318003,386760,466095,557040,
%U 660675,778128,910575,1059240,1225395,1410360,1615503,1842240,2092035
%N a(n) = 1*3*5 + 3*5*7 + 5*7*9 + ... (n terms).
%C All terms are multiples of 3.
%D Jolley, Summation of Series, Dover (1961), eq (43) page 8.
%H Vincenzo Librandi, <a href="/A196506/b196506.txt">Table of n, a(n) for n = 0..10000</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).
%F a(n) = ((4n^2 - 1)*(2n + 3)*(2n + 5) + 15)/ 8 = Sum_{i=1..n} (2i - 1)*(2i + 1)*(2i + 3).
%F G.f. -3*x*(5 + 15*x - 5*x^2 + x^3) / (x-1)^5 .
%F a(n) = 2 n^4 + 8 n^3 + 7 n^2 - 2 n. - _Harvey P. Dale_, Mar 14 2015, corrected by _Eric Rowland_, Aug 15 2017
%t LinearRecurrence[{5,-10,10,-5,1},{0,15,120,435,1128},40] (* or *) Accumulate[ Join[{0},Times@@@Partition[Range[1,111,2],3,1]]] (* or *) Table[2n^4-5n^2+3,{n,40}](* _Harvey P. Dale_, Mar 14 2015 *)
%o (Magma) [((4*n^2-1)*(2*n+3)*(2*n+5)+15)/ 8 : n in [0..30]]; // _Vincenzo Librandi_, Oct 05 2011
%Y Cf. A061550 (first differences).
%K nonn,easy
%O 0,2
%A _R. J. Mathar_, Oct 03 2011