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%I #15 Aug 09 2023 18:52:16
%S 1,9,46,164,460,1091,2289,4376,7779,13045,20856,32044,47606,68719,
%T 96755,133296,180149,239361,313234,404340,515536,649979,811141,
%U 1002824,1229175,1494701,1804284,2163196,2577114,3052135,3594791,4212064,4911401,5700729,6588470
%N a(n) = (8*n+13*n^3+3*n^5)/24; also the sum of triangular numbers taken in successive groups of increasing size (see Example).
%H Harvey P. Dale, <a href="/A260513/b260513.txt">Table of n, a(n) for n = 1..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 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>6. - _Colin Barker_, Aug 07 2015
%F G.f.: x*(x^4+3*x^3+7*x^2+3*x+1) / (x-1)^6. - _Colin Barker_, Aug 07 2015
%e The first ten triangular numbers are 1,3,6,10,15,21,28,36,45,and 55. Take them in groups, respectively, of 1, 2, 3, and 4 = (1), (3, 6), (10, 15, 21), and (28, 36, 45, 55). Summing each group separately = 1, 9, 46, 164.
%t Table[1/24*(8*x+13*x^3+3*x^5),{x,50}]
%t Module[{nn=40},Total/@TakeList[Accumulate[Range[(nn(nn+1))/2]],Range[nn]]] (* or *) LinearRecurrence[{6,-15,20,-15,6,-1},{1,9,46,164,460,1091},40] (* _Harvey P. Dale_, Aug 09 2023 *)
%o (PARI) Vec(x*(x^4+3*x^3+7*x^2+3*x+1)/(x-1)^6 + O(x^100)) \\ _Colin Barker_, Aug 07 2015
%K nonn,easy
%O 1,2
%A _Harvey P. Dale_, Jul 27 2015
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