A260513 - OEIS

%I #20 May 14 2024 17:00:49

%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 From _Colin Barker_, Aug 07 2015: (Start)

%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.

%F G.f.: x*(x^4+3*x^3+7*x^2+3*x+1) / (x-1)^6. (End)

%F E.g.f.: exp(x)*x*(24 + 84*x + 88*x^2 + 30*x^3 + 3*x^4)/24. - _Stefano Spezia_, May 14 2024

%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

%Y Cf. A000217.

%K nonn,easy

%O 1,2

%A _Harvey P. Dale_, Jul 27 2015