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%I #17 Aug 02 2023 13:51:43
%S 0,2,7,28,116,480,1968,8000,32320,130048,521984,2092032,8377344,
%T 33529856,134164480,536756224,2147237888,8589410304,34358624256,
%U 137436594176,549750833152,2199012769792,8796071002112,35184325951488,140737391886336,562949752094720
%N a(n) is the sum of the first 2*n nonzero n-bonacci numbers.
%C For our purposes, for n > 0 fixed we define the k-th n-bonacci number T(n,k) as equal to 0 for k <= 0, equal to 1 for k=1, and then equal to the sum of the previous n numbers for k > 1. For n=2, then, we get T(2,k) equal to F(n) = A000045(n), the Fibonacci numbers. For n=3, then, T(3,k) is the tribonacci numbers, and so on.
%C a(n) is thus defined as Sum_{k=1..2*n} T(n,k).
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (8,-20,16).
%F a(n) = (2*4^n - (n-1)*2^n)/4 for n>=1.
%F a(n) = Sum_{i=1..2*n} A092921(n,i).
%F G.f.: -x*(12*x^2-9*x+2)/((4*x-1)*(2*x-1)^2). - _Alois P. Heinz_, Jul 11 2023
%F E.g.f.: exp(2*x)*(1 - 2*x - cosh(2*x) + 5*sinh(2*x))/4. - _Stefano Spezia_, Jul 12 2023
%e For n=3, a(3) is the sum of the first 6 nonzero tribonacci numbers, found at A000073. This gives a(3) = 1 + 1 + 2 + 4 + 7 + 13 = 28.
%t T[n_, k_] := SeriesCoefficient[Series[x/(1 - Sum[x^i, {i, 1, n}]), {x, 0, k + 1}], k]; Table[Sum[T[n, k], {k, 1, 2n}], {n, 1, 30}]
%Y Cf. A000045, A000073, A000078, A092921, A144406.
%K nonn,easy
%O 0,2
%A Muhammad Adam Dombrowski and _Greg Dresden_, Jul 10 2023
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