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a(n) = n*(n + 1)*(n + 2)*(n^2 - n + 4)/24.
6

%I #32 Sep 08 2022 08:46:12

%S 1,6,25,80,210,476,966,1800,3135,5170,8151,12376,18200,26040,36380,

%T 49776,66861,88350,115045,147840,187726,235796,293250,361400,441675,

%U 535626,644931,771400,916980,1083760,1273976,1490016,1734425,2009910,2319345,2665776,3052426

%N a(n) = n*(n + 1)*(n + 2)*(n^2 - n + 4)/24.

%C This is the case k = n of b(n,k) = n*(n+1)*(n+2)*(k*(n-1)+4)/24, where b(n,k) is the n-th hypersolid number in 4 dimensions generated from an arithmetical progression with the first term 1 and common difference k. Therefore, the sequence is the main diagonal of the Table 3 in Sardelis et al. paper (see Links field).

%H G. C. Greubel, <a href="/A256859/b256859.txt">Table of n, a(n) for n = 1..1000</a>

%H D. A. Sardelis and T. M. Valahas, <a href="http://arxiv.org/abs/0805.4070v1">On Multidimensional Pythagorean Numbers</a>, arXiv:0805.4070v1 [math.GM], 2008.

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (6,-15,20,-15,6,-1).

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

%F a(n) = 4*A000389(n+2) + A000389(n+4). - _Bruno Berselli_, Apr 15 2015

%F E.g.f.: (24*x + 48*x^2 + 40*x^3 + 12*x^4 + x^5)*exp(x)/24. - _G. C. Greubel_, Nov 23 2017

%F a(n) = A261721(n,n-1). - _Alois P. Heinz_, Apr 15 2020

%t Table[n (n + 1) (n + 2) (n^2 - n + 4)/24, {n, 40}]

%t LinearRecurrence[{6,-15,20,-15,6,-1},{1,6,25,80,210,476},40] (* _Harvey P. Dale_, Mar 19 2022 *)

%o (PARI) vector(40, n, n*(n+1)*(n+2)*(n^2-n+4)/24) \\ _Bruno Berselli_, Apr 15 2015

%o (Magma) [n*(n + 1)*(n + 2)*(n^2 - n + 4)/24: n in [1..30]]; // _G. C. Greubel_, Nov 23 2017

%Y Cf. A000389, A261721.

%Y Cf. similar sequences of the form binomial(n+k-2,k-1)+n*binomial(n+k-2,k): A006000 (k=2); A257055 (k=3); this sequence (k=4); A256860 (k=5); A256861 (k=6).

%K nonn,easy

%O 1,2

%A _Luciano Ancora_, Apr 14 2015