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%I #22 Apr 10 2020 14:32:03
%S 1,7,33,119,350,882,1974,4026,7623,13585,23023,37401,58604,89012,
%T 131580,189924,268413,372267,507661,681835,903210,1181510,1527890,
%U 1955070,2477475,3111381,3875067,4788973,5875864,7161000,8672312,10440584,12499641,14886543
%N a(n) = n*(n + 1)*(n + 2)*(n + 3)*(n^2 - n + 5)/120.
%C This is the case k = n of b(n,k) = n*(n+1)*(n+2)*(n+3)*(k*(n-1)+5)/120, where b(n,k) is the n-th hypersolid number in 5 dimensions generated from an arithmetical progression with the first term 1 and common difference k (see Sardelis et al. paper).
%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_07">Index entries for linear recurrences with constant coefficients</a>, signature (7,-21,35,-35,21,-7,1).
%F G.f.: x*(1 + 5*x^2)/(1 - x)^7.
%F a(n) = 5*A000579(n+3) + A000579(n+5). [_Bruno Berselli_, Apr 15 2015]
%t Table[n (n + 1) (n + 2) (n + 3) (n^2 - n + 5)/120, {n, 40}]
%o (PARI) vector(40, n, n*(n+1)*(n+2)*(n+3)*(n^2-n+5)/120) \\ _Bruno Berselli_, Apr 15 2015
%Y Cf. A000579.
%Y Cf. similar sequences listed in A256859.
%K nonn,easy
%O 1,2
%A _Luciano Ancora_, Apr 14 2015
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