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%I #19 Sep 25 2019 14:23:47
%S 1,8,42,168,546,1512,3696,8184,16731,32032,58058,100464,167076,268464,
%T 418608,635664,942837,1369368,1951642,2734424,3772230,5130840,6888960,
%U 9140040,11994255,15580656,20049498,25574752,32356808,40625376,50642592,62706336
%N a(n) = n*(n + 1)*(n + 2)*(n + 3)*(n + 4)*(n^2 - n + 6)/720.
%C This is the case k = n of b(n,k) = n*(n+1)*(n+2)*(n+3)*(n+4)*(k*(n-1)+6)/120, where b(n,k) is the n-th hypersolid number in 6 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/0805.4070">On Multidimensional Pythagorean Numbers</a>, arXiv:0805.4070 [math.GM], 2008.
%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (8,-28,56,-70,56,-28,8,-1).
%F G.f.: x*(1 + 6*x^2)/(1 - x)^8.
%F a(n) = 6*A000580(n+4) + A000580(n+6). [_Bruno Berselli_, Apr 15 2015]
%t Table[n (1 + n) (2 + n) (3 + n) (4 + n) (6 - n + n^2)/720, {n, 40}]
%t Table[Times@@(n+Range[0,4])(n^2-n+6)/720,{n,40}] (* or *) LinearRecurrence[ {8,-28,56,-70,56,-28,8,-1},{1,8,42,168,546,1512,3696,8184},40] (* _Harvey P. Dale_, Sep 25 2019 *)
%o (PARI) vector(40, n, n*(n+1)*(n+2)*(n+3)*(n+4)*(n^2-n+6)/720) \\ _Bruno Berselli_, Apr 15 2015
%Y Cf. A000580.
%Y Cf. similar sequences listed in A256859.
%K nonn,easy
%O 1,2
%A _Luciano Ancora_, Apr 14 2015
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