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a(n) = n*(n+1)*(n+2)*(2*n^3 + 6*n^2 + 7*n - 3)/36.
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%I #18 Sep 08 2022 08:45:04

%S 0,2,34,210,830,2520,6412,14364,29220,55110,97790,165022,266994,

%T 416780,630840,929560,1337832,1885674,2608890,3549770,4757830,6290592,

%U 8214404,10605300,13549900,17146350,21505302,26750934,33022010

%N a(n) = n*(n+1)*(n+2)*(2*n^3 + 6*n^2 + 7*n - 3)/36.

%D L. Berzolari, Allgemeine Theorie der Höheren Ebenen Algebraischen Kurven, Encyclopädie der Mathematischen Wissenschaften mit Einschluss ihrer Anwendungen. Band III_2. Heft 3, Leipzig: B. G. Teubner, 1906. p. 352.

%H Vincenzo Librandi, <a href="/A064202/b064202.txt">Table of n, a(n) for n = 0..1000</a>

%F G.f.: 2*x*(1+10*x+7*x^2+2*x^3)/(1-x)^7. - _Colin Barker_, Feb 28 2012

%t Table[n*(n+1)*(n+2)*(2*n^3+6*n^2+7*n-3)/36,{n,0,40}] (* _Vincenzo Librandi_, Feb 29 2012 *)

%o (Magma) [n*(n+1)*(n+2)*(2*n^3 + 6*n^2 + 7*n -3)/36: n in [0..30]]; // _Vincenzo Librandi_, Feb 29 2012

%K nonn,easy

%O 0,2

%A Antreas P. Hatzipolakis (xpolakis(AT)otenet.gr), Sep 22 2001