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a(n) = n(n+7)(n+1)(n^2+2n+12)/120.
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%I #30 Sep 08 2022 08:44:59

%S 2,9,27,66,141,273,490,828,1332,2057,3069,4446,6279,8673,11748,15640,

%T 20502,26505,33839,42714,53361,66033,81006,98580,119080,142857,170289,

%U 201782,237771,278721,325128,377520,436458,502537,576387,658674,750101

%N a(n) = n(n+7)(n+1)(n^2+2n+12)/120.

%H Vincenzo Librandi, <a href="/A051746/b051746.txt">Table of n, a(n) for n = 1..1000</a> (Corrected by _Alois P. Heinz_, Mar 14 2016)

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

%F a(n) = binomial(n+4, n-1)+binomial(n+1, n-1).

%F G.f.: x*(2-x)*(1-x+x^2)/(1-x)^6. - _Colin Barker_, Mar 19 2012

%t Table[1/120*n*(n + 7)*(n + 1)*(n^2 + 2*n + 12), {n, 60}] (* _Vladimir Joseph Stephan Orlovsky_, Jun 14 2011 *)

%t LinearRecurrence[{6,-15,20,-15,6,-1},{2,9,27,66,141,273},40] (* _Harvey P. Dale_, Apr 28 2018 *)

%o (PARI) a(n)=((((n+10)*n+35)*n+110)*n+84)*n/120 \\ _Charles R Greathouse IV_, Jun 14 2011

%o (Magma) [1/120*n*(n+7)*(n+1)*(n^2+2*n+12): n in [1..50]]; // _Vincenzo Librandi_, Jun 15 2011

%K easy,nonn

%O 1,1

%A Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Dec 07 1999