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%I #15 Sep 08 2022 08:46:20
%S 0,26,588,5376,30660,129780,446292,1315776,3444012,8198190,18058040,
%T 37285248,72882264,135925608,243374040,420468480,703858344,1145608002,
%U 1818257364,2821132160,4288122300,6397170780,9381740940,13544556480,19273936500,27063076950,37532660256
%N a(n) = (1/2)*(n + 1)*(5*n^2 + 15*n + 6)*Pochhammer(n, 6) / 6!.
%H G. C. Greubel, <a href="/A293612/b293612.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (10,-45,120,-210,252,-210,120,-45,10,-1).
%F From _Colin Barker_, Jul 28 2019: (Start)
%F G.f.: 2*x*(13 + 164*x + 333*x^2 + 120*x^3) / (1 - x)^10.
%F a(n) = 10*a(n-1) - 45*a(n-2) + 120*a(n-3) - 210*a(n-4) + 252*a(n-5) - 210*a(n-6) + 120*a(n-7) - 45*a(n-8) + 10*a(n-9) - a(n-10) for n>9.
%F a(n) = ((n*(1+n)^2*(720 + 2724*n + 3336*n^2 + 1919*n^3 + 571*n^4 + 85*n^5 + 5*n^6))) / 1440.
%F (End)
%p A293612 := n -> (1/2)*(n + 1)*(5*n^2 + 15*n + 6)*pochhammer(n,6)/6!;
%p seq(A293612(n), n=0..29);
%t LinearRecurrence[{10,-45,120,-210,252,-210,120,-45,10,-1},{0, 26, 588, 5376, 30660, 129780, 446292, 1315776, 3444012, 8198190}, 40] (* or *)
%t a = (720 #1 + 4164 #1^2 + 9504 #1^3 + 11315 #1^4 + 7745 #1^5 + 3146 #1^6 + 746 #1^7 + 95 #1^8 + 5 #1^9)/1440 &; Table[a[n], {n, 0, 40}]
%t Table[(n + 1)*(5*n^2 + 15*n + 6)*Pochhammer[n, 6]/(2*6!), {n, 0, 50}] (* _G. C. Greubel_, Oct 22 2017 *)
%o (PARI) for(n=0,50, print1((n + 1)*(5*n^2 + 15*n + 6)*(n+5)*(n+4)*(n+3)*(n+2)*(n+1)*n/(2*6!), ", ")) \\ _G. C. Greubel_, Oct 22 2017
%o (PARI) concat(0, Vec(2*x*(13 + 164*x + 333*x^2 + 120*x^3) / (1 - x)^10 + O(x^40))) \\ _Colin Barker_, Jul 28 2019
%o (Magma) [(n + 1)*(5*n^2 + 15*n + 6)*(n+5)*(n+4)*(n+3)*(n+2)*(n+1)*n/(2*Factorial(6)):n in [0..50]]; // _G. C. Greubel_, Oct 22 2017
%K nonn,easy
%O 0,2
%A _Peter Luschny_, Oct 13 2017