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A253285 a(n) = RF(n+1,3)*C(n+2,n-1), where RF(a,n) is the rising factorial.

%I A253285 #25 Aug 05 2024 12:10:50
%S A253285 0,24,240,1200,4200,11760,28224,60480,118800,217800,377520,624624,
%T A253285 993720,1528800,2284800,3329280,4744224,6627960,9097200,12289200,
%U A253285 16364040,21507024,27931200,35880000,45630000,57493800,71823024,89011440,109498200,133771200,162370560
%N A253285 a(n) = RF(n+1,3)*C(n+2,n-1), where RF(a,n) is the rising factorial.
%H A253285 <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (7,-21,35,-35,21,-7,1).
%F A253285 G.f.: -24/(x-1)^4 - 144/(x-1)^5 - 240/(x-1)^6 - 120/(x-1)^7. See the comment in A253284 for the general case.
%F A253285 a(n) = n*((n+1)*(n+2))^2*(n+3)/6.
%F A253285 a(n) = (N^3 + 4*N^2 + 4*N)/6 = N*(N + 2)^2/6 with N = n^2 + 3*n.
%F A253285 From _Bruno Berselli_, Mar 06 2018: (Start)
%F A253285 a(n) = 24*A006542(n+3) for n>0.
%F A253285 a(n) = Sum_{i=0..n} i*(i+1)^3*(i+2). Therefore, the first differences are in A133754. (End)
%p A253285 seq(n*((n+1)*(n+2))^2*(n+3)/6,n=0..19);
%t A253285 Table[n ((n + 1) (n + 2))^2 (n + 3)/6, {n, 0, 40}] (* _Bruno Berselli_, Mar 06 2018 *)
%t A253285 LinearRecurrence[{7,-21,35,-35,21,-7,1},{0,24,240,1200,4200,11760,28224},40] (* _Harvey P. Dale_, Aug 05 2024 *)
%o A253285 (Sage) [n*((n+1)*(n+2))^2*(n+3)/6 for n in (0..40)] # _Bruno Berselli_, Mar 06 2018
%o A253285 (GAP) List([0..40], n -> n*((n+1)*(n+2))^2*(n+3)/6); # _Bruno Berselli_, Mar 06 2018
%o A253285 (Magma) [n*((n+1)*(n+2))^2*(n+3)/6: n in [0..40]]; // _Bruno Berselli_, Mar 06 2018
%o A253285 (Python) [n*((n+1)*(n+2))**2*(n+3)/6 for n in range(40)] # _Bruno Berselli_, Mar 06 2018
%Y A253285 Cf. A002378, A006542, A083374, A133754, A253284.
%K A253285 nonn,easy
%O A253285 0,2
%A A253285 _Peter Luschny_, Mar 23 2015