%I #35 Aug 31 2023 15:06:08

%S 0,3,15,48,120,255,483,840,1368,2115,3135,4488,6240,8463,11235,14640,

%T 18768,23715,29583,36480,44520,53823,64515,76728,90600,106275,123903,

%U 143640,165648,190095,217155,247008,279840,315843,355215,398160,444888

%N a(n) = n*(n+1)*(n^2 + n + 4)/4.

%H Harry J. Smith, <a href="/A061316/b061316.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).

%F a(n) = n*(n+1)*(n^2 + n + 4)/4.

%F a(n) = A005563(A000217(n)) = 3*A006007(n) = A061314(n, 2).

%F a(0)=0, a(1)=3, a(2)=15, a(3)=48, a(4)=120, a(n) = 5a(n-1) - 10a(n-2) + 10a(n-3) - 5a(n-4) + a(n-5).

%F G.f.: (-3 (x + x^3))/(-1 + x)^5. - _Harvey P. Dale_, May 03 2011

%F Sum_{n>=1} 1/a(n) = 5/4 - tanh(sqrt(15)*Pi/2)*Pi/sqrt(15). - _Amiram Eldar_, Aug 20 2022

%F E.g.f.: exp(x)*x*(12 + 18*x + 8*x^2 + x^3)/4. - _Stefano Spezia_, Aug 31 2023

%p a:=n->sum((n+j^3),j=0..n): seq(a(n),n=0..36); # _Zerinvary Lajos_, Jul 27 2006

%p with(combinat):a:=n->sum(fibonacci(4,i), i=0..n): seq(a(n), n=0..36); # _Zerinvary Lajos_, Mar 20 2008

%t s=0;lst={};Do[s+=n^3+n*2;AppendTo[lst,s],{n,0,5!}];lst (* _Vladimir Joseph Stephan Orlovsky_, Apr 04 2009 *)

%t Table[n(n+1)(n^2+n+4)/4,{n,0,40}] (* or *) LinearRecurrence[ {5,-10,10,-5,1}, {0,3,15,48,120},40] (* _Harvey P. Dale_, May 03 2011 *)

%o (PARI) for (n=0, 1000, write("b061316.txt", n, " ", n*(n + 1)*(n^2 + n + 4)/4) ) \\ _Harry J. Smith_, Jul 21 2009

%Y Cf. A000217, A005563, A006007, A061314.

%K nonn,easy

%O 0,2

%A _Henry Bottomley_, Apr 24 2001