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a(n) = n*(n+1)*(n+2)*(n+3)*(1+3*n+n^2)/120.
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%I #46 Sep 08 2022 08:45:15

%S 1,11,57,203,574,1386,2982,5874,10791,18733,31031,49413,76076,113764,

%T 165852,236436,330429,453663,612997,816431,1073226,1394030,1791010,

%U 2277990,2870595,3586401,4445091,5468617,6681368,8110344,9785336

%N a(n) = n*(n+1)*(n+2)*(n+3)*(1+3*n+n^2)/120.

%C Partial sums of A024166. Third partial sums of cubes (A000578).

%C Antidiagonal sums of the array A213564. - _Clark Kimberling_, Jun 18 2012

%H Danny Rorabaugh, <a href="/A101094/b101094.txt">Table of n, a(n) for n = 1..10000</a>

%H C. Rossiter, <a href="http://noticingnumbers.net/300SeriesCube.htm">Depictions, Explorations and Formulas of the Euler/Pascal Cube</a> [Dead link]

%H C. Rossiter, <a href="/A101094/a101094_1.pdf">Depictions, Explorations and Formulas of the Euler/Pascal Cube</a> [Cached copy, May 15 2013]

%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (7,-21,35,-35,21,-7,1).

%F This sequence could be obtained from the general formula n*(n+1)*(n+2)*(n+3)*...*(n+k)*(n*(n+k)+(k-1)*k/6)/((k+3)!/6) at k=3. - _Alexander R. Povolotsky_, May 17 2008

%F G.f. -x*(1+4*x+x^2) / (x-1)^7. - _R. J. Mathar_, Dec 06 2011

%F Sum_{n>0} 1/a(n) = (8/3)*(25-9*sqrt(5)*Pi*tan(sqrt(5)*Pi/2)). - _Enrique PĂ©rez Herrero_, Dec 02 2014

%F a(k) = MagicNKZ(3,k,7) where MagicNKZ(n,k,z) = Sum_{j=0..k+1} (-1)^j*binomial(n+1-z,j)*(k-j+1)^n. (Cf. A101104.) - _Danny Rorabaugh_, Apr 23 2015

%t s1=s2=s3=0; lst={}; Do[s1+=n^3; s2+=s1; s3+=s2; AppendTo[lst,s3],{n,0,6!}]; lst (* _Vladimir Joseph Stephan Orlovsky_, Jan 15 2009 *)

%t Table[n*(n + 1)*(n + 2)*(n + 3)*(1 + 3*n + n^2)/120, {n, 31}] (* _Michael De Vlieger_, Apr 20 2015 *)

%o (PARI) a(n)=sum(l=1,n,sum(j=1,l, sum(m=1, j, sum(i=m*(m+1)/2-m+1, m*(m+1)/2, (2*i-1))))) \\ _Alexander R. Povolotsky_, May 17 2008

%o (Sage) [n*(n+1)*(n+2)*(n+3)*(1+3*n+n^2)/120 for n in range(1,32)] # _Danny Rorabaugh_, Apr 20 2015

%o (Magma) [n*(n+1)*(n+2)*(n+3)*(1+3*n+n^2)/120 : n in [1..35]]; // _Vincenzo Librandi_, Apr 23 2015

%Y Cf. A000537, A101097, A101102, A101104.

%K nonn,easy

%O 1,2

%A Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 15 2004

%E Edited by _Ralf Stephan_, Dec 16 2004