%I A050534
%S A050534 0,0,0,3,15,45,105,210,378,630,990,1485,2145,3003,4095,5460,7140,9180,
%T A050534 11628,14535,17955,21945,26565,31878,37950,44850,52650,61425,71253,
%U A050534 82215,94395,107880,122760,139128,157080,176715,198135,221445,246753
%N A050534 Tritriangular numbers: a(n)=binomial(binomial(n,2),2), i.e. a(n) = (1/
               8)n(n + 1)(n - 1)(n - 2).
%C A050534 "There are n straight lines in a plane, no two of which are parallel 
               and no three of which are concurrent. Their points of interesection 
               being joined, show that the number of new lines drawn is (1/8)n(n-1)(n-2)(n-3)". 
               - The American Mathematical Monthly 22(1915) 130 by C. N. Schmall
%C A050534 Several different versions of this sequence are possible, beginning with 
               either one, two or three 0's.
%C A050534 a(n)=A052762(n+1)/8. - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), 
               Apr 26 2007
%C A050534 If Y is a 3-subset of an n-set X then, for n>=6, a(n-4) is the number 
               of (n-6)-subsets of X which have exactly one element in common with 
               Y. - Milan R. Janjic (agnus(AT)blic.net), Dec 28 2007
%D A050534 A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of 
               combinatorial proof, M.A.A. 2003, id. 154.
%D A050534 L. Comtet, Advanced Combinatorics, Reidel, 1974, Problem 1, page 72.
%D A050534 R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see 
               Problem 5.5, case k=2.
%H A050534 <a href="Sindx_Rea.html#recLCC">Index entries for sequences related to 
               linear recurrences with constant coefficients</a>
%F A050534 a(n) = 3*binomial(n+3, 4) = 3*A000332(n+3). [This produces 0, 3, 15, 
               45, ...]
%F A050534 Recurrence: a(n) = 5*a(n-1)-10*a(n-2)+10*a(n-3)-5*a(n-4)+a(n-5). G.f.: 
               3*x/(1-x)^5. - Vladeta Jovovic (vladeta(AT)eunet.rs), May 03 2002
%F A050534 Define T(n)=n*(n+1)/2, then a(n)=T(T(n))-T(n) and also a(n+1)=T(T(n)+n) 
               - Jon Perry (perry(AT)globalnet.co.uk), Jun 11 2003
%F A050534 Also a(n)=T(n)^2-T(T(n)) - Jon Perry (perry(AT)globalnet.co.uk), Jul 
               23 2003
%F A050534 a(n) = 3C(n, 4) + 3C(n, 3), for n>3.
%F A050534 a(n) = Sum[(k*(k-1)*(k-2)),{k,1,n}]/2. a(n) = A033487(n-2)/2, n>1. a(n) 
               = A107394(n-3)/2 = C(n-1,2)*C(n+1,2)/2, n>2. - Alexander Adamchuk 
               (alex(AT)kolmogorov.com), Apr 11 2006
%F A050534 a(n)=numbperm (n,4)/8, n>=1 . - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), 
               Apr 26 2007
%p A050534 [seq(binomial(n,4)*3,n=1..40)]; - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), 
               Jul 18 2006
%p A050534 seq(numbperm (n,4)/8, n=1..39); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), 
               Apr 26 2007
%p A050534 seq(sum(sum(sum(m, k=0..l),l=0..m),m=1..n), n=-2..36); - Zerinvary Lajos 
               (zerinvarylajos(AT)yahoo.com), Jan 26 2008
%p A050534 a:=n->sum((n-j)^3-n+j, j=1..n): seq(a(n)/2, n=0..38); - Zerinvary Lajos 
               (zerinvarylajos(AT)yahoo.com), May 26 2008
%p A050534 a:=n->add(binomial(n,2)+add(binomial(n,2), j=0..n),j=0..n):seq(a(n)/4, 
               n=-1..30);# [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), 
               Aug 26 2008]
%t A050534 Table[Binomial[Binomial[n, 2], 2], {n, 0, 50}] - Stefan Steinerberger 
               (stefan.steinerberger(AT)gmail.com), Apr 08 2006
%t A050534 Table[Sum[(k*(k-1)*(k-2)),{k,1,n}]/2,{n,0,60}] - Alexander Adamchuk (alex(AT)kolmogorov.com), 
               Apr 11 2006
%Y A050534 Cf. A000217, A000332.
%Y A050534 Second column of triangle A001498.
%Y A050534 Cf. A033487, A107394.
%Y A050534 Cf. A033487, A034827.
%Y A050534 Sequence in context: A161400 A112810 A094191 this_sequence A048099 A030505 
               A074355
%Y A050534 Adjacent sequences: A050531 A050532 A050533 this_sequence A050535 A050536 
               A050537
%K A050534 easy,nice,nonn
%O A050534 0,4
%A A050534 Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Dec 29 1999
%E A050534 Additional comments from Antreas P. Hatzipolakis, May 03, 2002