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Search: id:A050534
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| A050534 |
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Tritriangular numbers: a(n)=binomial(binomial(n,2),2), i.e. a(n) = (1/8)n(n + 1)(n - 1)(n - 2). |
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+0 20
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| 0, 0, 0, 3, 15, 45, 105, 210, 378, 630, 990, 1485, 2145, 3003, 4095, 5460, 7140, 9180, 11628, 14535, 17955, 21945, 26565, 31878, 37950, 44850, 52650, 61425, 71253, 82215, 94395, 107880, 122760, 139128, 157080, 176715, 198135, 221445, 246753
(list; graph; listen)
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OFFSET
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0,4
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COMMENT
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"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
Several different versions of this sequence are possible, beginning with either one, two or three 0's.
a(n)=A052762(n+1)/8. - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Apr 26 2007
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
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REFERENCES
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A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 154.
L. Comtet, Advanced Combinatorics, Reidel, 1974, Problem 1, page 72.
R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 5.5, case k=2.
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LINKS
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Index entries for sequences related to linear recurrences with constant coefficients
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FORMULA
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a(n) = 3*binomial(n+3, 4) = 3*A000332(n+3). [This produces 0, 3, 15, 45, ...]
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
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
Also a(n)=T(n)^2-T(T(n)) - Jon Perry (perry(AT)globalnet.co.uk), Jul 23 2003
a(n) = 3C(n, 4) + 3C(n, 3), for n>3.
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
a(n)=numbperm (n,4)/8, n>=1 . - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Apr 26 2007
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MAPLE
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[seq(binomial(n, 4)*3, n=1..40)]; - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jul 18 2006
seq(numbperm (n, 4)/8, n=1..39); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Apr 26 2007
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
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
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]
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MATHEMATICA
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Table[Binomial[Binomial[n, 2], 2], {n, 0, 50}] - Stefan Steinerberger (stefan.steinerberger(AT)gmail.com), Apr 08 2006
Table[Sum[(k*(k-1)*(k-2)), {k, 1, n}]/2, {n, 0, 60}] - Alexander Adamchuk (alex(AT)kolmogorov.com), Apr 11 2006
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PROGRAM
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(Other) sage: [(binomial(binomial(n, 2), 2)) for n in xrange(0, 39)] # [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Nov 30 2009]
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CROSSREFS
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Cf. A000217, A000332.
Second column of triangle A001498.
Cf. A033487, A107394.
Cf. A033487, A034827.
Sequence in context: A161400 A112810 A094191 this_sequence A048099 A030505 A074355
Adjacent sequences: A050531 A050532 A050533 this_sequence A050535 A050536 A050537
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KEYWORD
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easy,nice,nonn,new
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AUTHOR
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Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Dec 29 1999
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EXTENSIONS
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Additional comments from Antreas P. Hatzipolakis, May 03, 2002
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