|
%I #32 Sep 24 2020 03:33:25
%S 1,2,1,5,5,2,14,21,15,5,42,84,84,49,14,132,330,420,336,168,42,429,
%T 1287,1980,1980,1350,594,132,1430,5005,9009,10725,9075,5445,2145,429,
%U 4862,19448,40040,55055,55055,40898,22022,7865,1430,16796,75582
%N Triangle T(n,k) read by rows: number of 12312-avoiding matchings on [2n] with exactly k crossings (n >= 1, 0 <= k <= n-1).
%H Andrew Howroyd, <a href="/A108410/b108410.txt">Table of n, a(n) for n = 1..1275</a>
%H Alexander Burstein, Megan Martinez, <a href="https://permutationpatterns.com/files/2020/06/WednesdayA_Burstein.pdf">Pattern classes equinumerous to the class of ternary forests</a>, Permutation Patterns Virtual Workshop, Howard University (2020).
%H W. Y. C. Chen, T. Mansour and S. H. F. Yan, <a href="https://arxiv.org/abs/math/0504342">Matchings avoiding partial patterns</a>, arXiv:math/0504342 [math.CO], 2005.
%H W. Y. C. Chen, T. Mansour and S. H. F. Yan, <a href="http://www.combinatorics.org/ojs/index.php/eljc/article/view/v13i1r112">Matchings avoiding partial patterns</a>, The Electronic Journal of Combinatorics 13, 2006, #R112, Theorem 2.2.
%H D. S. Hough, <a href="http://www.combinatorics.org/ojs/index.php/eljc/article/view/v10i1n13">Descents in noncrossing trees</a>, Electronic J. Combinatorics 10 (2003), #N13, Theorem 2.2. [_Ira M. Gessel_, May 10 2010]
%F T(n, k) = Sum_{i=n..2*n-1} (-1)^(n+k+i)/i*C(i, n)*C(3*n, i+1+n)*C(i-n, k).
%F T(n,k) = C(n-1+k,n-1)*C(2*n-k,n+1)/n, (0 <= k <= n-1). [Chen et al.] - _Emeric Deutsch_, Dec 19 2006
%F O.g.f. equals the series reversion with respect to x of x*(1 + x*(1 - t))/(1 + x)^3. If R(n,t) is the n-th row polynomial of this triangle then R(n, 1+t) is the n-th row polynomial of A089434. - _Peter Bala_, Jul 15 2012
%e Triangle begins
%e 1;
%e 2, 1;
%e 5, 5, 2;
%e 14, 21, 15, 5;
%e 42, 84, 84, 49, 14;
%e 132, 330, 420, 336, 168, 42;
%e 429, 1287, 1980, 1980, 1350, 594, 132;
%e 1430, 5005, 9009, 10725, 9075, 5445, 2145, 429;
%e 4862, 19448, 40040, 55055, 55055, 40898, 22022, 7865, 1430;
%p T:=(n,k)->binomial(n-1+k,n-1)*binomial(2*n-k,n+1)/n: for n from 1 to 10 do seq(T(n,k),k=0..n-1) od; # yields sequence in triangular form - _Emeric Deutsch_, Dec 19 2006
%t T[n_, k_] := Binomial[n + k - 1, n - 1]*Binomial[2*n - k, n + 1]/n;
%t Table[T[n, k], {n, 1, 10}, {k, 0, n - 1}] // Flatten (* _Jean-François Alcover_, Nov 11 2017, after _Emeric Deutsch_ *)
%o (PARI) T(n,k) = binomial(n-1+k, n-1)*binomial(2*n-k, n+1)/n; \\ _Andrew Howroyd_, Nov 06 2017
%Y Left-hand columns include A000108 and A002054. Right-hand columns include A000108 and A007851+1. Row sums are A001764. A089434.
%K nonn,tabl
%O 1,2
%A _Ralf Stephan_, Jun 03 2005
|
