

A107131


A Motzkin related triangle.


12



1, 0, 1, 0, 1, 1, 0, 0, 3, 1, 0, 0, 2, 6, 1, 0, 0, 0, 10, 10, 1, 0, 0, 0, 5, 30, 15, 1, 0, 0, 0, 0, 35, 70, 21, 1, 0, 0, 0, 0, 14, 140, 140, 28, 1, 0, 0, 0, 0, 0, 126, 420, 252, 36, 1, 0, 0, 0, 0, 0, 42, 630, 1050, 420, 45, 1, 0, 0, 0, 0, 0, 0, 462, 2310, 2310, 660, 55, 1
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OFFSET

0,9


COMMENTS

Inverse binomial transform of Narayana number triangle A001263.  Paul Barry, May 15 2005
T(n,k)=number of Motzkin paths of length n with k steps U=(1,1) or H=(1,0). Example: T(3,2)=3 because we have HUD, UDH and UHD (here D=(1,1)). T(n,k) = number of bushes with n+1 edges and k+1 leaves (a bush is an ordered tree in which the outdegree of each nonroot node is at least two).  Emeric Deutsch, May 29 2005
Rows of A088617 are shifted columns of A107131, whose reversed rows are the Motzkin polynomials of A055151, which give the row polynomials (mod signs) of the o.g.f. that is the compositional inverse for an o.g.f. of the Fibonacci polynomials of A011973. The diagonals of A055151 give the rows of A088671, and the antidiagonals (top to bottom) of A088617 give the rows of A107131. The diagonals of A107131 give the columns of A055151. From the relation between A088617 and A107131, the o.g.f. of this entry is (1  t*x  sqrt((1t*x)^2  4*t*x^2))/(2*t*x^2).  Tom Copeland, Jan 21 2016


LINKS



FORMULA

Number triangle T(n, k) = binomial(k+1, nk+1)*binomial(n, k)/(k+1).
T(n, k) = Sum_{j=0..n} (1)^(nj)C(n, j)*C(j+1, k)*C(j+1, k+1)/(j+1).  Paul Barry, May 15 2005
G.f.: G = G(t, z) satisfies G = 1 + t*z*G + t*z^2*G^2.  Emeric Deutsch, May 29 2005
Coefficient array for the polynomials x^n*Hypergeometric2F1((1n)/2, n/2; 2; 4/x).  Paul Barry, Oct 04 2008
G.f.: 1/(1xy(1+x)/(1x^2*y/(1xy(1+x)/(1x^2y/(1xy(1+x).... (continued fraction).
T(n,k) = C(n, 2n2k)*A000108(nk). (End)


EXAMPLE

Triangle begins
1;
0, 1;
0, 1, 1;
0, 0, 3, 1;
0, 0, 2, 6, 1;
0, 0, 0, 10, 10, 1;
0, 0, 0, 5, 30, 15, 1;
0, 0, 0, 0, 35, 70, 21, 1;
0, 0, 0, 0, 14, 140, 140, 28, 1;
0, 0, 0, 0, 0, 126, 420, 252, 36, 1;


MAPLE

egf := exp(t*x)*hypergeom([], [2], t*x^2);
s := n > n!*coeff(series(egf, x, n+2), x, n);
seq(print(seq(coeff(s(n), t, j), j=0..n)), n=0..9); # Peter Luschny, Oct 29 2014


MATHEMATICA

T[n_, k_] := Binomial[k+1, nk+1] Binomial[n, k]/(k+1);


PROG

(Magma) [Binomial(n, 2*(nk))*Catalan(nk): k in [0..n], n in [0..13]]; // G. C. Greubel, May 22 2022
(SageMath) flatten([[binomial(n, 2*(nk))*catalan_number(nk) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 22 2022


CROSSREFS



KEYWORD



AUTHOR



STATUS

approved



