A143603 Triangle, read by rows, such that the g.f. of column k = G(x)^(2k+1) where G(x) = 1 + x*G(x)^3 is the g.f. of A001764 (ternary trees).

1, 1, 1, 3, 3, 1, 12, 12, 5, 1, 55, 55, 25, 7, 1, 273, 273, 130, 42, 9, 1, 1428, 1428, 700, 245, 63, 11, 1, 7752, 7752, 3876, 1428, 408, 88, 13, 1, 43263, 43263, 21945, 8379, 2565, 627, 117, 15, 1, 246675, 246675, 126500, 49588, 15939, 4235, 910, 150, 17, 1

Offset: 1

Author

Paul D. Hanna, Aug 29 2008

Keywords

• nonn
• tabl

Comments

From Peter Bala, Aug 07 2014: (Start)

Riordan array (G(x), x*G(x)). Let C(x) = 1 + x + 2*x^2 + 5*x^3 + 14*x^4 + ... be the o.g.f. of the Catalan numbers A000108. Then C(x*G(x)) = G(x).

This leads to a factorization of this array in the group of Riordan matrices as (1, x*G(x))*(C(x), x*C(x)) = (1 + A110616)*A033184 (here, in the final product, 1 refers to the 1 X 1 identity matrix and + means direct sum - see the Example section). (End)

Examples

			Triangle begins:
1;
1, 1;
3, 3, 1;
12, 12, 5, 1;
55, 55, 25, 7, 1;
273, 273, 130, 42, 9, 1;
1428, 1428, 700, 245, 63, 11, 1;
7752, 7752, 3876, 1428, 408, 88, 13, 1; ...
where g.f. of column k = G(x)^(2k+1) where G(x) = 1 + x*G(x)^3.
Matrix inverse begins:
1;
-1, 1;
0, -3, 1;
0, 3, -5, 1;
0, -1, 10, -7, 1;
0, 0, -10, 21, -9, 1;
0, 0, 5, -35, 36, -11, 1;
0, 0, -1, 35, -84, 55, -13, 1; ...
where g.f. of column k = (1-x)^(2k+1) for k>=0.
From _Peter Bala_, Aug 07 2014: (Start)
Matrix factorization as (1 + A110616)*A033184 begins
/1           \/ 1         \    / 1           \
|0  1        || 1  1       |   | 1  1        |
|0  1 1      || 2  2 1     | = | 3  3  1     |
|0  3 2 1    || 5  5 3 1   |   |12 12  5 1   |
|0 12 7 3 1  ||14 14 9 4 1 |   |55 55 25 7 1 |
(End)
		

Links

• Yuxuan Zhang and Yan Zhuang, A subfamily of skew Dyck paths related to k-ary trees, arXiv:2306.15778 [math.CO], 2023.

Crossrefs

Cf. columns: A001764, A102893, A102594; row sums: A006013. A033184, A110616.

Programs

PARI

{T(n,k)=binomial(3*n-k,n-k)*(2*k+1)/(2*n+1)}

Formula

T(n,k) = C(3n-k,n-k)*(2k+1)/(2n+1) for 0<=k<=n.

Let M = the production matrix:

1, 1

2, 2, 1

3, 3, 2, 1

4, 4, 3, 2, 1

5, 5, 4, 3, 2, 1

...

Top row of M^(n-1) gives n-th row. - Gary W. Adamson, Jul 07 2011