|
|
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).
|
|
4
|
|
|
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
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,4
|
|
COMMENTS
|
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)
|
|
LINKS
|
|
|
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
...
|
|
EXAMPLE
|
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.
/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)
|
|
PROG
|
(PARI) {T(n, k)=binomial(3*n-k, n-k)*(2*k+1)/(2*n+1)}
|
|
CROSSREFS
|
|
|
KEYWORD
|
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|