OFFSET
0,2
COMMENTS
FORMULA
T(n, k) = Sum_{j = k..n} binomial(n, j)*binomial(n+j, j)*binomial(j, k).
(n - k)*T(n, k) = 3*(2*n - 1)*T(n-1, k) - (n + k - 1)*T(n-2, k).
T(n, k) = (1/k!) * (d/dx)^k (P(n, 2*x+1)) evaluated at x = 1, where P(n,x) denotes the n-th Legendre polynomial.
G.f. for triangle: 1/sqrt(1 - 6*t + t^2 - 4*t*x) = 1 + (3 + 2*x)*t + (13 + 18*x + 6*x^2)*t^2 + ....
G.f. for column k: binomial(2*k, k) * x^k/(1 - 6*x + x^2)^(k+1/2).
T(n, k) is divisible by binomial(2*k, k) and the array ( T(n, k)/binomial(2*k, k) )n,k >= 0 is the Riordan array (1/sqrt(1 - 6*x + x^2), x/(1 - 6*x + x^2)).
T(n, k) is divisible by binomial(n+k, k) and the array ( T(n, k)/binomial(n+k, k) )n,k >= 0 is the Riordan array A118384.
T(n, n) = binomial(2*n, n); T(n, n-1) = 3*(2*n-1)!/(n-1)!^2 = 3 * A002457(n-1) for n >= 1.
The n-th row polynomial R(n, x) = Sum_{k = 0..n} binomial(n, k)*binomial(n+k, k)*(1 + x)^k = P(n, 2*x+3) = hypergeom([-n, n+1], [1], -1-x).
Recurrence: n*R(n, x) = (2*x + 3)*(2*n - 1)*R(n-1, x) - (n - 1)*R(n-2, x) with R(0, x) = 1.
If we set R(-1,x) = 1, we can run the recurrence backwards to give R(-n, x) = Sum_{k = 0..n} binomial(-n, k)*binomial(-n+k, k)*(1 + x)^k = R(n-1, x).
R(n, x) = (-1)^n * R(n, -x-3).
R(n, x) = 1/n! * (d/dx)^n( ((1 + x)*(2 + x))^n ).
R(1, x) = 3 + 2*x divides R(2*n+1, x) in the polynomial ring Z[x].
EXAMPLE
Triangle begins
n\k| 0 1 2 3 4 5 6 7
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
0 | 1
1 | 3 2
2 | 13 18 6
3 | 63 132 90 20
4 | 321 900 930 420 70
5 | 1683 5910 8190 5600 1890 252
6 | 8989 37926 65940 60480 30870 8316 924
7 | 48639 239624 501228 577080 395010 160776 36036 3432
...
MAPLE
CROSSREFS
KEYWORD
AUTHOR
Peter Bala, Sep 30 2024
STATUS
approved