A376467 Triangular array read by rows: A063007 * A007318.

1, 3, 2, 13, 18, 6, 63, 132, 90, 20, 321, 900, 930, 420, 70, 1683, 5910, 8190, 5600, 1890, 252, 8989, 37926, 65940, 60480, 30870, 8316, 924, 48639, 239624, 501228, 577080, 395010, 160776, 36036, 3432, 265729, 1497096, 3660300, 5072760, 4358970, 2378376, 804804, 154440, 12870, 1462563, 9274410, 25951860, 42060480, 43513470, 29801772, 13513500, 3912480, 656370, 48620

Offset: 0

Author

Peter Bala, Sep 30 2024

Keywords

• nonn
• tabl
• easy

Comments

Note that the n-th row generating polynomial of A063007 is equal to P(n,2*x + 1), where P(n,x) denotes the n-th Legendre polynomial.

The matrix product A063007 * A007318^(-1) is equal to a signed version of A063007 and A007318^(-1) * A063007 = A115951.

Examples

			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
  ...
		

Crossrefs

A000984 (main diagonal), A002457( (1/3)*first subdiagonal ), A001850 (Column 0), A002695 ( (1/2)*Column 1 ), A277660 ( (1/3)*Column 2 ), A006442 (row sums).

Cf. A063007, A007318, A115951, A118384.

Programs

Maple

A376467 := proc(n, k); add(binomial(n, j)*binomial(n+j, j)*binomial(j, k), j = k..n) end:
seq(print(seq(A376467(n, k) , k = 0..n)), n = 0..10);

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].