A280580 Triangle read by rows: T(n,k) = binomial(2*n,2*k)*binomial(2*n-2*k,n-k)/(n+1-k) for 0<=k<=n.

1, 1, 1, 2, 6, 1, 5, 30, 15, 1, 14, 140, 140, 28, 1, 42, 630, 1050, 420, 45, 1, 132, 2772, 6930, 4620, 990, 66, 1, 429, 12012, 42042, 42042, 15015, 2002, 91, 1, 1430, 51480, 240240, 336336, 180180, 40040, 3640, 120, 1, 4862, 218790, 1312740, 2450448, 1837836, 612612, 92820, 6120, 153, 1

Offset: 0

Author

Werner Schulte, Jan 05 2017

Keywords

• nonn
• tabl

Examples

			Triangle begins:
n\k:     0      1       2       3       4      5     6    7  8  . . .
  0:     1
  1:     1      1
  2:     2      6       1
  3:     5     30      15       1
  4:    14    140     140      28       1
  5:    42    630    1050     420      45      1
  6:   132   2772    6930    4620     990     66     1
  7:   429  12012   42042   42042   15015   2002    91    1
  8:  1430  51480  240240  336336  180180  40040  3640  120  1
  etc.
T(3,2) = binomial(6,4) * binomial(2,1) / (3+1-2) = 15 * 2 / 2 = 15. - _Indranil Ghosh_, Feb 15 2017
		

Links

• Indranil Ghosh, Rows 0..100 of triangle, flattened

Crossrefs

Row sums are A026945.

Triangle related to A000108, A001006, A001263, and A039599.

Cf. A000007, A001003, A005568, A103365, A238390.

Formula

T(n,k) = A001263(n+1,k+1)*A000108(n)/A000108(k) for 0 <= k <= n.

T(n,k) = binomial(2*n,2*k)*A000108(n-k) for 0 <= k <= n.

T(n,k) = A039599(n,k)*binomial(n+1+k,2*k+1)/(n+1-k) for 0 <= k <= n.

Recurrences: T(n,0) = A000108(n) and (1) T(n,k) = T(n,k-1)*(n+1-k)*(n+2-k)/ (2*k*(2*k-1)) for 0 < k <= n, (2) T(n,k) = T(n-1,k-1)*n*(2*n-1)/(k*(2*k-1)).

The row polynomials p(n,x) = Sum_{k=0..n} T(n,k)*x^(2*k) satisfy the recurrence equation p"(n,x) = 2*n*(2*n-1)*p(n-1,x) with initial value p(0,x) = 1 ( n > 0, p" is the second derivative of p ), and Sum_{n>=0} p(n,x)*t^(2*n)/((2*n)!) = cosh(x*t)*(Sum_{n>=0} A000108(n)*t^(2*n)/((2*n)!)).

Conjectures: (1) Sum_{k=0..n} (-1)^k*T(n,k)*A238390(k) = A000007(n);

(2) Antidiagonal sums equal A001003(n);

(3) Matrix inverse equals T(n,k)*A103365(n+1-k).

Sum_{k=0..n} (n+1-k)*T(n,k) = A002426(2*n) = A082758(n).

Sum_{k=0..n} T(n,k)*A000108(k) = A000108(n)*A000108(n+1) = A005568(n).

Matrix product: Sum_{i=0..n} T(n,i)*T(i,k) = T(n,k)*A000108(n+1-k) for 0<=k<=n.

T(n,k) = A097610(2*n,2*k) for 0 <= k <= n.

Sum_{k=0..n} (k+1)*T(n,k)*A000108(k) = binomial(2*n+1,n)*A000108(n).