A281000 Triangle read by rows: T(n,k) = binomial(2*n+1, 2*k+1)*binomial(2*n-2*k, n-k)/(n+1-k) for 0 <= k <= n.

1, 3, 1, 10, 10, 1, 35, 70, 21, 1, 126, 420, 252, 36, 1, 462, 2310, 2310, 660, 55, 1, 1716, 12012, 18018, 8580, 1430, 78, 1, 6435, 60060, 126126, 90090, 25025, 2730, 105, 1, 24310, 291720, 816816, 816816, 340340, 61880, 4760, 136, 1, 92378, 1385670, 4988412, 6651216, 3879876, 1058148, 135660, 7752, 171, 1

Offset: 0

Author

Werner Schulte, Jan 12 2017

Keywords

• nonn
• easy
• tabl

Examples

			Triangle begins:
n\k:      0       1       2       3       4      5     6    7  8  . . .
  0:      1
  1:      3       1
  2:     10      10       1
  3:     35      70      21       1
  4:    126     420     252      36       1
  5:    462    2310    2310     660      55      1
  6:   1716   12012   18018    8580    1430     78     1
  7:   6435   60060  126126   90090   25025   2730   105    1
  8:  24310  291720  816816  816816  340340  61880  4760  136  1
  etc.
T(3,2) = binomial(7,5) * binomial(2,1) / (3+1-2) = 21 * 2 / 2 = 21. - _Indranil Ghosh_, Feb 15 2017
		

Links

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

Crossrefs

Row sums are A099250.

Triangle related to A000108, A097610, A280580.

Cf. A000007, A001003, A001006, A002426, A103365, A125558, A273055.

Programs

Mathematica

Table[Binomial[2n+1,2k+1] Binomial[2n-2k,n-k]/(n+1-k),{n,0,10},{k,0,n}]// Flatten (* Harvey P. Dale, Nov 25 2018 *)

Formula

T(n,k) = A097610(2*n+1, 2*k+1) = binomial(2*n+1, 2*k+1)*A000108(n-k) = A280580(n,k)*(2*n+1)/(2*k+1) for 0 <= k <= n.

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

Conjectures:

(1) Antidiagonal sums equal A001003(n+1);

(2) Sum_{k=0..n} (-1)^k*T(n,k)*(2*k+1)*A000108(k)*A103365(k) = A000007(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+1) = A273055(n).

Sum_{k=0..n} T(n,k)*(2*k+1)*A000108(k) = (2*n+1)*A000108(n)*A000108(n+1) = A125558(n+1).

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