OFFSET
0,2
LINKS
Indranil Ghosh, Rows 0..100 of triangle, flattened
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);
(3) Matrix inverse equals T(n,k)*A103365(n+1-k).
Matrix product: Sum_{i=0..n} T(n,i)*T(i,k) = T(n,k)*A000108(n+1-k) for 0<=k<=n.
EXAMPLE
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
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 *)
CROSSREFS
KEYWORD
AUTHOR
Werner Schulte, Jan 12 2017
STATUS
approved