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A280580
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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.
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2
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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
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OFFSET
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0,4
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LINKS
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FORMULA
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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)!)).
(2) Antidiagonal sums equal A001003(n);
(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.
T(n,k) = A097610(2*n,2*k) for 0 <= k <= n.
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EXAMPLE
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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
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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