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A189231
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Extended Catalan triangle read by rows.
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8
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1, 1, 1, 1, 2, 1, 3, 2, 3, 1, 2, 8, 3, 4, 1, 10, 5, 15, 4, 5, 1, 5, 30, 9, 24, 5, 6, 1, 35, 14, 63, 14, 35, 6, 7, 1, 14, 112, 28, 112, 20, 48, 7, 8, 1, 126, 42, 252, 48, 180, 27, 63, 8, 9, 1, 42, 420, 90, 480, 75, 270, 35, 80, 9, 10, 1, 462, 132, 990, 165, 825, 110, 385, 44, 99, 10, 11, 1
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OFFSET
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0,5
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COMMENTS
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Let S(n,k) denote the coefficients of the positive powers of the Laurent polynomials C_n(x) = (x+1/x)^(n-1)*(x-1/x)*(x+1/x+n) (if n>0) and C_0(x) = 0.
Then T(n,k) = S(n+1,k+1) for n>=0, k>=0.
The classical Catalan triangle A053121(n,k) can be recovered from this triangle by setting T(n,k) = 0 if n-k is odd.
The complementary Catalan triangle A189230(n,k) can be recovered from this triangle by setting T(n,k) = 0 if n-k is even.
T(n,0) are the extended Catalan numbers A057977(n).
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LINKS
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FORMULA
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Recurrence: If k>n or k<0 then T(n,k) = 0 else if n=k then T(n,k) = 1; otherwise T(n,k) = T(n-1,k-1) + ((n-k) mod 2)*T(n-1,k) + T(n-1,k+1).
S(n,k) = (k/n)* A162246(n,k) for n>0 where S(n,k) are the coefficients from the definition provided the triangle A162246 is indexed in Laurent style by the recurrence: if abs(k) > n then A162246(n,k) = 0 else if n = k then A162246(n,k) = 1 and otherwise A162246(n,k) = A162246(n-1,k-1)+ modp(n-k,2) * A162246(n-1,k) + A162246(n-1,k+1).
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EXAMPLE
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The Laurent polynomials:
C(0,x) = 0
C(1,x) = x - 1/x
C(2,x) = x^2 + x - 1/x - 1/x^2
C(3,x) = x^3 + 2 x^2 + x - 1/x - 2/x^2 -1/x^3
Triangle T(n,k) = S(n+1,k+1) starts
[0] 1,
[1] 1, 1,
[2] 1, 2, 1,
[3] 3, 2, 3, 1,
[4] 2, 8, 3, 4, 1,
[5] 10, 5, 15, 4, 5, 1,
[6] 5, 30, 9, 24, 5, 6, 1,
[7] 35, 14, 63, 14, 35, 6, 7, 1,
[0],[1],[2],[3],[4],[5],[6],[7]
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MAPLE
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A189231_poly := (n, x)-> `if`(n=0, 0, (x+1/x)^(n-2)*(x-1/x)*(x+1/x+n-1)):
seq(print([n], seq(coeff(expand(A189231_poly(n, x)), x, k), k=1..n)), n=1..9);
A189231 := proc(n, k) option remember; `if`(k>n or k<0, 0, `if`(n=k, 1, A189231(n-1, k-1)+modp(n-k, 2)*A189231(n-1, k)+A189231(n-1, k+1))) end:
seq(print(seq(A189231(n, k), k=0..n)), n=0..9);
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MATHEMATICA
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t[n_, k_] /; (k > n || k < 0) = 0; t[n_, n_] = 1; t[n_, k_] := t[n, k] = t[n-1, k-1] + Mod[n-k, 2]*t[n-1, k] + t[n-1, k+1]; Table[t[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 30 2013 *)
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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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