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A271025
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A(n, k) is the n-th binomial transform of the Catalan sequence (A000108) evaluated at k. Array read by descending antidiagonals for n >= 0 and k >= 0.
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2
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1, 1, 1, 2, 2, 1, 5, 5, 3, 1, 14, 15, 10, 4, 1, 42, 51, 37, 17, 5, 1, 132, 188, 150, 77, 26, 6, 1, 429, 731, 654, 371, 141, 37, 7, 1, 1430, 2950, 3012, 1890, 798, 235, 50, 8, 1, 4862, 12235, 14445, 10095, 4706, 1539, 365, 65, 9, 1, 16796, 51822, 71398, 56040, 28820, 10392, 2726, 537, 82, 10, 1
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OFFSET
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0,4
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COMMENTS
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Interestingly, the determinant of the n X n array of entries of the form A(i,j) is equal to the (n-1)-th superfactorial number (see A000178).
As indicated in A104455, the k-th binomial transform of A000108 will have:
o.g.f.: (1-sqrt((1-(k+4)*x)/(1-k*x)))/(2*x),
e.g.f.: exp((k+2)*x)*(BesselI(0,2x) - BesselI(1,2x)) and
a(n) = Sum_{i=0..n} binomial(n, i)*CatalanNumber(i)*k^(n-i).
The columns of this array are polynomial integer sequences. The successive polynomials corresponding to the columns of this array are: p0(n) = 1, p1(n) = n + 1, p2(n) = n^2 + 2n + 2, p3(n) = n^3 + 3*n^2 + 6*n + 5, p4(n) = n^4 + 4*n^3 + 12*n^2 + 20*n + 14, and so forth. The coefficients of these successive polynomials form a number triangle, which is given by A098474.
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LINKS
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FORMULA
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A(i,j) = Sum_{k=0..j} binomial(j,k)*A(i-1,k) for i >= 1.
A(i,j) = Sum_{k=0..j} binomial(j,k)*A000108(k)*i^(j-k).
A(n,k) = n^k*hypergeom([1/2, -k], [2], -4/n) for n >= 1.
A(n,k) = (2/Pi)*Integral_{x=-1..1}(k + 4*x^2)^(n - k)*sqrt(1 - x^2). (End)
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EXAMPLE
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The array given by integers of the form A(n,k) is illustrated below:
[0] 1, 1, 2, 5, 14, 42, 132, 429, 1430, ...
[1] 1, 2, 5, 15, 51, 188, 731, 2950, 12235, ...
[2] 1, 3, 10, 37, 150, 654, 3012, 14445, 71398, ...
[3] 1, 4, 17, 77, 371, 1890, 10095, 56040, 320795, ...
[4] 1, 5, 26, 141, 798, 4706, 28820, 182461, 1188406, ...
[5] 1, 6, 37, 235, 1539, 10392, 72267, 516474, 3783115, ...
[6] 1, 7, 50, 365, 2726, 20838, 162996, 1303485, 10642310, ...
[7] 1, 8, 65, 537, 4515, 38654, 337007, 2991340, 27013723, ...
[8] 1, 9, 82, 757, 7086, 67290, 648420, 6340365, 62893270, ...
[9] 1, 10, 101, 1031, 10643, 111156, 1174875, 12568686, 136080971, ...
Seen as a triangle:
1
1, 1
2, 2, 1
5, 5, 3, 1
14, 15, 10, 4, 1
42, 51, 37, 17, 5, 1
132, 188, 150, 77, 26, 6, 1
429, 731, 654, 371, 141, 37, 7, 1
1430, 2950, 3012, 1890, 798, 235, 50, 8, 1
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MAPLE
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A := (n, k) -> (2/Pi)*int((k+4*x^2)^(n-k)*sqrt(1 - x^2), x=-1..1):
for n from 0 to 9 do seq(A(n, k), k=0..n) od; # Peter Luschny, Jan 27 2020
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MATHEMATICA
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A000108[n_]:= Binomial[2*n, n]/(n+1) ;
T[i_, j_]: Sum[Binomial(j, k)*A000108(k)*i^(j-k), {k, 0, j}] ;
A[0, k_] := CatalanNumber[k]; A[n_, k_] := n^k*Hypergeometric2F1[1/2, -k, 2, -4/n];
Table[A[n, k], {n, 0, 6}, {k, 0, 8}] (* Peter Luschny, Jan 27 2020 *)
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PROG
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(Sage) def A000108(n): return binomial(2*n, n)/(n+1) ;
def T(i, j): return sum(binomial(j, k)*A000108(k)*i^(j-k) for k in range(j+1))
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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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