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A212267
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Array A(i,j) read by antidiagonals: A(i,j) is the (2*i-1)-th derivative of tan(tan(tan(...tan(x)))) nested j times evaluated at 0.
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0
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1, 1, 2, 1, 4, 16, 1, 6, 72, 272, 1, 8, 168, 2896, 7936, 1, 10, 304, 10672, 203904, 353792, 1, 12, 480, 26400, 1198080, 22112000, 22368256, 1, 14, 696, 52880, 4071040, 208521728, 3412366336, 1903757312, 1, 16, 952, 92912, 10373760, 976629760, 51874413568, 709998153728, 209865342976
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OFFSET
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1,3
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COMMENTS
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The determinant of the n X n such matrix has a closed form given in the Mathematica code below.
Rows appear to be given by polynomials (see formula section).
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LINKS
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FORMULA
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A(i,j) = ((d/dx)^(2i-1) tan^j(x))_{x=0}.
Third row: n*(5*n - 1)*4 = 8*A005476(n).
Fourth row: 8/3*n*(11 - 84*n + 175*n^2).
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EXAMPLE
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Array A(i,j) begins:
. 1, 1, 1, 1, 1, ...
. 2, 4, 6, 8, 10, ...
. 16, 72, 168, 304, 480, ...
. 272, 2896, 10672, 26400, 52880, ...
. 7936, 203904, 1198080, 4071040, 10373760, ...
. 353792, 22112000, 208521728, 976629760, 3172514560, ...
Evaluate the (2*3-1)th derivate of tan(tan(tan(x))) at 0, which is 168. Thus A(3,3)=168.
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MAPLE
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A:= (i, j)-> (D@@(2*i-1))(tan@@j)(0):
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MATHEMATICA
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A[a_, b_] :=
A[a, b] =
Array[D[Nest[Tan, x, #2], {x, 2*#1 - 1}] /. x -> 0 &, {a, b}];
Print[A[7, 7] // MatrixForm];
Table2 = {};
k = 1;
While[k < 8, Table1 = {};
i = 1;
j = k;
While[0 < j,
AppendTo[Table1, First[Take[First[Take[A[7, 7], {i, i}]], {j, j}]]];
j = j - 1;
i = i + 1];
AppendTo[Table2, Table1];
k++];
Print[Flatten[Table2]];
Print[Table[Det[A[n, n]], {n, 1, 7}]];
Table[(2^(11/12 +
1/2 (5 + 3 (-1 + n)) (-1 + n)) 3^(-(1/2) (-1 +
n) n) Glaisher^3 \[Pi]^-n BarnesG[1/2 + n] BarnesG[1 + n] BarnesG[3/2 + n])/E^(1/4), {n, 1, 7}]
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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