A212267 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.

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

Offset: 1

Author

John M. Campbell, May 12 2012

Keywords

• nonn
• tabl
• hard

Comments

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).

Examples

			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.
		

Links

• Eric Weisstein's World of Mathematics, Nested Function

Crossrefs

Columns j=1-3 give: A000182, A003718, A003720.

Programs

Maple

A:= (i, j)-> (D@@(2*i-1))(tan@@j)(0):
seq(seq(A(i, 1+d-i), i=1..d), d=1..8); # Alois P. Heinz, May 13 2012

Mathematica

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}]

Formula

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).

Extensions

More terms from Alois P. Heinz, May 13 2012