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%I #19 Nov 30 2015 09:35:38
%S 1,1,-1,1,-2,1,1,-3,12,-1,1,-4,33,-128,1,1,-5,64,-731,1872,-1,1,-6,
%T 105,-2160,25857,-37600,1,1,-7,156,-4765,121600,-1311379,990784,-1,1,
%U -8,217,-8896,368145,-10138880,89060065,-32333824,1
%N Array A(i,j) read by antidiagonals: A(i,j) is the (2i-1)-th derivative of sin(sin(sin(...sin(x)))) nested j times evaluated at 0.
%C The determinant of the n X n such matrix has a closed form given in the formula section (and the Mathematica code below).
%C Rows appear to be given by polynomials (see formula section).
%F A(i,j) = ((d/dx)^(2i-1) sin^j(x))_{x=0}.
%F Let A_n denote the n X n such matrix. Then:
%F det(A_n)=(i^(n + n^2) 2^(-(1/12) + n^2) 3^(n/2 - n^2/2) G^3 (-(1/pi))^n B(1/2 + n) B(1 + n) B(3/2 + n))/e^(1/4), where B is the Barnes G-function and G is the Glaisher-Kinkelin constant (and i is the imaginary unit). (This can be shown by evaluating recurrence relations for det(A_n)). See Mathematica code below.
%F First row: 1.
%F Second row: -x.
%F Third row: x (5 x - 4).
%F Fourth row: -(1/3) x (164 + 7 x (-48 + 25 x)).
%F Fifth row: (8 - 7 x)^2 x (-24 + 25 x).
%F Sixth row: -(1/3) x (213568 - 766656 x + 1004696 x^2 - 572880 x^3 + 121275 x^4).
%F Seventh row: 1/3 x (-14371328 + 65012064 x - 111160192 x^2 + 91291200 x^3 - 36552516 x^4 + 5780775 x^5).
%F Second column: A003712.
%F Third column: A003715.
%e Evaluate the fifth derivative of sin(sin(sin(x))) at 0, which is 33. So the (3,3) entry of the array is 33. The array begins as:
%e | 1 1 1 1 1 1 |
%e | -1 -2 -3 -4 -5 -6 |
%e | 1 12 33 64 105 156 |
%e | -1 -128 -731 -2160 -4765 -8896 |
%e | 1 1872 25857 121600 368145 873936 |
%e | -1 -37600 -1311379 -10138880 -42807605 -130426016 |
%p A:= (i, j)-> (D@@(2*i-1))(sin@@j)(0):
%p seq(seq(A(i, 1+d-i), i=1..d), d=1..9); # _Alois P. Heinz_, May 14 2012
%t A[a_, b_] :=
%t A[a, b] =
%t Array[D[Nest[Sin, x, #2], {x, 2*#1 - 1}] /. x -> 0 &, {a, b}];
%t Print[A[7, 7] // MatrixForm];
%t Table2 = {};
%t k = 1;
%t While[k < 8, Table1 = {};
%t i = 1;
%t j = k;
%t While[0 < j,
%t AppendTo[Table1, First[Take[First[Take[A[7, 7], {i, i}]], {j, j}]]];
%t j = j - 1;
%t i = i + 1];
%t AppendTo[Table2, Table1];
%t k++];
%t Print[Flatten[Table2]]
%t Print[Table[Det[A[n, n]], {n, 1, 7}]];
%t Print[Table[(
%t I^(n + n^2) 2^(-(1/12) + n^2) 3^(n/2 - n^2/2)
%t Glaisher^3 (-(1/\[Pi]))^
%t n BarnesG[1/2 + n] BarnesG[1 + n] BarnesG[3/2 + n])/E^(1/4), {n, 1, 7}]]
%Y Cf. A003712, A003715.
%K sign,tabl,hard,nice
%O 1,5
%A _John M. Campbell_, May 12 2012
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