A212261 - OEIS

%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