A377657 Array read by ascending antidiagonals: A(n, k) = Sum_{j=0..k} tan(j*Pi/(1 + 2*k))^(2*n).

1, 0, 2, 0, 3, 3, 0, 9, 10, 4, 0, 27, 90, 21, 5, 0, 81, 850, 371, 36, 6, 0, 243, 8050, 7077, 1044, 55, 7, 0, 729, 76250, 135779, 33300, 2365, 78, 8, 0, 2187, 722250, 2606261, 1070244, 113311, 4654, 105, 9, 0, 6561, 6841250, 50028755, 34420356, 5476405, 312390, 8295, 136, 10

Offset: 0

Author

Peter Luschny, Nov 10 2024

Keywords

• nonn
• tabl

Comments

Based on an observation made by Fredrik Johansson about A376777, which is the main diagonal of this array.

Examples

			Array begins
  [0] 1,    2,       3,         4,           5,            6, ... A000027
  [1] 0,    3,      10,        21,          36,           55, ... A014105
  [2] 0,    9,      90,       371,        1044,         2365, ... A377858
  [3] 0,   27,     850,      7077,       33300,       113311, ... A376778
  [4] 0,   81,    8050,    135779,     1070244,      5476405, ...
  [5] 0,  243,   76250,   2606261,    34420356,    264893255, ...
  [6] 0,  729,  722250,  50028755,  1107069876,  12813875437, ...
  [7] 0, 2187, 6841250, 960335173, 35607151476, 619859803695, ...
  .
Seen as a triangle T(n, k) = A(n-k, k):
  [0] 1;
  [1] 0,    2;
  [2] 0,    3,      3;
  [3] 0,    9,     10,       4;
  [4] 0,   27,     90,      21,       5;
  [5] 0,   81,    850,     371,      36,      6;
  [6] 0,  243,   8050,    7077,    1044,     55,    7;
  [7] 0,  729,  76250,  135779,   33300,   2365,   78,   8;
  [8] 0, 2187, 722250, 2606261, 1070244, 113311, 4654, 105, 9;
		

Crossrefs

Rows: A000027, A014105, A377858, A376778.

Columns: A376478.

Cf. A376777 (main diagonal), A377658 (antidiagonal sums).

Cf. A091042.

Programs

Maple

A := (n, k) -> add(tan(j*Pi/(1 + 2*k))^(2*n), j = 0..k):
seq(print(seq(round(evalf(A(n, k), 32)), k = 0..6)), n = 0..7);

PARI

A(n, k) = {trace(matcompanion(sum(m=0, k, x^m*binomial(2*k+1, 2*(k-m))*(-1)^(m+1)))^n)+(n==0) } \\ Thomas Scheuerle, Nov 11 2024

Formula

Row n of A091042(n, k) = binomial(2*n+1, 2*k) gives the polynomial Pe(n, x), with zeros in -tan(Pi/2*n+1)^2, -tan(2*Pi/2*n+1)^2, ..., -tan(n*Pi/2*n+1)^2. Let Pm(n, k, x) be the polynomial with zeros in (-tan(Pi/2*n+1)^2)^k, (-tan(2*Pi/2*n+1)^2)^k, ..., (-tan(n*Pi/2*n+1)^2)^k, then A(k, n) is the coefficient of X^(n-1) in the polynomial Pm(n, k, x). A way to do this calculation without evaluation of irrational numbers is to obtain the companion matrix M of the polynomial Pe(n, x), then A(k, n) = tr(M^k) (the trace of M^k). - Thomas Scheuerle, Nov 11 2024