A377657 - OEIS

%I #23 Nov 13 2024 17:22:00

%S 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,

%T 1044,55,7,0,729,76250,135779,33300,2365,78,8,0,2187,722250,2606261,

%U 1070244,113311,4654,105,9,0,6561,6841250,50028755,34420356,5476405,312390,8295,136,10

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

%C Based on an observation made by _Fredrik Johansson_ about A376777, which is the main diagonal of this array.

%F 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

%e Array begins

%e   [0] 1,    2,       3,         4,           5,            6, ... A000027

%e   [1] 0,    3,      10,        21,          36,           55, ... A014105

%e   [2] 0,    9,      90,       371,        1044,         2365, ... A377858

%e   [3] 0,   27,     850,      7077,       33300,       113311, ... A376778

%e   [4] 0,   81,    8050,    135779,     1070244,      5476405, ...

%e   [5] 0,  243,   76250,   2606261,    34420356,    264893255, ...

%e   [6] 0,  729,  722250,  50028755,  1107069876,  12813875437, ...

%e   [7] 0, 2187, 6841250, 960335173, 35607151476, 619859803695, ...

%e   .

%e Seen as a triangle T(n, k) = A(n-k, k):

%e   [0] 1;

%e   [1] 0,    2;

%e   [2] 0,    3,      3;

%e   [3] 0,    9,     10,       4;

%e   [4] 0,   27,     90,      21,       5;

%e   [5] 0,   81,    850,     371,      36,      6;

%e   [6] 0,  243,   8050,    7077,    1044,     55,    7;

%e   [7] 0,  729,  76250,  135779,   33300,   2365,   78,   8;

%e   [8] 0, 2187, 722250, 2606261, 1070244, 113311, 4654, 105, 9;

%p A := (n, k) -> add(tan(j*Pi/(1 + 2*k))^(2*n), j = 0..k):

%p seq(print(seq(round(evalf(A(n, k), 32)), k = 0..6)), n = 0..7);

%o (PARI)

%o 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

%Y Rows: A000027, A014105, A377858, A376778.

%Y Columns: A376478.

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

%Y Cf. A091042.

%K nonn,tabl

%O 0,3

%A _Peter Luschny_, Nov 10 2024