A336239 - OEIS

%I #14 Jul 13 2020 17:54:03

%S 0,16,-144,36,2304,-960,96,-57600,24000,-3600,200,2073600,-864000,

%T 146880,-11520,360,-101606400,42336000,-7197120,635040,-29400,588,

%U 6502809600,-2709504000,460615680,-41932800,2204160,-64512,896

%N Triangle read by rows, coefficients of polynomials in Pi^2, given by trigonometric double integrals over the unit square.

%C Formulas given by Srinivasa Raghava, see the picture in the links section.

%H Srinivasa Raghava, <a href="/A336239/a336239.jpg">Trigonometric double integrals</a>.

%F T(n, k) = Pi^(2*n-1) * [Pi^(2*k)] Integral_{x=0..1, y=0..1} sin(-Pi*(x^(1/n) + y^(1/n)) for n >= 2 and k = 0..n-2.

%e Triangle starts:

%e                              [1] 0

%e                             [2] 16

%e                          [3] -144, 36

%e                       [4] 2304, -960, 96

%e                  [5] -57600, 24000, -3600, 200

%e            [6] 2073600, -864000, 146880, -11520, 360

%e     [7] -101606400, 42336000, -7197120, 635040, -29400, 588

%e [8] 6502809600, -2709504000, 460615680, -41932800, 2204160, -64512, 896

%e .

%e The first few polynomials are:

%e p_1(Pi) = 0;

%e p_2(Pi) = 16;

%e p_3(Pi) = 36*(Pi^2 - 4);

%e p_4(Pi) = 96*(Pi^4 - 10*Pi^2 + 24);

%e p_5(Pi) = 200*(Pi^6 - 18*Pi^4 + 120*Pi^2 - 288);

%e p_6(Pi) = 360*(Pi^8 - 32*Pi^6 + 408*Pi^4 - 2400*Pi^2 + 5760);

%p dblint := n -> Pi^(2*n-1)*int(sin(-Pi*(x^(1/n) + y^(1/n))), [x=0..1, y=0..1]):

%p Trow := proc(n) if n = 1 then return 0 fi; dblint(n);

%p seq(coeff(%, Pi, 2*k), k=0..n-2) end: seq(Trow(n), n=1..8);

%K sign,tabf

%O 1,2

%A _Peter Luschny_, Jul 13 2020