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%I #21 Jan 09 2021 07:35:11
%S 1,1,1,1,4,1,1,19,11,1,1,91,176,29,1,1,436,2911,1471,76,1,1,2089,
%T 48301,79808,11989,199,1,1,10009,801701,4375897,2091817,97021,521,1,1,
%U 47956,13307111,240378643,372713728,53924597,783511,1364,1
%N Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where T(n,k) = Product_{a=1..n} Product_{b=1..k} (4*sin(a*Pi/(2*n+1))^2 + 4*cos(b*Pi/(2*k+1))^2).
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Chebyshev_polynomials">Chebyshev polynomials</a>
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Resultant">Resultant</a>
%F T(n,k) = 2^k * sqrt(Resultant(T_{2*n+1}(i*x/2), U_{2*k}(x/2))), where T_n(x) is a Chebyshev polynomial of the first kind, U_n(x) is a Chebyshev polynomial of the second kind and i = sqrt(-1).
%e Square array begins:
%e 1, 1, 1, 1, 1, ...
%e 1, 4, 19, 91, 436, ...
%e 1, 11, 176, 2911, 48301, ...
%e 1, 29, 1471, 79808, 4375897, ...
%e 1, 76, 11989, 2091817, 372713728, ...
%o (PARI) default(realprecision, 120);
%o {T(n, k) = round(prod(a=1, n, prod(b=1, k, 4*sin(a*Pi/(2*n+1))^2+4*cos(b*Pi/(2*k+1))^2)))}
%o (PARI) {T(n, k) = sqrtint(4^k*polresultant(polchebyshev(2*n+1, 1, I*x/2), polchebyshev(2*k, 2, x/2)))}
%Y Column k=0..1 give A000012, A002878.
%Y Row n=0..7 give A000012, A004253(n+1), A003729, A028478, A028480, A028482, A028484, A028486.
%Y Main diagonal gives A127606.
%Y Cf. A187617, A340475.
%K nonn,tabl
%O 0,5
%A _Seiichi Manyama_, Jan 09 2021