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A127606
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a(n) = 2^(2*n*n) * Product_{1<=i,j<=n} (cos(i*Pi/(2*n+1))^2 + sin(j*Pi/(2*n+1))^2).
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3
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1, 4, 176, 79808, 372713728, 17931360207872, 8887976555024756736, 45390122553039546330628096, 2388340820825093234015277927170048, 1294826675280341699389150405743029631844352
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OFFSET
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0,2
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COMMENTS
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LINKS
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FORMULA
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a(n) = 2^n * sqrt(Resultant(T_{2*n+1}(i*x/2), U_{2*n}(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). - Seiichi Manyama, Jan 09 2021
a(n) ~ 2^(1/8) * exp(G*(2*n + 1)^2/Pi) / (1 + sqrt(2))^(n + 1/2), where G is Catalan's constant A006752. - Vaclav Kotesovec, Mar 18 2023
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MAPLE
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for n from 0 to 12 do a[n]:=2^(2*n*n)*product(product(cos(i*Pi/(2*n+1))^2+ sin(j*Pi/(2*n+1))^2, j=1..n), i=1..n) od: seq(round(evalf(a[n], 300)), n=0..12);
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MATHEMATICA
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Table[2^(2*n^2) * Product[Product[Cos[i*Pi/(2*n + 1)]^2 + Sin[j*Pi/(2*n + 1)]^2, {i, 1, n}], {j, 1, n}], {n, 0, 15}] // Round (* Vaclav Kotesovec, Mar 18 2023 *)
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PROG
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(PARI) default(realprecision, 120);
{a(n) = round(prod(i=1, n, prod(j=1, n, 4*cos(i*Pi/(2*n+1))^2+4*sin(j*Pi/(2*n+1))^2)))} \\ Seiichi Manyama, Dec 31 2020
(PARI) {a(n) = sqrtint(4^n*polresultant(polchebyshev(2*n+1, 1, I*x/2), polchebyshev(2*n, 2, x/2)))} \\ Seiichi Manyama, Jan 09 2021
(Python)
from math import isqrt
from sympy.abc import x
from sympy import resultant, chebyshevt, chebyshevu, I
def A127606(n): return isqrt(resultant(chebyshevt((n<<1)+1, I*x/2), chebyshevu(n<<1, x/2)))<<n if n else 1 # Chai Wah Wu, Nov 07 2023
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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