A340182 a(n) = Product_{1<=j,k,m<=n} (4*cos(j*Pi/(2*n+1))^2 + 4*cos(k*Pi/(2*n+1))^2 + 4*cos(m*Pi/(2*n+1))^2).
1, 3, 61731, 220157391087140625, 3109768877542258728107559478225309328087616
Offset: 0
Keywords
Programs
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Mathematica
Round[Table[4^(n^3) * Product[Cos[j*Pi/(2*n + 1)]^2 + Cos[k*Pi/(2*n + 1)]^2 + Cos[m*Pi/(2*n + 1)]^2, {j, 1, n}, {k, 1, n}, {m, 1, n}], {n, 0, 5}]] (* or *) Round[Table[2^(n^3) * Product[3 + Cos[2*j*Pi/(2*n + 1)] + Cos[2*k*Pi/(2*n + 1)] + Cos[2*m*Pi/(2*n + 1)], {j, 1, n}, {k, 1, n}, {m, 1, n}], {n, 0, 5}]] (* or *) Round[Table[Product[u = Sqrt[Cos[j*Pi/(2*n + 1)]^2 + Cos[k*Pi/(2*n + 1)]^2]; (((u + Sqrt[1 + u^2])^(2*n + 1) - (u - Sqrt[1 + u^2])^(2*n + 1))/(2*Sqrt[1 + u^2])), {j, 1, n}, {k, 1, n}], {n, 0, 5}]] (* Vaclav Kotesovec, Jan 04 2021 *) -
PARI
default(realprecision, 500); {a(n) = round(prod(j=1, n, prod(k=1, n, prod(m=1, n, 4*cos(j*Pi/(2*n+1))^2+4*cos(k*Pi/(2*n+1))^2+4*cos(m*Pi/(2*n+1))^2))))}
Formula
From Vaclav Kotesovec, Jan 04 2021: (Start)
a(n) ~ c * d^n * s^(n^2) * r^(n^3), where
r = exp(8*A340322/Pi^3) = exp((8/Pi^3) * Integral_{x=0..Pi/2, y=0..Pi/2, z=0..Pi/2} log(4*cos(x)^2 + 4*cos(y)^2 + 4*cos(z)^2) dx dy dz) = 5.3302028892051674211345979966496595201084467305922855029660919024805225841...
s = 0.57208914727550556482486188829703578692890272003698306852389010626941042...
d = 0.91012013388841787275362130594290903074302493828277326742531159...
c = 1.057086458532774496412062406469810663638243576302292119... (End)
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