Summary

You searched for: inst=-96

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1

New Number: 2.17 |  AESZ: 111  |  Superseeker: 32 1440  |  Hash: d8535e0f3d0bfd4ebcc9c042df43c218  

Degree: 2

\(\theta^4-2^{4} x(2\theta+1)^2(8\theta^2+8\theta+3)+2^{12} x^{2}(2\theta+1)^2(2\theta+3)^2\)

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Coefficients of the holomorphic solution: 1, 48, 5904, 940800, 169520400, ...
--> OEIS
Normalized instanton numbers (n0=1): 32, -96, 1440, 19704, -14496, ... ; Common denominator:...

Discriminant

\((256z-1)^2\)

Local exponents

\(0\)\(\frac{ 1}{ 256}\)\(\infty\)
\(0\)\(0\)\(\frac{ 1}{ 2}\)
\(0\)\(\frac{ 1}{ 2}\)\(\frac{ 1}{ 2}\)
\(0\)\(\frac{ 1}{ 2}\)\(\frac{ 3}{ 2}\)
\(0\)\(1\)\(\frac{ 3}{ 2}\)

Note:

B-Incarnations:
Fibre product 81111- x 18--21, 4*11-- x 53211,
Double Octics: D.O.8, D.O.36, D.O.73, D.O.249, D.O.258,D.O.265

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2

New Number: 8.13 |  AESZ: 163  |  Superseeker: 12 3020/3  |  Hash: e21fd830a9dca03305deb8363a26fcf2  

Degree: 8

\(\theta^4-2^{2} 3 x\left((3\theta^2+3\theta+1)^2\right)+2^{4} 3^{2} x^{2}\left(21\theta^4+156\theta^3+219\theta^2+126\theta+29\right)+2^{7} 3^{4} x^{3}(3\theta^2+3\theta+1)(3\theta^2-21\theta-35)-2^{10} 3^{5} x^{4}\left(27\theta^4+54\theta^3-114\theta^2-141\theta-49\right)+2^{12} 3^{7} x^{5}(3\theta^2+3\theta+1)(3\theta^2+27\theta-11)+2^{14} 3^{8} x^{6}\left(21\theta^4-72\theta^3-123\theta^2-72\theta-13\right)-2^{17} 3^{10} x^{7}\left((3\theta^2+3\theta+1)^2\right)+2^{20} 3^{12} x^{8}\left((\theta+1)^4\right)\)

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Coefficients of the holomorphic solution: 1, 12, 180, 2352, 6084, ...
--> OEIS
Normalized instanton numbers (n0=1): 12, -96, 3020/3, -71493/4, 319584, ... ; Common denominator:...

Discriminant

\((1728z^2-72z+1)(432z^2-36z+1)(-1+864z^2)^2\)

Local exponents

\(-\frac{ 1}{ 72}\sqrt{ 6}\)\(0\)\(\frac{ 1}{ 48}-\frac{ 1}{ 144}\sqrt{ 3}I\)\(\frac{ 1}{ 48}+\frac{ 1}{ 144}\sqrt{ 3}I\)\(\frac{ 1}{ 72}\sqrt{ 6}\)\(\frac{ 1}{ 24}-\frac{ 1}{ 72}\sqrt{ 3}I\)\(\frac{ 1}{ 24}+\frac{ 1}{ 72}\sqrt{ 3}I\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(1\)
\(1\)\(0\)\(1\)\(1\)\(1\)\(1\)\(1\)\(1\)
\(3\)\(0\)\(1\)\(1\)\(3\)\(1\)\(1\)\(1\)
\(4\)\(0\)\(2\)\(2\)\(4\)\(2\)\(2\)\(1\)

Note:

Hadamard product $d \ast f$. This operator has a second MUM-point at infinity with the same instanton numbers. Itg can be reduced to an operator of degree 4 with a single MUM-point defined over $Q(\sqrt{?})$.

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