Summary

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New Number: 8.3 |  AESZ: 105  |  Superseeker: 8 -104  |  Hash: 7b27135451cf2016217211c633b7ab83  

Degree: 8

\(\theta^4-2^{2} x(3\theta^2+3\theta+1)(7\theta^2+7\theta+2)+2^{5} 3 x^{2}\left(15\theta^4+28\theta^3+39\theta^2+22\theta+4\right)-2^{10} x^{3}\left(21\theta^4-126\theta^3-386\theta^2-291\theta-76\right)+2^{14} x^{4}\left(37\theta^4+74\theta^3+50\theta^2+13\theta+6\right)+2^{18} x^{5}\left(21\theta^4+210\theta^3+118\theta^2-19\theta-24\right)+2^{21} 3 x^{6}\left(15\theta^4+32\theta^3+45\theta^2+32\theta+8\right)+2^{26} x^{7}(3\theta^2+3\theta+1)(7\theta^2+7\theta+2)+2^{32} x^{8}\left((\theta+1)^4\right)\)

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Coefficients of the holomorphic solution: 1, 8, 200, 6272, 233896, ...
--> OEIS
Normalized instanton numbers (n0=1): 8, 71/2, -104, 4202, 50112, ... ; Common denominator:...

Discriminant

\((8z+1)(64z-1)(4z+1)(32z-1)(1+256z^2)^2\)

Local exponents

\(-\frac{ 1}{ 4}\)\(-\frac{ 1}{ 8}\)\(0-\frac{ 1}{ 16}I\)\(0\)\(0+\frac{ 1}{ 16}I\)\(\frac{ 1}{ 64}\)\(\frac{ 1}{ 32}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(1\)
\(1\)\(1\)\(1\)\(0\)\(1\)\(1\)\(1\)\(1\)
\(1\)\(1\)\(3\)\(0\)\(3\)\(1\)\(1\)\(1\)
\(2\)\(2\)\(4\)\(0\)\(4\)\(2\)\(2\)\(1\)

Note:

Hadamard product $a \ast d$. This operator has a second MUM-point at infinity with the same instanton numbers.
It can be reduced to an operator of degree 4 with a single MUM-point defined over
$Q(\sqrt{-1})$.

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