Summary

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New Number: 8.1 |  AESZ: 102  |  Superseeker: 8 1053  |  Hash: e928905653beb9d844e6a942f50d94ac  

Degree: 8

\(\theta^4-x(7\theta^2+7\theta+2)(11\theta^2+11\theta+3)-x^{2}\left(1049\theta^4+4100\theta^3+5689\theta^2+3178\theta+640\right)+2^{3} x^{3}\left(77\theta^4-462\theta^3-1420\theta^2-1053\theta-252\right)+2^{4} x^{4}\left(1041\theta^4+2082\theta^3-1406\theta^2-2447\theta-746\right)+2^{6} x^{5}\left(77\theta^4+770\theta^3+428\theta^2-93\theta-80\right)-2^{6} x^{6}\left(1049\theta^4+96\theta^3-317\theta^2+96\theta+100\right)-2^{9} x^{7}(7\theta^2+7\theta+2)(11\theta^2+11\theta+3)+2^{12} x^{8}\left((\theta+1)^4\right)\)

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Coefficients of the holomorphic solution: 1, 6, 190, 8232, 432846, ...
--> OEIS
Normalized instanton numbers (n0=1): 8, 153/2, 1053, 49101/2, 670214, ... ; Common denominator:...

Discriminant

\((64z^2+88z-1)(z^2-11z-1)(-1+8z^2)^2\)

Local exponents

\(-\frac{ 11}{ 16}-\frac{ 5}{ 16}\sqrt{ 5}\)\(-\frac{ 1}{ 4}\sqrt{ 2}\)\(\frac{ 11}{ 2}-\frac{ 5}{ 2}\sqrt{ 5}\)\(0\)\(-\frac{ 11}{ 16}+\frac{ 5}{ 16}\sqrt{ 5}\)\(\frac{ 1}{ 4}\sqrt{ 2}\)\(\frac{ 11}{ 2}+\frac{ 5}{ 2}\sqrt{ 5}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(1\)
\(1\)\(1\)\(1\)\(0\)\(1\)\(1\)\(1\)\(1\)
\(1\)\(3\)\(1\)\(0\)\(1\)\(3\)\(1\)\(1\)
\(2\)\(4\)\(2\)\(0\)\(2\)\(4\)\(2\)\(1\)

Note:

Hadamard product $a \ast b$. The operator has a second MUM-point at infinity with the same instanton numbers. In fact, there is a symmetry in the operator. It can be reduced to an operator with a single MUM point of degree 4, defined over $Q(\sqrt{2})$.

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