Summary

You searched for: superseeker=9,-1229/3

Your search produced exactly one match

1

New Number: 8.8 |  AESZ: 161  |  Superseeker: 9 -1229/3  |  Hash: 641d1de9a6564241575c5db52faef694  

Degree: 8

\(\theta^4-3 x(3\theta^2+3\theta+1)(11\theta^2+11\theta+3)+3^{2} x^{2}\left(366\theta^4+1428\theta^3+1980\theta^2+1104\theta+221\right)-3^{4} x^{3}\left(33\theta^4-198\theta^3-607\theta^2-456\theta-117\right)+3^{5} x^{4}\left(726\theta^4+1452\theta^3-978\theta^2-1704\theta-515\right)+3^{7} x^{5}\left(33\theta^4+330\theta^3+185\theta^2-32\theta-37\right)+3^{8} x^{6}\left(366\theta^4+36\theta^3-108\theta^2+36\theta+35\right)+3^{10} x^{7}(3\theta^2+3\theta+1)(11\theta^2+11\theta+3)+3^{12} x^{8}\left((\theta+1)^4\right)\)

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Coefficients of the holomorphic solution: 1, 9, 171, 3087, 11259, ...
--> OEIS
Normalized instanton numbers (n0=1): 9, -81/4, -1229/3, -4644, -26685, ... ; Common denominator:...

Discriminant

\((729z^4+2673z^3+3240z^2-99z+1)(1+27z^2)^2\)

Local exponents

≈\(-1.848362\) ≈\(-1.848362\)\(0-\frac{ 1}{ 9}\sqrt{ 3}I\)\(0\)\(0+\frac{ 1}{ 9}\sqrt{ 3}I\) ≈\(0.015028\) ≈\(0.015028\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(1\)
\(1\)\(1\)\(1\)\(0\)\(1\)\(1\)\(1\)\(1\)
\(1\)\(1\)\(3\)\(0\)\(3\)\(1\)\(1\)\(1\)
\(2\)\(2\)\(4\)\(0\)\(4\)\(2\)\(2\)\(1\)

Note:

Hadamard product $b \qst f$. 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 $Q(\sqrt{\})$

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