Summary

You searched for: superseeker=9,242/3

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1

New Number: 8.11 |  AESZ: 162  |  Superseeker: 9 242/3  |  Hash: 542708b59b898c35f43e00120897ff8d  

Degree: 8

\(\theta^4-3 x(3\theta^2+3\theta+1)(10\theta^2+10\theta+3)+3^{3} x^{2}\left(91\theta^4+472\theta^3+659\theta^2+374\theta+81\right)+3^{6} x^{3}\left(30\theta^4-180\theta^3-551\theta^2-417\theta-111\right)-3^{8} x^{4}\left(200\theta^4+400\theta^3-514\theta^2-714\theta-237\right)+3^{11} x^{5}\left(30\theta^4+300\theta^3+169\theta^2-25\theta-35\right)+3^{13} x^{6}\left(91\theta^4-108\theta^3-211\theta^2-108\theta-15\right)-3^{16} x^{7}(3\theta^2+3\theta+1)(10\theta^2+10\theta+3)+3^{20} x^{8}\left((\theta+1)^4\right)\)

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Coefficients of the holomorphic solution: 1, 9, 135, 1953, 5751, ...
--> OEIS
Normalized instanton numbers (n0=1): 9, -153/4, 242/3, -4923, 34245, ... ; Common denominator:...

Discriminant

\((27z^2-9z+1)(2187z^2-81z+1)(-1+243z^2)^2\)

Local exponents

\(-\frac{ 1}{ 27}\sqrt{ 3}\)\(0\)\(\frac{ 1}{ 54}-\frac{ 1}{ 162}\sqrt{ 3}I\)\(\frac{ 1}{ 54}+\frac{ 1}{ 162}\sqrt{ 3}I\)\(\frac{ 1}{ 27}\sqrt{ 3}\)\(\frac{ 1}{ 6}-\frac{ 1}{ 18}\sqrt{ 3}I\)\(\frac{ 1}{ 6}+\frac{ 1}{ 18}\sqrt{ 3}I\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(1\)
\(1\)\(0\)\(1\)\(1\)\(1\)\(1\)\(1\)\(1\)
\(3\)\(0\)\(1\)\(1\)\(3\)\(1\)\(1\)\(1\)
\(4\)\(0\)\(2\)\(2\)\(4\)\(2\)\(2\)\(1\)

Note:

Hadamard product $c \ast f$. 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 over
$Q(\sqrt{?})$

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