Summary

You searched for: inst=162

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New Number: 5.71 |  AESZ: 290  |  Superseeker: 162 751026  |  Hash: 5552195a371df176b84ac2c2d791be7e  

Degree: 5

\(\theta^4+3 x\left(279\theta^4-252\theta^3-160\theta^2-34\theta-3\right)+2 3^{5} x^{2}\left(423\theta^4-468\theta^3+457\theta^2+215\theta+37\right)+2 3^{9} x^{3}\left(531\theta^4+1296\theta^3+1243\theta^2+567\theta+104\right)+3^{15} 5 x^{4}\left(51\theta^4+120\theta^3+126\theta^2+66\theta+14\right)+3^{20} 5^{2} x^{5}\left((\theta+1)^4\right)\)

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Coefficients of the holomorphic solution: 1, 9, -837, -32553, 4787019, ...
--> OEIS
Normalized instanton numbers (n0=1): 162, -8829, 751026, -163125009/2, 10343901204, ... ; Common denominator:...

Discriminant

\((27z+1)(19683z^2+1)(1+405z)^2\)

Local exponents

\(-\frac{ 1}{ 27}\)\(-\frac{ 1}{ 405}\)\(0-\frac{ 1}{ 243}\sqrt{ 3}I\)\(0\)\(0+\frac{ 1}{ 243}\sqrt{ 3}I\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(1\)
\(1\)\(1\)\(1\)\(0\)\(1\)\(1\)
\(1\)\(3\)\(1\)\(0\)\(1\)\(1\)
\(2\)\(4\)\(2\)\(0\)\(2\)\(1\)

Note:

There is a second MUM-point at infinity,
corresponding to Operator AESZ 17/5.1

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