Summary

You searched for: sol=-37248

Your search produced exactly one match

1

New Number: 8.80 |  AESZ:  |  Superseeker: -28/3 2764/3  |  Hash: 01b1872abfd55652952ae535920a40fe  

Degree: 8

\(3^{2} \theta^4+2^{2} 3 x\left(148\theta^4+248\theta^3+223\theta^2+99\theta+18\right)+2^{7} x^{2}\left(1124\theta^4+3080\theta^3+4211\theta^2+2709\theta+675\right)+2^{12} x^{3}\left(1684\theta^4+4872\theta^3+7059\theta^2+5373\theta+1530\right)+2^{17} x^{4}\left(1828\theta^4+4952\theta^3+5125\theta^2+2799\theta+599\right)+2^{23} x^{5}\left(720\theta^4+1992\theta^3+2102\theta^2+691\theta-13\right)+2^{29} x^{6}\left(200\theta^4+504\theta^3+669\theta^2+390\theta+83\right)+2^{35} x^{7}\left(40\theta^4+104\theta^3+118\theta^2+66\theta+15\right)+2^{43} x^{8}\left((\theta+1)^4\right)\)

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Coefficients of the holomorphic solution: 1, -24, 872, -37248, 1740456, ...
--> OEIS
Normalized instanton numbers (n0=1): -28/3, 49/3, 2764/3, 13414, 44384, ... ; Common denominator:...

Discriminant

\((16z+1)(32z+1)(64z+1)^2(2048z^2+32z+3)^2\)

Local exponents

\(-\frac{ 1}{ 16}\)\(-\frac{ 1}{ 32}\)\(-\frac{ 1}{ 64}\)\(-\frac{ 1}{ 128}-\frac{ 1}{ 128}\sqrt{ 23}I\)\(-\frac{ 1}{ 128}+\frac{ 1}{ 128}\sqrt{ 23}I\)\(0\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(1\)
\(1\)\(1\)\(\frac{ 1}{ 2}\)\(1\)\(1\)\(0\)\(1\)
\(1\)\(1\)\(\frac{ 1}{ 2}\)\(3\)\(3\)\(0\)\(1\)
\(2\)\(2\)\(1\)\(4\)\(4\)\(0\)\(1\)

Note:

This is operator "8.80" from ...

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