Summary

You searched for: sol=28160

Your search produced exactly one match

1

New Number: 8.77 |  AESZ:  |  Superseeker: 91/5 25991/5  |  Hash: fa37d863a8d0cc4b7a34e7d9b5e3a1a5  

Degree: 8

\(5^{2} \theta^4-5 x\left(693\theta^4+1242\theta^3+931\theta^2+310\theta+40\right)-2^{4} x^{2}\left(659\theta^4+9977\theta^3+17174\theta^2+10200\theta+2160\right)-2^{5} x^{3}\left(7235\theta^4-19374\theta^3-34715\theta^2-7290\theta+1560\right)-2^{8} x^{4}\left(14861\theta^4+40168\theta^3-70511\theta^2-88342\theta-26424\right)-2^{10} x^{5}\left(6973\theta^4+29386\theta^3+99859\theta^2+58446\theta+9864\right)-2^{14} x^{6}\left(6951\theta^4-25713\theta^3-34544\theta^2-14472\theta-1680\right)-2^{15} 11 x^{7}\left(2029\theta^4+5030\theta^3+5139\theta^2+2570\theta+520\right)+2^{18} 3 11^{2} x^{8}(\theta+1)^2(3\theta+2)(3\theta+4)\)

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Coefficients of the holomorphic solution: 1, 8, 408, 28160, 2360440, ...
--> OEIS
Normalized instanton numbers (n0=1): 91/5, 1158/5, 25991/5, 192163, 42855113/5, ... ; Common denominator:...

Discriminant

\((z-1)(8z+1)(864z^2+136z-1)(5-24z+352z^2)^2\)

Local exponents

\(-\frac{ 17}{ 216}-\frac{ 7}{ 216}\sqrt{ 7}\)\(-\frac{ 1}{ 8}\)\(0\)\(-\frac{ 17}{ 216}+\frac{ 7}{ 216}\sqrt{ 7}\)\(\frac{ 3}{ 88}-\frac{ 1}{ 88}\sqrt{ 101}I\)\(\frac{ 3}{ 88}+\frac{ 1}{ 88}\sqrt{ 101}I\)\(1\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(\frac{ 2}{ 3}\)
\(1\)\(1\)\(0\)\(1\)\(1\)\(1\)\(1\)\(1\)
\(1\)\(1\)\(0\)\(1\)\(3\)\(3\)\(1\)\(1\)
\(2\)\(2\)\(0\)\(2\)\(4\)\(4\)\(2\)\(\frac{ 4}{ 3}\)

Note:

This is operator "8.77" from ...

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