Summary

You searched for: sol=258039

Your search produced exactly one match

1

New Number: 9.3 |  AESZ:  |  Superseeker: 10 3394/3  |  Hash: 40e3715abcc5c4cb07e700ca79f80abf  

Degree: 9

\(\theta^4-x\left(57\theta^4+116\theta^3+84\theta^2+26\theta+3\right)-2 x^{2}\left(894\theta^4+3208\theta^3+4571\theta^2+2771\theta+651\right)-2 x^{3}\left(7322\theta^4+56368\theta^3+124783\theta^2+101099\theta+29757\right)+2 3^{2} x^{4}\left(6967\theta^4-27080\theta^3-139991\theta^2-138507\theta-45297\right)+2 3^{4} x^{5}\left(17617\theta^4+49068\theta^3-31255\theta^2-79893\theta-34578\right)+2 3^{8} x^{6}\left(1082\theta^4+8360\theta^3+7967\theta^2+1439\theta-773\right)-2 3^{11} x^{7}\left(198\theta^4-864\theta^3-1545\theta^2-909\theta-155\right)-3^{15} x^{8}\left(69\theta^4+144\theta^3+126\theta^2+54\theta+10\right)-3^{20} x^{9}\left((\theta+1)^4\right)\)

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Coefficients of the holomorphic solution: 1, 3, 135, 5349, 258039, ...
--> OEIS
Normalized instanton numbers (n0=1): 10, 77, 3394/3, 24029, 640402, ... ; Common denominator:...

Discriminant

\(-(-1+81z)(-1+9z)^2(81z^2+14z+1)^3\)

Local exponents

\(-\frac{ 7}{ 81}-\frac{ 4}{ 81}\sqrt{ 2}I\)\(-\frac{ 7}{ 81}+\frac{ 4}{ 81}\sqrt{ 2}I\)\(0\)\(\frac{ 1}{ 81}\)\(\frac{ 1}{ 9}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(1\)
\(\frac{ 1}{ 2}\)\(\frac{ 1}{ 2}\)\(0\)\(1\)\(1\)\(1\)
\(\frac{ 3}{ 2}\)\(\frac{ 3}{ 2}\)\(0\)\(1\)\(3\)\(1\)
\(2\)\(2\)\(0\)\(2\)\(4\)\(1\)

Note:

This is operator "9.3" from ...

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