Summary

You searched for: inst=1659840

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1

New Number: 6.39 |  AESZ:  |  Superseeker: 8 3784/3  |  Hash: 6429f42cbe18bee944ac13edab1fbbcc  

Degree: 6

\(\theta^4+2^{2} x\left(49\theta^4+98\theta^3+86\theta^2+37\theta+6\right)+2^{5} x^{2}\left(593\theta^4+2372\theta^3+3521\theta^2+2298\theta+504\right)+2^{10} 3 x^{3}\left(332\theta^4+1992\theta^3+4194\theta^2+3618\theta+945\right)+2^{14} 3^{2} x^{4}\left(204\theta^4+1632\theta^3+4449\theta^2+4740\theta+1400\right)+2^{18} 3^{3} x^{5}(16\theta^2+80\theta+35)(2\theta+5)^2+2^{21} 3^{4} x^{6}(2\theta+11)(2\theta+7)(2\theta+5)(2\theta+1)\)

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Coefficients of the holomorphic solution: 1, -24, 648, -11520, -123480, ...
--> OEIS
Normalized instanton numbers (n0=1): 8, 39/2, 3784/3, 51036, 1659840, ... ; Common denominator:...

Discriminant

\((24z+1)(110592z^3+6912z^2+108z+1)(1+32z)^2\)

Local exponents

≈\(-0.045368\)\(-\frac{ 1}{ 24}\)\(-\frac{ 1}{ 32}\) ≈\(-0.008566-0.011222I\) ≈\(-0.008566+0.011222I\)\(0\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(\frac{ 1}{ 2}\)
\(1\)\(1\)\(\frac{ 1}{ 2}\)\(1\)\(1\)\(0\)\(\frac{ 5}{ 2}\)
\(1\)\(1\)\(\frac{ 1}{ 2}\)\(1\)\(1\)\(0\)\(\frac{ 7}{ 2}\)
\(2\)\(2\)\(1\)\(2\)\(2\)\(0\)\(\frac{ 11}{ 2}\)

Note:

This is operator "6.39" from ...

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