Summary

You searched for: inst=-575948640

Your search produced exactly one match

1

New Number: 7.13 |  AESZ:  |  Superseeker: -32 -107936  |  Hash: 80eaab6a34199e98f88d8472c115c4df  

Degree: 7

\(\theta^4+2^{4} x\left(44\theta^4+72\theta^3+64\theta^2+28\theta+5\right)+2^{11} x^{2}\left(60\theta^4+328\theta^3+420\theta^2+228\theta+51\right)-2^{18} x^{3}\left(52\theta^4-328\theta^3-885\theta^2-663\theta-181\right)-2^{25} x^{4}\left(148\theta^4+344\theta^3-403\theta^2-559\theta-199\right)-2^{32} x^{5}\left(24\theta^4+544\theta^3+519\theta^2+147\theta-12\right)+2^{39} x^{6}\left(80\theta^4+32\theta^3-147\theta^2-159\theta-46\right)+2^{47} x^{7}(4\theta+3)(\theta+1)^2(4\theta+5)\)

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Coefficients of the holomorphic solution: 1, -80, 10512, -1703168, 309951760, ...
--> OEIS
Normalized instanton numbers (n0=1): -32, -2840, -107936, -7514224, -575948640, ... ; Common denominator:...

Discriminant

\((64z+1)(128z+1)(128z-1)^2(256z+1)^3\)

Local exponents

\(-\frac{ 1}{ 64}\)\(-\frac{ 1}{ 128}\)\(-\frac{ 1}{ 256}\)\(0\)\(\frac{ 1}{ 128}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(\frac{ 3}{ 4}\)
\(1\)\(1\)\(\frac{ 1}{ 4}\)\(0\)\(1\)\(1\)
\(1\)\(1\)\(\frac{ 7}{ 4}\)\(0\)\(3\)\(1\)
\(2\)\(2\)\(2\)\(0\)\(4\)\(\frac{ 5}{ 4}\)

Note:

This is operator "7.13" from ...

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