Summary

You searched for: inst=-2100

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1

New Number: 5.62 |  AESZ: 270  |  Superseeker: -76/5 -2100  |  Hash: 256e3b3a92e3fd332be8b01f71853ea4  

Degree: 5

\(5^{2} \theta^4-2^{2} 5 x\left(48\theta^4+192\theta^3+251\theta^2+155\theta+35\right)-2^{4} x^{2}\left(8704\theta^4+28672\theta^3+43664\theta^2+31760\theta+9265\right)+2^{11} x^{3}\left(1792\theta^4+15360\theta^3+36248\theta^2+33240\theta+10795\right)+2^{16} x^{4}\left(2304\theta^4+12288\theta^3+20816\theta^2+14672\theta+3719\right)+2^{30} x^{5}\left((\theta+1)^4\right)\)

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Coefficients of the holomorphic solution: 1, 28, 1324, 63856, 3489004, ...
--> OEIS
Normalized instanton numbers (n0=1): -76/5, 367/5, -2100, 43436, -6582256/5, ... ; Common denominator:...

Discriminant

\((1+64z)(64z+5)^2(64z-1)^2\)

Local exponents

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

Note:

There is a second MUM-point at infinity,
corresponding to Operator AESZ 271/ 5.63

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