Summary

You searched for: inst=102374160

Your search produced exactly one match

1

New Number: 4.65 |  AESZ:  |  Superseeker: 48 -9104  |  Hash: 5ec2790b5eda514313634b7aeb0a295c  

Degree: 4

\(\theta^4-2^{4} x\left(5\theta^4+34\theta^3+25\theta^2+8\theta+1\right)+2^{11} x^{2}\left(5\theta^4+47\theta^3+90\theta^2+47\theta+8\right)+2^{16} x^{3}\left(51\theta^4+192\theta^3+155\theta^2+48\theta+5\right)+2^{23} x^{4}\left((2\theta+1)^4\right)\)

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Coefficients of the holomorphic solution: 1, 16, 144, -70400, -9858800, ...
--> OEIS
Normalized instanton numbers (n0=1): 48, -1298, -9104, 387230, 102374160, ... ; Common denominator:...

Discriminant

\((32768z^2-208z+1)(1+64z)^2\)

Local exponents

\(-\frac{ 1}{ 64}\)\(0\)\(s_1\)\(s_2\)\(\frac{ 13}{ 4096}-\frac{ 7}{ 4096}\sqrt{ 7}I\)\(\frac{ 13}{ 4096}+\frac{ 7}{ 4096}\sqrt{ 7}I\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(\frac{ 1}{ 2}\)
\(1\)\(0\)\(1\)\(1\)\(1\)\(1\)\(\frac{ 1}{ 2}\)
\(3\)\(0\)\(1\)\(1\)\(1\)\(1\)\(\frac{ 1}{ 2}\)
\(4\)\(0\)\(2\)\(2\)\(2\)\(2\)\(\frac{ 1}{ 2}\)

Note:

Sporadic Operator. There is a second MUM-point hiding at infinity,
corresponding to Operator AESZ 295/4.64

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