Summary

You searched for: superseeker=69/5,29081/5

Your search produced exactly one match

1

New Number: 4.42 |  AESZ: 222  |  Superseeker: 69/5 29081/5  |  Hash: aad7a72e711c9c463396d319e0bf7603  

Degree: 4

\(5^{2} \theta^4-5 x\left(407\theta^4+1198\theta^3+909\theta^2+310\theta+40\right)-2^{7} x^{2}\left(2103\theta^4+6999\theta^3+8358\theta^2+4050\theta+680\right)-2^{12} x^{3}\left(1387\theta^4+3840\theta^3+3081\theta^2+960\theta+100\right)-2^{21} x^{4}\left((2\theta+1)^4\right)\)

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Coefficients of the holomorphic solution: 1, 8, 504, 36800, 3518200, ...
--> OEIS
Normalized instanton numbers (n0=1): 69/5, 1383/4, 29081/5, 346080, 72023607/5, ... ; Common denominator:...

Discriminant

\(-(8192z^2+107z-1)(5+64z)^2\)

Local exponents

\(-\frac{ 5}{ 64}\)\(-\frac{ 107}{ 16384}-\frac{ 51}{ 16384}\sqrt{ 17}\)\(0\)\(s_1\)\(s_2\)\(-\frac{ 107}{ 16384}+\frac{ 51}{ 16384}\sqrt{ 17}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(\frac{ 1}{ 2}\)
\(1\)\(1\)\(0\)\(1\)\(1\)\(1\)\(\frac{ 1}{ 2}\)
\(3\)\(1\)\(0\)\(1\)\(1\)\(1\)\(\frac{ 1}{ 2}\)
\(4\)\(2\)\(0\)\(2\)\(2\)\(2\)\(\frac{ 1}{ 2}\)

Note:

Sporadic Operator. There is a second MUM-point hiding at infinity, corresponding to Operator AESZ225/4.43

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