Summary

You searched for: sol=991152400

Your search produced exactly one match

1

New Number: 4.50 |  AESZ: 256  |  Superseeker: -128 -800384  |  Hash: 05e172cfdecc836685981a2b01b75d1d  

Degree: 4

\(\theta^4+2^{5} x\left(24\theta^4+42\theta^3+30\theta^2+9\theta+1\right)+2^{8} x^{2}\left(164\theta^4+104\theta^3-144\theta^2-100\theta-17\right)+2^{14} x^{3}\left(28\theta^4-48\theta^3-44\theta^2-12\theta-1\right)-2^{18} x^{4}\left((2\theta+1)^4\right)\)

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Coefficients of the holomorphic solution: 1, -32, 7056, -2393600, 991152400, ...
--> OEIS
Normalized instanton numbers (n0=1): -128, 6884, -800384, 143245314, -31691939200, ... ; Common denominator:...

Discriminant

\(-(4096z^2-704z-1)(1+32z)^2\)

Local exponents

\(-\frac{ 1}{ 32}\)\(\frac{ 11}{ 128}-\frac{ 5}{ 128}\sqrt{ 5}\)\(0\)\(s_1\)\(s_2\)\(\frac{ 11}{ 128}+\frac{ 5}{ 128}\sqrt{ 5}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(\frac{ 1}{ 2}\)
\(1\)\(1\)\(0\)\(1\)\(1\)\(1\)\(\frac{ 1}{ 2}\)
\(3\)\(1\)\(0\)\(1\)\(1\)\(1\)\(\frac{ 1}{ 2}\)
\(4\)\(2\)\(0\)\(2\)\(2\)\(2\)\(\frac{ 1}{ 2}\)

Note:

Sporadic Operator. There is a second MUM-point hiding at infinity, corresponding to Operator AESZ 257/4.51
B-Incarnation:
Fibre product 4*11-- x 25311,
Double octic; D.O.257

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