Summary

You searched for: sol=17472

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1

New Number: 8.5 |  AESZ: 173  |  Superseeker: 11 -2434/3  |  Hash: afa82ed9ee239bb5fcac960f8884db01  

Degree: 8

\(\theta^4-x(7\theta^2+7\theta+2)(17\theta^2+17\theta+6)+2^{6} x^{2}\left(55\theta^4+112\theta^3+155\theta^2+86\theta+15\right)-2^{6} 3^{2} x^{3}\left(119\theta^4-714\theta^3-2185\theta^2-1656\theta-444\right)+2^{12} 3^{2} x^{4}\left(92\theta^4+184\theta^3+98\theta^2+6\theta+9\right)+2^{12} 3^{4} x^{5}\left(119\theta^4+1190\theta^3+671\theta^2-96\theta-140\right)+2^{18} 3^{4} x^{6}\left(55\theta^4+108\theta^3+149\theta^2+108\theta+27\right)+2^{18} 3^{6} x^{7}(7\theta^2+7\theta+2)(17\theta^2+17\theta+6)+2^{24} 3^{8} x^{8}\left((\theta+1)^4\right)\)

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Coefficients of the holomorphic solution: 1, 12, 420, 17472, 828324, ...
--> OEIS
Normalized instanton numbers (n0=1): 11, 229/4, -2434/3, 7512, 54801, ... ; Common denominator:...

Discriminant

\((72z-1)(8z+1)(64z-1)(9z+1)(1+576z^2)^2\)

Local exponents

\(-\frac{ 1}{ 8}\)\(-\frac{ 1}{ 9}\)\(0-\frac{ 1}{ 24}I\)\(0\)\(0+\frac{ 1}{ 24}I\)\(\frac{ 1}{ 72}\)\(\frac{ 1}{ 64}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(1\)
\(1\)\(1\)\(1\)\(0\)\(1\)\(1\)\(1\)\(1\)
\(1\)\(1\)\(3\)\(0\)\(3\)\(1\)\(1\)\(1\)
\(2\)\(2\)\(4\)\(0\)\(4\)\(2\)\(2\)\(1\)

Note:

Hadamard product $a \ast g$. This operator has a second MUM-point at infinity with the same instanton numbers.
It can be reduced to an operator of degree 4 with a single MUM-point defined over
$Q(\sqrt{?})$.

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