Summary

You searched for: inst=161840

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1

New Number: 8.7 |  AESZ: 106  |  Superseeker: 12 356  |  Hash: fe1c90929d18b81637eaaa93366409ed  

Degree: 8

\(\theta^4-2^{2} x(3\theta^2+3\theta+1)(11\theta^2+11\theta+3)+2^{4} x^{2}\left(241\theta^4+940\theta^3+1303\theta^2+726\theta+145\right)-2^{7} x^{3}\left(33\theta^4-198\theta^3-607\theta^2-456\theta-117\right)+2^{10} x^{4}\left(239\theta^4+478\theta^3-322\theta^2-561\theta-169\right)+2^{12} x^{5}\left(33\theta^4+330\theta^3+185\theta^2-32\theta-37\right)+2^{14} x^{6}\left(241\theta^4+24\theta^3-71\theta^2+24\theta+23\right)+2^{17} x^{7}(3\theta^2+3\theta+1)(11\theta^2+11\theta+3)+2^{20} x^{8}\left((\theta+1)^4\right)\)

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Coefficients of the holomorphic solution: 1, 12, 380, 16464, 845676, ...
--> OEIS
Normalized instanton numbers (n0=1): 12, 20, 356, 34561/4, 161840, ... ; Common denominator:...

Discriminant

\((64z^2+88z-1)(16z^2+44z-1)(1+32z^2)^2\)

Local exponents

\(-\frac{ 11}{ 8}-\frac{ 5}{ 8}\sqrt{ 5}\)\(-\frac{ 11}{ 16}-\frac{ 5}{ 16}\sqrt{ 5}\)\(0-\frac{ 1}{ 8}\sqrt{ 2}I\)\(0\)\(0+\frac{ 1}{ 8}\sqrt{ 2}I\)\(-\frac{ 11}{ 16}+\frac{ 5}{ 16}\sqrt{ 5}\)\(-\frac{ 11}{ 8}+\frac{ 5}{ 8}\sqrt{ 5}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(1\)
\(1\)\(1\)\(1\)\(0\)\(1\)\(1\)\(1\)\(1\)
\(1\)\(1\)\(3\)\(0\)\(3\)\(1\)\(1\)\(1\)
\(2\)\(2\)\(4\)\(0\)\(4\)\(2\)\(2\)\(1\)

Note:

Hadamard product $b\ast d$. 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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