Summary

You searched for: sol=431964

Your search produced exactly one match

1

New Number: 8.10 |  AESZ: 123  |  Superseeker: 12 1828/3  |  Hash: f0d76ab2b6b8808f4faa4ab8ecadff2c  

Degree: 8

\(\theta^4-2^{2} x(3\theta^2+3\theta+1)(10\theta^2+10\theta+3)+2^{4} x^{2}\left(209\theta^4+1052\theta^3+1471\theta^2+838\theta+183\right)+2^{7} 3^{2} x^{3}\left(30\theta^4-180\theta^3-551\theta^2-417\theta-111\right)-2^{10} 3^{2} x^{4}\left(227\theta^4+454\theta^3-550\theta^2-777\theta-261\right)+2^{12} 3^{4} x^{5}\left(30\theta^4+300\theta^3+169\theta^2-25\theta-35\right)+2^{14} 3^{4} x^{6}\left(209\theta^4-216\theta^3-431\theta^2-216\theta-27\right)-2^{17} 3^{6} x^{7}(3\theta^2+3\theta+1)(10\theta^2+10\theta+3)+2^{20} 3^{8} x^{8}\left((\theta+1)^4\right)\)

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Coefficients of the holomorphic solution: 1, 12, 300, 10416, 431964, ...
--> OEIS
Normalized instanton numbers (n0=1): 12, -47/2, 1828/3, -10813/4, 127948, ... ; Common denominator:...

Discriminant

\((36z-1)(8z-1)(72z-1)(4z-1)(-1+288z^2)^2\)

Local exponents

\(-\frac{ 1}{ 24}\sqrt{ 2}\)\(0\)\(\frac{ 1}{ 72}\)\(\frac{ 1}{ 36}\)\(\frac{ 1}{ 24}\sqrt{ 2}\)\(\frac{ 1}{ 8}\)\(\frac{ 1}{ 4}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(1\)
\(1\)\(0\)\(1\)\(1\)\(1\)\(1\)\(1\)\(1\)
\(3\)\(0\)\(1\)\(1\)\(3\)\(1\)\(1\)\(1\)
\(4\)\(0\)\(2\)\(2\)\(4\)\(2\)\(2\)\(1\)

Note:

Hadamard product $c \ast d$. 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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