Summary

You searched for: inst=29091017/5

Your search produced exactly one match

1

New Number: 5.39 |  AESZ: 224  |  Superseeker: 59/5 22503/5  |  Hash: ba17e8cb074bba75e7a27206be530698  

Degree: 5

\(5^{2} \theta^4-5 x\left(1057\theta^4+1058\theta^3+819\theta^2+290\theta+40\right)+2^{5} x^{2}\left(10123\theta^4+11419\theta^3+5838\theta^2+1510\theta+180\right)-2^{8} x^{3}\left(30981\theta^4+46560\theta^3+48211\theta^2+25500\theta+5100\right)+2^{14} 11 x^{4}(2\theta+1)(234\theta^3+591\theta^2+581\theta+202)-2^{20} 11^{2} x^{5}(2\theta+1)(\theta+1)^2(2\theta+3)\)

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Coefficients of the holomorphic solution: 1, 8, 312, 19520, 1475320, ...
--> OEIS
Normalized instanton numbers (n0=1): 59/5, 186, 22503/5, 718052/5, 29091017/5, ... ; Common denominator:...

Discriminant

\(-(128z-1)(128z^2-13z+1)(-5+176z)^2\)

Local exponents

\(0\)\(\frac{ 1}{ 128}\)\(\frac{ 5}{ 176}\)\(\frac{ 13}{ 256}-\frac{ 7}{ 256}\sqrt{ 7}I\)\(\frac{ 13}{ 256}+\frac{ 7}{ 256}\sqrt{ 7}I\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(\frac{ 1}{ 2}\)
\(0\)\(1\)\(1\)\(1\)\(1\)\(1\)
\(0\)\(1\)\(3\)\(1\)\(1\)\(1\)
\(0\)\(2\)\(4\)\(2\)\(2\)\(\frac{ 3}{ 2}\)

Note:

This is operator "5.39" from ...

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