Summary

You searched for: sol=69

Your search produced exactly one match

1

New Number: 5.18 |  AESZ: 124  |  Superseeker: 163/61 4795/61  |  Hash: 394b401a3162e31c79ede5b46973791d  

Degree: 5

\(61^{2} \theta^4-61 x\left(3029\theta^4+5572\theta^3+4677\theta^2+1891\theta+305\right)+x^{2}\left(1215215\theta^4+3428132\theta^3+4267228\theta^2+2572675\theta+611586\right)-3^{4} x^{3}\left(39370\theta^4+140178\theta^3+206807\theta^2+142191\theta+37332\right)+3^{8} x^{4}\left(566\theta^4+2230\theta^3+3356\theta^2+2241\theta+558\right)-3^{13} x^{5}\left((\theta+1)^4\right)\)

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Coefficients of the holomorphic solution: 1, 5, 69, 1427, 35749, ...
--> OEIS
Normalized instanton numbers (n0=1): 163/61, 630/61, 4795/61, 48422/61, 599809/61, ... ; Common denominator:...

Discriminant

\(-(243z^3-200z^2+47z-1)(-61+81z)^2\)

Local exponents

\(0\) ≈\(0.023574\) ≈\(0.399736-0.121575I\) ≈\(0.399736+0.121575I\)\(\frac{ 61}{ 81}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(1\)
\(0\)\(1\)\(1\)\(1\)\(1\)\(1\)
\(0\)\(1\)\(1\)\(1\)\(3\)\(1\)
\(0\)\(2\)\(2\)\(2\)\(4\)\(1\)

Note:

This is operator "5.18" from ...

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