Summary

You searched for: sol=30772

Your search produced exactly one match

1

New Number: 5.79 |  AESZ: 310  |  Superseeker: 181/11 47171/11  |  Hash: 2b9b103b1c8f0d3175cd1fb9ef5aacc2  

Degree: 5

\(11^{2} \theta^4-11 x\left(1673\theta^4+3046\theta^3+2337\theta^2+814\theta+110\right)+2 5 x^{2}\left(19247\theta^4+28298\theta^3+13285\theta^2+3454\theta+660\right)-2^{2} x^{3}\left(167497\theta^4+245982\theta^3+227451\theta^2+115434\theta+22968\right)+2^{3} 5^{2} x^{4}\left(4079\theta^4+10270\theta^3+11427\theta^2+6226\theta+1340\right)-2^{5} 5^{4} x^{5}(4\theta+3)(\theta+1)^2(4\theta+5)\)

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Coefficients of the holomorphic solution: 1, 10, 450, 30772, 2551810, ...
--> OEIS
Normalized instanton numbers (n0=1): 181/11, 2018/11, 47171/11, 3261479/22, 69313270/11, ... ; Common denominator:...

Discriminant

\(-(z-1)(128z^2-142z+1)(-11+50z)^2\)

Local exponents

\(0\)\(\frac{ 71}{ 128}-\frac{ 17}{ 128}\sqrt{ 17}\)\(\frac{ 11}{ 50}\)\(1\)\(\frac{ 71}{ 128}+\frac{ 17}{ 128}\sqrt{ 17}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(\frac{ 3}{ 4}\)
\(0\)\(1\)\(1\)\(1\)\(1\)\(1\)
\(0\)\(1\)\(3\)\(1\)\(1\)\(1\)
\(0\)\(2\)\(4\)\(2\)\(2\)\(\frac{ 5}{ 4}\)

Note:

This is operator "5.79" from ...

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