Summary

You searched for: sol=112140

Your search produced exactly one match

1

New Number: 8.79 |  AESZ:  |  Superseeker: 22/5 68  |  Hash: 064e5b590dd8b6a4daa1e905fbe693c2  

Degree: 8

\(5^{2} \theta^4-2 5 x\left(338\theta^4+412\theta^3+371\theta^2+165\theta+30\right)+2^{2} x^{2}\left(46396\theta^4+103408\theta^3+125291\theta^2+76370\theta+19080\right)-2^{4} 3 x^{3}\left(115508\theta^4+357896\theta^3+524149\theta^2+375205\theta+106530\right)+2^{6} 3^{2} x^{4}\left(173456\theta^4+669024\theta^3+1118292\theta^2+883484\theta+269049\right)-2^{11} 3^{3} x^{5}\left(20272\theta^4+91616\theta^3+168594\theta^2+142006\theta+45053\right)+2^{14} 3^{4} x^{6}\left(5792\theta^4+29504\theta^3+58300\theta^2+51220\theta+16641\right)-2^{21} 3^{5} x^{7}(\theta+1)^2(58\theta^2+208\theta+201)+2^{26} 3^{6} x^{8}(\theta+1)^2(\theta+2)^2\)

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Coefficients of the holomorphic solution: 1, 12, 204, 4368, 112140, ...
--> OEIS
Normalized instanton numbers (n0=1): 22/5, 8, 68, 3292/5, 38826/5, ... ; Common denominator:...

Discriminant

\((-1+48z)(16z-1)^2(48z-5)^2(12z-1)^3\)

Local exponents

\(0\)\(\frac{ 1}{ 48}\)\(\frac{ 1}{ 16}\)\(\frac{ 1}{ 12}\)\(\frac{ 5}{ 48}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(1\)
\(0\)\(1\)\(\frac{ 1}{ 2}\)\(\frac{ 1}{ 2}\)\(1\)\(1\)
\(0\)\(1\)\(\frac{ 1}{ 2}\)\(\frac{ 3}{ 2}\)\(3\)\(2\)
\(0\)\(2\)\(1\)\(2\)\(4\)\(2\)

Note:

This is operator "8.79" from ...

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