Summary

You searched for: sol=13080

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1

New Number: 8.55 |  AESZ:  |  Superseeker: 1 68/9  |  Hash: 39ed8ce7572bc79a333f77c892033bcf  

Degree: 8

\(\theta^4-x\left(33\theta^4+98\theta^3+105\theta^2+56\theta+12\right)+2^{3} x^{2}\left(34\theta^4+276\theta^3+609\theta^2+582\theta+216\right)+2^{4} 3 x^{3}\left(11\theta^4-170\theta^3-941\theta^2-1520\theta-846\right)-2^{7} 3^{2} x^{4}(2\theta^2+6\theta+5)(4\theta^2+12\theta-31)+2^{8} 3 x^{5}\left(11\theta^4+302\theta^3+1183\theta^2+1652\theta+726\right)+2^{11} x^{6}\left(34\theta^4+132\theta^3-39\theta^2-708\theta-747\right)-2^{12} x^{7}\left(33\theta^4+298\theta^3+1005\theta^2+1492\theta+816\right)+2^{16} x^{8}\left((\theta+3)^4\right)\)

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Coefficients of the holomorphic solution: 1, 12, 120, 1216, 13080, ...
--> OEIS
Normalized instanton numbers (n0=1): 1, -2, 68/9, -30, 150, ... ; Common denominator:...

Discriminant

\((z-1)(16z-1)(16z^2-16z+1)(4z-1)^2(4z+1)^2\)

Local exponents

\(-\frac{ 1}{ 4}\)\(0\)\(\frac{ 1}{ 16}\)\(\frac{ 1}{ 2}-\frac{ 1}{ 4}\sqrt{ 3}\)\(\frac{ 1}{ 4}\)\(\frac{ 1}{ 2}+\frac{ 1}{ 4}\sqrt{ 3}\)\(1\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(3\)
\(1\)\(0\)\(1\)\(1\)\(0\)\(1\)\(1\)\(3\)
\(3\)\(0\)\(1\)\(1\)\(-1\)\(1\)\(1\)\(3\)
\(4\)\(0\)\(2\)\(2\)\(1\)\(2\)\(2\)\(3\)

Note:

This is operator "8.55" from ...

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