Summary

You searched for: sol=19221300

Your search produced exactly one match

1

New Number: 5.45 |  AESZ: 242  |  Superseeker: -18 1568/3  |  Hash: 562c18d54c0080ebb0bb01b14a8241ce  

Degree: 5

\(\theta^4+2 3 x\left(72\theta^4+108\theta^3+91\theta^2+37\theta+6\right)+2^{2} 3^{3} x^{2}\left(648\theta^4+1800\theta^3+2211\theta^2+1248\theta+260\right)+2^{4} 3^{5} x^{3}\left(1344\theta^4+4968\theta^3+7320\theta^2+4749\theta+1072\right)+2^{6} 3^{7} x^{4}(2\theta+1)(630\theta^3+2241\theta^2+2617\theta+971)+2^{8} 3^{10} x^{5}(2\theta+1)(6\theta+5)(6\theta+7)(2\theta+3)\)

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Coefficients of the holomorphic solution: 1, -36, 2484, -208080, 19221300, ...
--> OEIS
Normalized instanton numbers (n0=1): -18, 99/2, 1568/3, 22698, -165960, ... ; Common denominator:...

Discriminant

\((1+144z)(36z+1)^2(108z+1)^2\)

Local exponents

\(-\frac{ 1}{ 36}\)\(-\frac{ 1}{ 108}\)\(-\frac{ 1}{ 144}\)\(0\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(\frac{ 1}{ 2}\)
\(1\)\(\frac{ 1}{ 2}\)\(1\)\(0\)\(\frac{ 5}{ 6}\)
\(3\)\(\frac{ 1}{ 2}\)\(1\)\(0\)\(\frac{ 7}{ 6}\)
\(4\)\(1\)\(2\)\(0\)\(\frac{ 3}{ 2}\)

Note:

This is operator "5.45" from ...

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