Summary

You searched for: sol=270

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1

New Number: 8.50 |  AESZ:  |  Superseeker: 2/23 27/23  |  Hash: 68499833fa99ef8841a3d64e042d4a6e  

Degree: 8

\(23^{2} \theta^4-2 23 x\theta^2(136\theta^2+2\theta+1)-2^{2} x^{2}\left(7589\theta^4+54926\theta^3+89975\theta^2+69828\theta+21160\right)+x^{3}\left(573259\theta^4+2342274\theta^3+3791849\theta^2+3070914\theta+1010160\right)-2 5 x^{4}\left(122351\theta^4+62266\theta^3-795547\theta^2-1404486\theta-669744\right)-2^{3} 3 5^{2} x^{5}(\theta+1)(16105\theta^3+133047\theta^2+320040\theta+245740)+2^{4} 3^{2} 5^{3} x^{6}(\theta+1)(\theta+2)(3107\theta^2+16911\theta+22834)-2^{4} 3^{4} 5^{4} x^{7}(\theta+3)(\theta+2)(\theta+1)(133\theta+404)+2^{5} 3^{6} 5^{5} x^{8}(\theta+1)(\theta+2)(\theta+3)(\theta+4)\)

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Coefficients of the holomorphic solution: 1, 0, 10, 0, 270, ...
--> OEIS
Normalized instanton numbers (n0=1): 2/23, 18/23, 27/23, 136/23, 395/23, ... ; Common denominator:...

Discriminant

\((3z-1)(2z-1)(10z-1)(6z+1)(25z^2-5z-1)(-23+90z)^2\)

Local exponents

\(-\frac{ 1}{ 6}\)\(\frac{ 1}{ 10}-\frac{ 1}{ 10}\sqrt{ 5}\)\(0\)\(\frac{ 1}{ 10}\)\(\frac{ 23}{ 90}\)\(\frac{ 1}{ 10}+\frac{ 1}{ 10}\sqrt{ 5}\)\(\frac{ 1}{ 3}\)\(\frac{ 1}{ 2}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(1\)
\(1\)\(1\)\(0\)\(1\)\(1\)\(1\)\(1\)\(1\)\(2\)
\(1\)\(1\)\(0\)\(1\)\(3\)\(1\)\(1\)\(1\)\(3\)
\(2\)\(2\)\(0\)\(2\)\(4\)\(2\)\(2\)\(2\)\(4\)

Note:

This is operator "8.50" from ...

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