Summary

You searched for: sol=5040

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1

New Number: 6.21 |  AESZ:  |  Superseeker: 1 11  |  Hash: 319d6b2f1541de5252840442cc6f8dcd  

Degree: 6

\(\theta^4+x\left(6+27\theta+47\theta^2+40\theta^3+20\theta^4\right)-x^{2}(143\theta^2+286\theta+120)(\theta+1)^2-2 3^{2} x^{3}(\theta+2)(\theta+1)(291\theta^2+873\theta+766)-2^{2} 3^{3} 5 x^{4}(\theta+3)(\theta+1)(41\theta^2+164\theta+196)+2^{3} 3^{3} 5^{2} x^{5}(\theta+4)(\theta+1)(29\theta^2+145\theta+150)+2^{5} 3^{5} 5^{3} x^{6}(\theta+5)(\theta+4)(\theta+2)(\theta+1)\)

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Coefficients of the holomorphic solution: 1, -6, 60, -480, 5040, ...
--> OEIS
Normalized instanton numbers (n0=1): 1, 13/4, 11, 50, 1674/5, ... ; Common denominator:...

Discriminant

\((6z-1)(15z-1)(9z+1)(12z+1)(10z+1)^2\)

Local exponents

\(-\frac{ 1}{ 9}\)\(-\frac{ 1}{ 10}\)\(-\frac{ 1}{ 12}\)\(0\)\(\frac{ 1}{ 15}\)\(\frac{ 1}{ 6}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(1\)
\(1\)\(\frac{ 1}{ 2}\)\(1\)\(0\)\(1\)\(1\)\(2\)
\(1\)\(\frac{ 1}{ 2}\)\(1\)\(0\)\(1\)\(1\)\(4\)
\(2\)\(1\)\(2\)\(0\)\(2\)\(2\)\(5\)

Note:

This is operator "6.21" from ...

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