Summary

You searched for: sol=10440

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1

New Number: 8.73 |  AESZ:  |  Superseeker: 161/13 26946/13  |  Hash: 13db5d8c98a3d4f31589970217896191  

Degree: 8

\(13^{2} \theta^4-13 x\theta(614\theta^3+1804\theta^2+1149\theta+247)-x^{2}\left(775399\theta^4+2692636\theta^3+3693483\theta^2+2450110\theta+648960\right)-2^{2} x^{3}\left(5408420\theta^4+24616488\theta^3+45163287\theta^2+38795913\theta+12838410\right)-2^{5} x^{4}\left(9763642\theta^4+55386224\theta^3+123097843\theta^2+124066416\theta+46600563\right)-2^{9} 3 x^{5}(\theta+1)(1717504\theta^3+9940776\theta^2+20063523\theta+13933966)-2^{13} 3^{2} x^{6}(\theta+1)(\theta+2)(178975\theta^2+874119\theta+1112486)-2^{19} 3^{4} x^{7}(\theta+3)(\theta+2)(\theta+1)(857\theta+2533)-2^{23} 3^{6} 7 x^{8}(\theta+1)(\theta+2)(\theta+3)(\theta+4)\)

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Coefficients of the holomorphic solution: 1, 0, 240, 10440, 679104, ...
--> OEIS
Normalized instanton numbers (n0=1): 161/13, 1406/13, 26946/13, 742982/13, 25168759/13, ... ; Common denominator:...

Discriminant

\(-(-1+96z+896z^2)(9z+1)^2(96z+13)^2(8z+1)^2\)

Local exponents

\(-\frac{ 13}{ 96}\)\(-\frac{ 1}{ 8}\)\(-\frac{ 3}{ 56}-\frac{ 5}{ 112}\sqrt{ 2}\)\(-\frac{ 1}{ 9}\)\(0\)\(-\frac{ 3}{ 56}+\frac{ 5}{ 112}\sqrt{ 2}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(1\)
\(1\)\(0\)\(1\)\(0\)\(0\)\(1\)\(2\)
\(3\)\(1\)\(1\)\(1\)\(0\)\(1\)\(3\)
\(4\)\(1\)\(2\)\(1\)\(0\)\(2\)\(4\)

Note:

This is operator "8.73" from ...

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