Summary

You searched for: sol=16846

Your search produced exactly one match

1

New Number: 8.33 |  AESZ: 322  |  Superseeker: 4/3 95/3  |  Hash: a19da26bf1a7748e3b7e6151e803da30  

Degree: 8

\(3^{2} \theta^4+3 x\left(5\theta^4-122\theta^3-100\theta^2-39\theta-6\right)-x^{2}\left(5052+23736\theta+41729\theta^2+32600\theta^3+8603\theta^4\right)-2^{2} x^{3}\left(33304\theta^4+108297\theta^3+122347\theta^2+61470\theta+11712\right)-2^{2} x^{4}\left(180401\theta^4+547606\theta^3+638125\theta^2+339248\theta+69036\right)-2^{4} x^{5}\left(94934\theta^4+298745\theta^3+355667\theta^2+189660\theta+38224\right)-2^{4} x^{6}\left(73291\theta^4+204216\theta^3+190453\theta^2+68916\theta+6964\right)-2^{7} 3 x^{7}\left(811\theta^4+1886\theta^3+1804\theta^2+861\theta+174\right)-2^{10} 3^{2} x^{8}\left((\theta+1)^4\right)\)

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Coefficients of the holomorphic solution: 1, 2, 46, 632, 16846, ...
--> OEIS
Normalized instanton numbers (n0=1): 4/3, 18, 95/3, 14575/12, 18158/3, ... ; Common denominator:...

Discriminant

\(-(-1+13z+827z^2+1928z^3+64z^4)(3+22z+12z^2)^2\)

Local exponents

\(-\frac{ 11}{ 12}-\frac{ 1}{ 12}\sqrt{ 85}\)\(-\frac{ 11}{ 12}+\frac{ 1}{ 12}\sqrt{ 85}\)\(0\)\(#ND+#NDI\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(1\)
\(1\)\(1\)\(0\)\(1\)\(1\)
\(3\)\(3\)\(0\)\(1\)\(1\)
\(4\)\(4\)\(0\)\(2\)\(1\)

Note:

This operator has a second MUM-point at infininty corresponding to operator 8.34

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