Summary

You searched for: superseeker=16,7568/3

Your search produced exactly one match

1

New Number: 8.41 |  AESZ:  |  Superseeker: 16 7568/3  |  Hash: 051d7068f49d14c45c3c3369d63d56b5  

Degree: 8

\(3^{2} \theta^4-2^{2} 3^{2} x\left(23\theta^4+58\theta^3+44\theta^2+15\theta+2\right)-2^{5} 3 x^{2}\left(254\theta^4+662\theta^3+623\theta^2+309\theta+66\right)-2^{8} 3 x^{3}\left(569\theta^4+1092\theta^3+602\theta^2+285\theta+78\right)-2^{11} x^{4}\left(2266\theta^4+4076\theta^3+2167\theta^2+537\theta+18\right)-2^{16} x^{5}\left(519\theta^4+798\theta^3+821\theta^2+391\theta+62\right)-2^{19} x^{6}\left(305\theta^4+558\theta^3+625\theta^2+360\theta+82\right)-2^{24} x^{7}\left(26\theta^4+70\theta^3+83\theta^2+48\theta+11\right)-2^{29} x^{8}\left((\theta+1)^4\right)\)

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Coefficients of the holomorphic solution: 1, 8, 328, 18944, 1324456, ...
--> OEIS
Normalized instanton numbers (n0=1): 16, 751/6, 7568/3, 229516/3, 8099456/3, ... ; Common denominator:...

Discriminant

\(-(4z+1)(2048z^3+768z^2+112z-1)(3+24z+256z^2)^2\)

Local exponents

\(-\frac{ 1}{ 4}\) ≈\(-0.191715-0.145483I\) ≈\(-0.191715+0.145483I\)\(-\frac{ 3}{ 64}-\frac{ 1}{ 64}\sqrt{ 39}I\)\(-\frac{ 3}{ 64}+\frac{ 1}{ 64}\sqrt{ 39}I\)\(0\) ≈\(0.00843\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(1\)
\(1\)\(1\)\(1\)\(1\)\(1\)\(0\)\(1\)\(1\)
\(1\)\(1\)\(1\)\(3\)\(3\)\(0\)\(1\)\(1\)
\(2\)\(2\)\(2\)\(4\)\(4\)\(0\)\(2\)\(1\)

Note:

This is operator "8.41" from ...

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