Summary

You searched for: sol=163636

Your search produced exactly one match

1

New Number: 11.20 |  AESZ:  |  Superseeker: 21/4 1285/2  |  Hash: b07e191c8c5d8b6a2c25e842f85fcaf0  

Degree: 11

\(2^{4} \theta^4-2^{2} x\left(278\theta^4+394\theta^3+309\theta^2+112\theta+16\right)-x^{2}\left(11952+57616\theta+96951\theta^2+56722\theta^3+4615\theta^4\right)+2 x^{3}\left(129366\theta^4+473682\theta^3+531879\theta^2+282576\theta+62656\right)-x^{4}\left(1430728+5365104\theta+7153953\theta^2+3814866\theta^3+1139565\theta^4\right)-2 3 x^{5}\left(286602\theta^4-694990\theta^3-3072025\theta^2-2917584\theta-895328\right)+2^{2} x^{6}\left(1338547\theta^4+4488552\theta^3+821964\theta^2-3171240\theta-1633306\right)+2^{4} x^{7}\left(17380\theta^4-1361536\theta^3-2049918\theta^2-1043692\theta-152703\right)-2^{6} x^{8}\left(106051\theta^4+123172\theta^3+23589\theta^2-28382\theta-10873\right)+2^{10} x^{9}\left(4885\theta^4+15033\theta^3+20559\theta^2+13908\theta+3737\right)-2^{12} x^{10}\left(335\theta^4+1270\theta^3+1875\theta^2+1240\theta+307\right)+2^{17} x^{11}\left((\theta+1)^4\right)\)

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Coefficients of the holomorphic solution: 1, 4, 116, 3856, 163636, ...
--> OEIS
Normalized instanton numbers (n0=1): 21/4, 1965/32, 1285/2, 103095/8, 1157421/4, ... ; Common denominator:...

Discriminant

\((z-1)(16z^2-16z-1)(32z^2-71z+1)(4-27z-50z^2+16z^3)^2\)

Local exponents

≈\(-0.573963\)\(\frac{ 1}{ 2}-\frac{ 1}{ 4}\sqrt{ 5}\)\(0\)\(\frac{ 71}{ 64}-\frac{ 17}{ 64}\sqrt{ 17}\) ≈\(0.121762\)\(1\)\(\frac{ 1}{ 2}+\frac{ 1}{ 4}\sqrt{ 5}\)\(\frac{ 71}{ 64}+\frac{ 17}{ 64}\sqrt{ 17}\) ≈\(3.577201\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(1\)
\(1\)\(1\)\(0\)\(1\)\(1\)\(1\)\(1\)\(1\)\(1\)\(1\)
\(3\)\(1\)\(0\)\(1\)\(3\)\(1\)\(1\)\(1\)\(3\)\(1\)
\(4\)\(2\)\(0\)\(2\)\(4\)\(2\)\(2\)\(2\)\(4\)\(1\)

Note:

This is operator "11.20" from ...

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