Summary

You searched for: sol=2500

Your search produced exactly one match

1

New Number: 10.6 |  AESZ:  |  Superseeker: 8 2200/9  |  Hash: b5aa0abf76ddfbd280ec220a43822aa4  

Degree: 10

\(\theta^4+2^{2} x\left(21\theta^4-6\theta^3+3\theta+1\right)+2^{4} x^{2}\left(126\theta^4-96\theta^3-16\theta^2-56\theta-33\right)+2^{6} x^{3}\left(84\theta^4-336\theta^3-226\theta^2-366\theta-163\right)+2^{11} 3 x^{4}\left(39\theta^4+500\theta^3+1230\theta^2+1160\theta+407\right)+2^{12} x^{5}\left(7029\theta^4+50118\theta^3+125086\theta^2+129149\theta+48902\right)+2^{14} x^{6}\left(38550\theta^4+294456\theta^3+806428\theta^2+911232\theta+368273\right)+2^{16} x^{7}\left(77544\theta^4+708720\theta^3+2233434\theta^2+2804346\theta+1214177\right)+2^{20} x^{8}\left(9171\theta^4+117228\theta^3+467444\theta^2+684316\theta+324572\right)-2^{23} x^{9}(2\theta+3)(2114\theta^3+16713\theta^2+37111\theta+22497)+2^{26} 3 5^{2} x^{10}(\theta+1)(2\theta+5)(2\theta+3)(\theta+3)\)

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Coefficients of the holomorphic solution: 1, -4, 52, -688, 2500, ...
--> OEIS
Normalized instanton numbers (n0=1): 8, -75/2, 2200/9, -8117/2, 47936, ... ; Common denominator:...

Discriminant

\((12z+1)(6400z^3+192z^2-24z+1)(16z+1)^2(32z^2-32z-1)^2\)

Local exponents

≈\(-0.090507\)\(-\frac{ 1}{ 12}\)\(-\frac{ 1}{ 16}\)\(\frac{ 1}{ 2}-\frac{ 3}{ 8}\sqrt{ 2}\)\(0\) ≈\(0.030254-0.02848I\) ≈\(0.030254+0.02848I\)\(\frac{ 1}{ 2}+\frac{ 3}{ 8}\sqrt{ 2}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(1\)
\(1\)\(1\)\(\frac{ 1}{ 2}\)\(1\)\(0\)\(1\)\(1\)\(1\)\(\frac{ 3}{ 2}\)
\(1\)\(1\)\(\frac{ 1}{ 2}\)\(3\)\(0\)\(1\)\(1\)\(3\)\(\frac{ 5}{ 2}\)
\(2\)\(2\)\(1\)\(4\)\(0\)\(2\)\(2\)\(4\)\(3\)

Note:

This is operator "10.6" from ...

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