Summary

You searched for: superseeker=82450,22323908689400

Your search produced exactly one match

1

New Number: 5.53 |  AESZ: 259  |  Superseeker: 82450 22323908689400  |  Hash: 8b20756bb52131d41c44fd699c9e3a24  

Degree: 5

\(\theta^4+2 5 x\left(40000\theta^4-17500\theta^3-8125\theta^2+625\theta+238\right)+2^{2} 5^{6} x^{2}\left(835000\theta^4-365000\theta^3+371125\theta^2+58500\theta+2116\right)+2^{4} 5^{11} x^{3}\left(3130000\theta^4+1815000\theta^3+1662000\theta^2+625875\theta+96914\right)+2^{6} 5^{19} 13 x^{4}(625\theta^2+745\theta+351)(2\theta+1)^2+2^{8} 5^{25} 13^{2} x^{5}(2\theta+1)^2(2\theta+3)^2\)

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Coefficients of the holomorphic solution: 1, -2380, 14400900, -112575082000, 993749164922500, ...
--> OEIS
Normalized instanton numbers (n0=1): 82450, -976323150, 22323908689400, -680892969306394000, 24398212781075814030620, ... ; Common denominator:...

Discriminant

\((1+50000z)(12500z+1)^2(162500z+1)^2\)

Local exponents

\(-\frac{ 1}{ 12500}\)\(-\frac{ 1}{ 50000}\)\(-\frac{ 1}{ 162500}\)\(0\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(\frac{ 1}{ 2}\)
\(\frac{ 1}{ 2}\)\(1\)\(1\)\(0\)\(\frac{ 1}{ 2}\)
\(\frac{ 1}{ 2}\)\(1\)\(3\)\(0\)\(\frac{ 3}{ 2}\)
\(1\)\(2\)\(4\)\(0\)\(\frac{ 3}{ 2}\)

Note:

This is operator "5.53" from ...

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