Summary

You searched for: superseeker=7/3,148

Your search produced exactly one match

1

New Number: 5.49 |  AESZ: 248  |  Superseeker: 7/3 148  |  Hash: 0c9ccff1cb4f5096e455a9026799ed5a  

Degree: 5

\(3^{2} \theta^4-3 x\left(106\theta^4+146\theta^3+115\theta^2+42\theta+6\right)-x^{2}\left(4511\theta^4+24314\theta^3+37829\theta^2+23598\theta+5286\right)+2^{2} x^{3}\left(10457\theta^4+32184\theta^3+24449\theta^2+3627\theta-1317\right)-2^{2} 11 x^{4}\left(1596\theta^4+2040\theta^3-101\theta^2-1085\theta-386\right)-2^{4} 11^{2} x^{5}(4\theta+3)(\theta+1)^2(4\theta+5)\)

Maple   LaTex

Coefficients of the holomorphic solution: 1, 2, 54, 1028, 29110, ...
--> OEIS
Normalized instanton numbers (n0=1): 7/3, 551/24, 148, 8241/4, 86854/3, ... ; Common denominator:...

Discriminant

\(-(16z+1)(16z^2+44z-1)(-3+11z)^2\)

Local exponents

\(-\frac{ 11}{ 8}-\frac{ 5}{ 8}\sqrt{ 5}\)\(-\frac{ 1}{ 16}\)\(0\)\(-\frac{ 11}{ 8}+\frac{ 5}{ 8}\sqrt{ 5}\)\(\frac{ 3}{ 11}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(\frac{ 3}{ 4}\)
\(1\)\(1\)\(0\)\(1\)\(1\)\(1\)
\(1\)\(1\)\(0\)\(1\)\(3\)\(1\)
\(2\)\(2\)\(0\)\(2\)\(4\)\(\frac{ 5}{ 4}\)

Note:

This is operator "5.49" from ...

Show more...  or download as   plain text  |  PDF  |  Maple  |  LaTex