Summary

You searched for: inst=40/27

Your search produced exactly one match

1

New Number: 6.9 |  AESZ:  |  Superseeker: 31/81 29/9  |  Hash: 98af5121f39c27098356e3ade277f975  

Degree: 6

\(3^{8} \theta^4-3^{4} x\left(1234\theta^4+2168\theta^3+1975\theta^2+891\theta+162\right)-x^{2}\left(428004+1521180\theta+2033921\theta^2+1177556\theta^3+205589\theta^4\right)+x^{3}\left(2310517\theta^4+12882402\theta^3+26939429\theta^2+25052328\theta+8683524\right)-2^{2} 5^{2} x^{4}\left(51526\theta^4+332687\theta^3+804453\theta^2+849398\theta+325796\right)+2^{2} 5^{4} x^{5}(\theta+1)(1593\theta^3+8667\theta^2+15104\theta+8516)-2^{4} 5^{6} x^{6}(\theta+2)(\theta+1)(2\theta+3)^2\)

Maple   LaTex

Coefficients of the holomorphic solution: 1, 2, 14, 104, 1030, ...
--> OEIS
Normalized instanton numbers (n0=1): 31/81, 40/27, 29/9, 1532/81, 6551/81, ... ; Common denominator:...

Discriminant

\(-(16z-1)(25z^3-17z^2+2z+1)(-81+50z)^2\)

Local exponents

≈\(-0.17455\)\(0\)\(\frac{ 1}{ 16}\) ≈\(0.427275-0.215865I\) ≈\(0.427275+0.215865I\)\(\frac{ 81}{ 50}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(1\)
\(1\)\(0\)\(1\)\(1\)\(1\)\(1\)\(\frac{ 3}{ 2}\)
\(1\)\(0\)\(1\)\(1\)\(1\)\(3\)\(\frac{ 3}{ 2}\)
\(2\)\(0\)\(2\)\(2\)\(2\)\(4\)\(2\)

Note:

This is operator "6.9" from ...

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