Summary

You searched for: superseeker=126/17,11700/17

Your search produced exactly one match

1

New Number: 5.23 |  AESZ: 194  |  Superseeker: 126/17 11700/17  |  Hash: 6bf19665aa6705f30ef88df42bc4eac4  

Degree: 5

\(17^{2} \theta^4-17 x\left(1465\theta^4+2768\theta^3+2200\theta^2+816\theta+119\right)+2 x^{2}\left(62015\theta^4+131582\theta^3+125017\theta^2+65926\theta+15300\right)-2 3^{3} x^{3}\left(4325\theta^4+10914\theta^3+12803\theta^2+7446\theta+1700\right)+3^{6} x^{4}\left(265\theta^4+836\theta^3+1118\theta^2+700\theta+168\right)-3^{10} x^{5}\left((\theta+1)^4\right)\)

Maple   LaTex

Coefficients of the holomorphic solution: 1, 7, 183, 7225, 345079, ...
--> OEIS
Normalized instanton numbers (n0=1): 126/17, 848/17, 11700/17, 229808/17, 5539258/17, ... ; Common denominator:...

Discriminant

\(-(-1+81z)(27z-17)^2(z-1)^2\)

Local exponents

\(0\)\(\frac{ 1}{ 81}\)\(\frac{ 17}{ 27}\)\(1\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(1\)
\(0\)\(1\)\(1\)\(\frac{ 1}{ 2}\)\(1\)
\(0\)\(1\)\(3\)\(\frac{ 1}{ 2}\)\(1\)
\(0\)\(2\)\(4\)\(1\)\(1\)

Note:

There is a second MUM-point at infinity, corresponding to Operator AESZ 199/5.26

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