Summary

You searched for: superseeker=647/13,942613/13

Your search produced exactly one match

1

New Number: 4.34 |  AESZ: 99  |  Superseeker: 647/13 942613/13  |  Hash: f6c6b846edc829f336d8e4ae1dcb5618  

Degree: 4

\(13^{2} \theta^4-13 x\left(4569\theta^4+9042\theta^3+6679\theta^2+2158\theta+260\right)+2^{4} x^{2}\left(6386\theta^4-1774\theta^3-17898\theta^2-11596\theta-2119\right)+2^{8} x^{3}\left(67\theta^4+1248\theta^3+1091\theta^2+312\theta+26\right)-2^{12} x^{4}\left((2\theta+1)^4\right)\)

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Coefficients of the holomorphic solution: 1, 20, 2196, 369200, 75562900, ...
--> OEIS
Normalized instanton numbers (n0=1): 647/13, 16166/13, 942613/13, 80218296/13, 8418215008/13, ... ; Common denominator:...

Discriminant

\(-(256z^2+349z-1)(-13+16z)^2\)

Local exponents

\(-\frac{ 349}{ 512}-\frac{ 85}{ 512}\sqrt{ 17}\)\(0\)\(s_1\)\(s_2\)\(-\frac{ 349}{ 512}+\frac{ 85}{ 512}\sqrt{ 17}\)\(\frac{ 13}{ 16}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(\frac{ 1}{ 2}\)
\(1\)\(0\)\(1\)\(1\)\(1\)\(1\)\(\frac{ 1}{ 2}\)
\(1\)\(0\)\(1\)\(1\)\(1\)\(3\)\(\frac{ 1}{ 2}\)
\(2\)\(0\)\(2\)\(2\)\(2\)\(4\)\(\frac{ 1}{ 2}\)

Note:

Sporadic Operator.
There is a second MUM point hidden at infinity. That is operator AESZ 207/4.38
A-Incarnation: $5 \times 5$-Pfaffian in P^5

A-Incarnation: 5 \times 5 Pfaffian in P^5

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