Summary

You searched for: superseeker=7/13,21/13

Your search produced exactly one match

1

New Number: 5.104 |  AESZ: 357  |  Superseeker: 7/13 21/13  |  Hash: afee0651c9b3b8e98079f5c2d5bfa8a5  

Degree: 5

\(13^{2} \theta^4-13 x\left(441\theta^4+690\theta^3+631\theta^2+286\theta+52\right)+2^{4} x^{2}\left(5121\theta^4+15576\theta^3+21215\theta^2+13702\theta+3445\right)-2^{10} x^{3}\left(640\theta^4+2847\theta^3+5078\theta^2+4056\theta+1196\right)+2^{14} x^{4}\left(125\theta^4+562\theta^3+905\theta^2+624\theta+157\right)-2^{21} x^{5}\left((\theta+1)^4\right)\)

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Coefficients of the holomorphic solution: 1, 4, 20, 112, 916, ...
--> OEIS
Normalized instanton numbers (n0=1): 7/13, -10/13, 21/13, 296/13, 608/13, ... ; Common denominator:...

Discriminant

\(-(16z-1)(128z^2-13z+1)(-13+32z)^2\)

Local exponents

\(0\)\(\frac{ 13}{ 256}-\frac{ 7}{ 256}\sqrt{ 7}I\)\(\frac{ 13}{ 256}+\frac{ 7}{ 256}\sqrt{ 7}I\)\(\frac{ 1}{ 16}\)\(\frac{ 13}{ 32}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(1\)
\(0\)\(1\)\(1\)\(1\)\(1\)\(1\)
\(0\)\(1\)\(1\)\(1\)\(3\)\(1\)
\(0\)\(2\)\(2\)\(2\)\(4\)\(1\)

Note:

There is a second MUM-point at infinity,
corresponding to Operator AESZ 358/5.105

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