Summary

You searched for: inst=13/3

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1

New Number: 5.6 |  AESZ: 23  |  Superseeker: 4/3 44/3  |  Hash: 65760d446ba9c3da587ce5bd9912745e  

Degree: 5

\(3^{2} \theta^4-2^{2} 3 x\left(64\theta^4+80\theta^3+73\theta^2+33\theta+6\right)+2^{7} x^{2}\left(194\theta^4+440\theta^3+527\theta^2+315\theta+75\right)-2^{12} x^{3}\left(94\theta^4+288\theta^3+397\theta^2+261\theta+66\right)+2^{17} x^{4}\left(22\theta^4+80\theta^3+117\theta^2+77\theta+19\right)-2^{23} x^{5}\left((\theta+1)^4\right)\)

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Coefficients of the holomorphic solution: 1, 8, 104, 1664, 30376, ...
--> OEIS
Normalized instanton numbers (n0=1): 4/3, 13/3, 44/3, 278/3, 2336/3, ... ; Common denominator:...

Discriminant

\(-(-1+32z)(16z-1)^2(32z-3)^2\)

Local exponents

\(0\)\(\frac{ 1}{ 32}\)\(\frac{ 1}{ 16}\)\(\frac{ 3}{ 32}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(1\)
\(0\)\(1\)\(\frac{ 1}{ 2}\)\(1\)\(1\)
\(0\)\(1\)\(\frac{ 1}{ 2}\)\(3\)\(1\)
\(0\)\(2\)\(1\)\(4\)\(1\)

Note:

There is a second MUM-point at infinity,corresponding to Operator AESZ 56/5.9
A-Incarnation: (2,0),(2.0),(0,2),(0,2),(1,1).intersection in $P^4 \times P^4$

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2

New Number: 6.31 |  AESZ:  |  Superseeker: 2/3 13/3  |  Hash: fcd8db2a3ad7e58151e501b5872652df  

Degree: 6

\(3^{6} \theta^4+2 3^{5} x\left(7\theta^2+7\theta+2\right)-2^{2} 3^{4} x^{2}\left(465\theta^4+1860\theta^3+3069\theta^2+2418\theta+752\right)-3^{3} x^{3}(\theta+2)(\theta+1)(19327\theta^2+57981\theta+52674)+2^{5} 3^{2} x^{4}(17298\theta^2+69192\theta+54655)(\theta+2)^2+2^{4} 3^{2} 11 31 251 x^{5}(\theta+1)(\theta+2)(\theta+3)(\theta+4)-2^{3} 11^{2} 251^{2} x^{6}(\theta+5)(\theta+4)(\theta+2)(\theta+1)\)

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Coefficients of the holomorphic solution: 1, -4/3, 196/9, -604/27, 83956/81, ...
--> OEIS
Normalized instanton numbers (n0=1): 2/3, 5/3, 13/3, 59/3, 119, ... ; Common denominator:...

Discriminant

\(-(11z-3)(22z+3)(1004z^2+66z-9)(251z^2-33z-9)\)

Local exponents

\(-\frac{ 3}{ 22}\)\(\frac{ 33}{ 502}-\frac{ 45}{ 502}\sqrt{ 5}\)\(-\frac{ 33}{ 1004}-\frac{ 45}{ 1004}\sqrt{ 5}\)\(0\)\(-\frac{ 33}{ 1004}+\frac{ 45}{ 1004}\sqrt{ 5}\)\(\frac{ 33}{ 502}+\frac{ 45}{ 502}\sqrt{ 5}\)\(\frac{ 3}{ 11}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(1\)
\(1\)\(1\)\(1\)\(0\)\(1\)\(1\)\(1\)\(2\)
\(1\)\(1\)\(1\)\(0\)\(1\)\(1\)\(1\)\(4\)
\(2\)\(2\)\(2\)\(0\)\(2\)\(2\)\(2\)\(5\)

Note:

This is operator "6.31" from ...

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3

New Number: 7.7 |  AESZ:  |  Superseeker: 2/3 13/3  |  Hash: c7abb9c42d46f14955f0f23351082bef  

Degree: 7

\(3^{2} \theta^4-2 3 x\left(88\theta^4+110\theta^3+103\theta^2+48\theta+9\right)+2^{2} x^{2}\left(2923\theta^4+6610\theta^3+8041\theta^2+4908\theta+1206\right)-x^{3}\left(123365\theta^4+374814\theta^3+519741\theta^2+346176\theta+89676\right)+2 x^{4}\left(309657\theta^4+1102938\theta^3+1591157\theta^2+1032920\theta+249740\right)-2^{3} 11 x^{5}(\theta+1)(12897\theta^3+35469\theta^2+31181\theta+8042)-2^{3} 11^{2} x^{6}(\theta+1)(\theta+2)(355\theta^2+1047\theta+806)-2^{4} 11^{3} x^{7}(\theta+1)(\theta+2)^2(\theta+3)\)

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Coefficients of the holomorphic solution: 1, 6, 56, 636, 8196, ...
--> OEIS
Normalized instanton numbers (n0=1): 2/3, 5/3, 13/3, 59/3, 119, ... ; Common denominator:...

Discriminant

\(-(11z-1)(4z^2+22z-1)(z^2+11z-1)(-3+22z)^2\)

Local exponents

\(-\frac{ 11}{ 2}-\frac{ 5}{ 2}\sqrt{ 5}\)\(-\frac{ 11}{ 4}-\frac{ 5}{ 4}\sqrt{ 5}\)\(0\)\(-\frac{ 11}{ 4}+\frac{ 5}{ 4}\sqrt{ 5}\)\(-\frac{ 11}{ 2}+\frac{ 5}{ 2}\sqrt{ 5}\)\(\frac{ 1}{ 11}\)\(\frac{ 3}{ 22}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(1\)
\(1\)\(1\)\(0\)\(1\)\(1\)\(1\)\(1\)\(2\)
\(1\)\(1\)\(0\)\(1\)\(1\)\(1\)\(3\)\(2\)
\(2\)\(2\)\(0\)\(2\)\(2\)\(2\)\(4\)\(3\)

Note:

This is operator "7.7" from ...

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