Summary

You searched for: inst=11

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1

New Number: 6.21 |  AESZ:  |  Superseeker: 1 11  |  Hash: 319d6b2f1541de5252840442cc6f8dcd  

Degree: 6

\(\theta^4+x\left(6+27\theta+47\theta^2+40\theta^3+20\theta^4\right)-x^{2}(143\theta^2+286\theta+120)(\theta+1)^2-2 3^{2} x^{3}(\theta+2)(\theta+1)(291\theta^2+873\theta+766)-2^{2} 3^{3} 5 x^{4}(\theta+3)(\theta+1)(41\theta^2+164\theta+196)+2^{3} 3^{3} 5^{2} x^{5}(\theta+4)(\theta+1)(29\theta^2+145\theta+150)+2^{5} 3^{5} 5^{3} x^{6}(\theta+5)(\theta+4)(\theta+2)(\theta+1)\)

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Coefficients of the holomorphic solution: 1, -6, 60, -480, 5040, ...
--> OEIS
Normalized instanton numbers (n0=1): 1, 13/4, 11, 50, 1674/5, ... ; Common denominator:...

Discriminant

\((6z-1)(15z-1)(9z+1)(12z+1)(10z+1)^2\)

Local exponents

\(-\frac{ 1}{ 9}\)\(-\frac{ 1}{ 10}\)\(-\frac{ 1}{ 12}\)\(0\)\(\frac{ 1}{ 15}\)\(\frac{ 1}{ 6}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(1\)
\(1\)\(\frac{ 1}{ 2}\)\(1\)\(0\)\(1\)\(1\)\(2\)
\(1\)\(\frac{ 1}{ 2}\)\(1\)\(0\)\(1\)\(1\)\(4\)
\(2\)\(1\)\(2\)\(0\)\(2\)\(2\)\(5\)

Note:

This is operator "6.21" from ...

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2

New Number: 7.10 |  AESZ:  |  Superseeker: 1 11  |  Hash: b1c277f62ba740f9f7e0371ba53e4194  

Degree: 7

\(\theta^4-x\left(76\theta^4+80\theta^3+73\theta^2+33\theta+6\right)+x^{2}\left(2209\theta^4+4228\theta^3+4745\theta^2+2726\theta+648\right)-2 3^{2} x^{3}\left(1735\theta^4+4646\theta^3+6099\theta^2+4072\theta+1124\right)+2^{2} 3^{3} x^{4}\left(2085\theta^4+7388\theta^3+11695\theta^2+9140\theta+2844\right)-2^{3} 3^{3} x^{5}(\theta+1)(3707\theta^3+14055\theta^2+20242\theta+10704)+2^{6} 3^{5} x^{6}(\theta+1)(\theta+2)(86\theta^2+285\theta+262)-2^{7} 3^{8} x^{7}(\theta+1)(\theta+2)^2(\theta+3)\)

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Coefficients of the holomorphic solution: 1, 6, 60, 816, 13104, ...
--> OEIS
Normalized instanton numbers (n0=1): 1, 13/4, 11, 50, 1674/5, ... ; Common denominator:...

Discriminant

\(-(3z-1)(18z-1)(27z-1)(12z-1)^2(-1+2z)^2\)

Local exponents

\(0\)\(\frac{ 1}{ 27}\)\(\frac{ 1}{ 18}\)\(\frac{ 1}{ 12}\)\(\frac{ 1}{ 3}\)\(\frac{ 1}{ 2}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(1\)
\(0\)\(1\)\(1\)\(1\)\(1\)\(\frac{ 1}{ 2}\)\(2\)
\(0\)\(1\)\(1\)\(3\)\(1\)\(\frac{ 1}{ 2}\)\(2\)
\(0\)\(2\)\(2\)\(4\)\(2\)\(1\)\(3\)

Note:

This is operator "7.10" from ...

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3

New Number: 8.5 |  AESZ: 173  |  Superseeker: 11 -2434/3  |  Hash: afa82ed9ee239bb5fcac960f8884db01  

Degree: 8

\(\theta^4-x(7\theta^2+7\theta+2)(17\theta^2+17\theta+6)+2^{6} x^{2}\left(55\theta^4+112\theta^3+155\theta^2+86\theta+15\right)-2^{6} 3^{2} x^{3}\left(119\theta^4-714\theta^3-2185\theta^2-1656\theta-444\right)+2^{12} 3^{2} x^{4}\left(92\theta^4+184\theta^3+98\theta^2+6\theta+9\right)+2^{12} 3^{4} x^{5}\left(119\theta^4+1190\theta^3+671\theta^2-96\theta-140\right)+2^{18} 3^{4} x^{6}\left(55\theta^4+108\theta^3+149\theta^2+108\theta+27\right)+2^{18} 3^{6} x^{7}(7\theta^2+7\theta+2)(17\theta^2+17\theta+6)+2^{24} 3^{8} x^{8}\left((\theta+1)^4\right)\)

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Coefficients of the holomorphic solution: 1, 12, 420, 17472, 828324, ...
--> OEIS
Normalized instanton numbers (n0=1): 11, 229/4, -2434/3, 7512, 54801, ... ; Common denominator:...

Discriminant

\((72z-1)(8z+1)(64z-1)(9z+1)(1+576z^2)^2\)

Local exponents

\(-\frac{ 1}{ 8}\)\(-\frac{ 1}{ 9}\)\(0-\frac{ 1}{ 24}I\)\(0\)\(0+\frac{ 1}{ 24}I\)\(\frac{ 1}{ 72}\)\(\frac{ 1}{ 64}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(1\)
\(1\)\(1\)\(1\)\(0\)\(1\)\(1\)\(1\)\(1\)
\(1\)\(1\)\(3\)\(0\)\(3\)\(1\)\(1\)\(1\)
\(2\)\(2\)\(4\)\(0\)\(4\)\(2\)\(2\)\(1\)

Note:

Hadamard product $a \ast g$. This operator has a second MUM-point at infinity with the same instanton numbers.
It can be reduced to an operator of degree 4 with a single MUM-point defined over
$Q(\sqrt{?})$.

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4

New Number: 8.6 |  AESZ: 113  |  Superseeker: 11 1200  |  Hash: 3754b3cce7930e99efa8acb802e524bb  

Degree: 8

\(\theta^4-x(10\theta^2+10\theta+3)(11\theta^2+11\theta+3)+x^{2}\left(1025\theta^4+3992\theta^3+5533\theta^2+3082\theta+615\right)-3^{2} x^{3}\left(110\theta^4-660\theta^3-2027\theta^2-1509\theta-369\right)+3^{2} x^{4}\left(2032\theta^4+4064\theta^3-2726\theta^2-4758\theta-1431\right)+3^{4} x^{5}\left(110\theta^4+1100\theta^3+613\theta^2-125\theta-117\right)+3^{4} x^{6}\left(1025\theta^4+108\theta^3-293\theta^2+108\theta+99\right)+3^{6} x^{7}(10\theta^2+10\theta+3)(11\theta^2+11\theta+3)+3^{8} x^{8}\left((\theta+1)^4\right)\)

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Coefficients of the holomorphic solution: 1, 9, 285, 13671, 799389, ...
--> OEIS
Normalized instanton numbers (n0=1): 11, 66, 1200, 28201, 802124, ... ; Common denominator:...

Discriminant

\((81z^2+99z-1)(z^2+11z-1)(1+9z^2)^2\)

Local exponents

\(-\frac{ 11}{ 2}-\frac{ 5}{ 2}\sqrt{ 5}\)\(-\frac{ 11}{ 18}-\frac{ 5}{ 18}\sqrt{ 5}\)\(0-\frac{ 1}{ 3}I\)\(0\)\(0+\frac{ 1}{ 3}I\)\(-\frac{ 11}{ 18}+\frac{ 5}{ 18}\sqrt{ 5}\)\(-\frac{ 11}{ 2}+\frac{ 5}{ 2}\sqrt{ 5}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(1\)
\(1\)\(1\)\(1\)\(0\)\(1\)\(1\)\(1\)\(1\)
\(1\)\(1\)\(3\)\(0\)\(3\)\(1\)\(1\)\(1\)
\(2\)\(2\)\(4\)\(0\)\(4\)\(2\)\(2\)\(1\)

Note:

Hadamard product $b \ast c$.This operator has a second MUM-point at infinity with the same instanton numbers.
If can be reduced to an operator of degree 4 with a single MUM-point defined over
$Q(\sqrt{?})$.

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