### Summary

You searched for: superseeker=11,1200

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1

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
$Q(\sqrt{?})$.