Summary

You searched for: inst=35/2

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1

New Number: 6.22 |  AESZ: 376  |  Superseeker: 5/4 35/2  |  Hash: 24fee66b67ff1a8a852f7a562f7665c1  

Degree: 6

\(2^{12} \theta^4-2^{10} x\left(106\theta^4+212\theta^3+183\theta^2+77\theta+13\right)+2^{8} x^{2}\left(19\theta^4+76\theta^3-43\theta^2-238\theta-141\right)+2^{6} 3 x^{3}\left(2988\theta^4+17928\theta^3+39970\theta^2+39234\theta+14267\right)+2^{4} 3^{3} x^{4}\left(309\theta^4+2472\theta^3+7618\theta^2+10696\theta+5311\right)-2^{2} 3^{3} 5 7^{2} x^{5}(\theta+4)(\theta+1)(22\theta^2+110\theta+123)-3^{5} 5^{2} 7^{2} x^{6}(\theta+5)(\theta+4)(\theta+2)(\theta+1)\)

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Coefficients of the holomorphic solution: 1, 13/4, 489/16, 25429/64, 1591057/256, ...
--> OEIS
Normalized instanton numbers (n0=1): 5/4, 57/16, 35/2, 459/4, 3615/4, ... ; Common denominator:...

Discriminant

\(-(7z+4)(5z-4)(105z-4)(9z-4)(4+3z)^2\)

Local exponents

\(-\frac{ 4}{ 3}\)\(-\frac{ 4}{ 7}\)\(0\)\(\frac{ 4}{ 105}\)\(\frac{ 4}{ 9}\)\(\frac{ 4}{ 5}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(1\)
\(\frac{ 1}{ 3}\)\(1\)\(0\)\(1\)\(1\)\(1\)\(2\)
\(\frac{ 2}{ 3}\)\(1\)\(0\)\(1\)\(1\)\(1\)\(4\)
\(1\)\(2\)\(0\)\(2\)\(2\)\(2\)\(5\)

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2

New Number: 8.40 |  AESZ:  |  Superseeker: 5/4 35/2  |  Hash: d3cb0fbbc65d6c5dace733d3d1ca181b  

Degree: 8

\(2^{4} \theta^4-2^{2} x\theta(2\theta^3+82\theta^2+53\theta+12)-x^{2}\left(4895\theta^4+18410\theta^3+26199\theta^2+18308\theta+5120\right)-x^{3}\left(60679\theta^4+272424\theta^3+497452\theta^2+430092\theta+143808\right)-x^{4}\left(344527\theta^4+1870838\theta^3+4034628\theta^2+3987101\theta+1478544\right)-x^{5}(\theta+1)(1076509\theta^3+5847783\theta^2+11226106\theta+7492832)-2 x^{6}(\theta+1)(\theta+2)(944887\theta^2+4249317\theta+5045304)-2^{8} 13 x^{7}(\theta+3)(\theta+2)(\theta+1)(518\theta+1381)-2^{5} 5 13^{2} 23 x^{8}(\theta+1)(\theta+2)(\theta+3)(\theta+4)\)

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Coefficients of the holomorphic solution: 1, 0, 20, 168, 2652, ...
--> OEIS
Normalized instanton numbers (n0=1): 5/4, 57/16, 35/2, 459/4, 3615/4, ... ; Common denominator:...

Discriminant

\(-(23z-1)(5z+1)(2z+1)(z+1)(13z+4)^2(4z+1)^2\)

Local exponents

\(-1\)\(-\frac{ 1}{ 2}\)\(-\frac{ 4}{ 13}\)\(-\frac{ 1}{ 4}\)\(-\frac{ 1}{ 5}\)\(0\)\(\frac{ 1}{ 23}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(1\)
\(1\)\(1\)\(1\)\(\frac{ 1}{ 3}\)\(1\)\(0\)\(1\)\(2\)
\(1\)\(1\)\(3\)\(\frac{ 2}{ 3}\)\(1\)\(0\)\(1\)\(3\)
\(2\)\(2\)\(4\)\(1\)\(2\)\(0\)\(2\)\(4\)

Note:

This is operator "8.40" from ...

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