Summary

You searched for: inst=51036

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1

New Number: 6.39 |  AESZ:  |  Superseeker: 8 3784/3  |  Hash: 6429f42cbe18bee944ac13edab1fbbcc  

Degree: 6

\(\theta^4+2^{2} x\left(49\theta^4+98\theta^3+86\theta^2+37\theta+6\right)+2^{5} x^{2}\left(593\theta^4+2372\theta^3+3521\theta^2+2298\theta+504\right)+2^{10} 3 x^{3}\left(332\theta^4+1992\theta^3+4194\theta^2+3618\theta+945\right)+2^{14} 3^{2} x^{4}\left(204\theta^4+1632\theta^3+4449\theta^2+4740\theta+1400\right)+2^{18} 3^{3} x^{5}(16\theta^2+80\theta+35)(2\theta+5)^2+2^{21} 3^{4} x^{6}(2\theta+11)(2\theta+7)(2\theta+5)(2\theta+1)\)

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Coefficients of the holomorphic solution: 1, -24, 648, -11520, -123480, ...
--> OEIS
Normalized instanton numbers (n0=1): 8, 39/2, 3784/3, 51036, 1659840, ... ; Common denominator:...

Discriminant

\((24z+1)(110592z^3+6912z^2+108z+1)(1+32z)^2\)

Local exponents

≈\(-0.045368\)\(-\frac{ 1}{ 24}\)\(-\frac{ 1}{ 32}\) ≈\(-0.008566-0.011222I\) ≈\(-0.008566+0.011222I\)\(0\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(\frac{ 1}{ 2}\)
\(1\)\(1\)\(\frac{ 1}{ 2}\)\(1\)\(1\)\(0\)\(\frac{ 5}{ 2}\)
\(1\)\(1\)\(\frac{ 1}{ 2}\)\(1\)\(1\)\(0\)\(\frac{ 7}{ 2}\)
\(2\)\(2\)\(1\)\(2\)\(2\)\(0\)\(\frac{ 11}{ 2}\)

Note:

This is operator "6.39" from ...

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2

New Number: 7.11 |  AESZ:  |  Superseeker: -8 -3784/3  |  Hash: cb1bf6566f9c1a0dbfe98fb55f81944c  

Degree: 7

\(\theta^4+2^{2} x\left(23\theta^4-34\theta^3-30\theta^2-13\theta-2\right)+2^{5} x^{2}\left(177\theta^4+108\theta^3+577\theta^2+518\theta+116\right)+2^{10} x^{3}\left(355\theta^4+960\theta^3+1178\theta^2+139\theta-44\right)+2^{15} x^{4}\left(451\theta^4+1228\theta^3+997\theta^2+489\theta+103\right)+2^{20} x^{5}\left(285\theta^4+720\theta^3+766\theta^2+410\theta+83\right)+2^{26} x^{6}(2\theta+1)(20\theta^3+50\theta^2+49\theta+17)+2^{31} x^{7}(2\theta+1)(\theta+1)^2(2\theta+3)\)

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Coefficients of the holomorphic solution: 1, 8, -120, -4480, 55720, ...
--> OEIS
Normalized instanton numbers (n0=1): -8, 43/2, -3784/3, 51036, -1659840, ... ; Common denominator:...

Discriminant

\((8z+1)(32768z^3+3072z^2-12z+1)(32z+1)^3\)

Local exponents

\(-\frac{ 1}{ 8}\) ≈\(-0.100423\)\(-\frac{ 1}{ 32}\)\(0\) ≈\(0.003336-0.01711I\) ≈\(0.003336+0.01711I\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(\frac{ 1}{ 2}\)
\(1\)\(1\)\(2\)\(0\)\(1\)\(1\)\(1\)
\(1\)\(1\)\(3\)\(0\)\(1\)\(1\)\(1\)
\(2\)\(2\)\(5\)\(0\)\(2\)\(2\)\(\frac{ 3}{ 2}\)

Note:

This is operator "7.11" from ...

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