Summary

You searched for: sol=140505

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1

New Number: 8.31 |  AESZ: 315  |  Superseeker: 38 26135  |  Hash: 44be55b95bb1c725c5aaa2c9a6635e89  

Degree: 8

\(5^{2} \theta^4-5^{2} x\left(239\theta^4+496\theta^3+368\theta^2+120\theta+15\right)-2 3 5 x^{2}\left(1727\theta^4+3206\theta^3+2341\theta^2+1090\theta+245\right)-3^{2} 5 x^{3}\left(1519\theta^4+7338\theta^3+14271\theta^2+8340\theta+1690\right)+3^{3} x^{4}\left(10358\theta^4-16622\theta^3-49763\theta^2-37900\theta-10210\right)+3^{4} 5 x^{5}\left(922\theta^4+3526\theta^3-1357\theta^2-3028\theta-1031\right)-3^{5} x^{6}\left(1219\theta^4-6030\theta^3-6441\theta^2-1740\theta+160\right)-2^{2} 3^{6} x^{7}\left(162\theta^4+234\theta^3+65\theta^2-52\theta-25\right)-2^{4} 3^{8} x^{8}\left((\theta+1)^4\right)\)

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Coefficients of the holomorphic solution: 1, 15, 1179, 140505, 20362059, ...
--> OEIS
Normalized instanton numbers (n0=1): 38, 3068/5, 26135, 7871998/5, 117518569, ... ; Common denominator:...

Discriminant

\(-(z+1)(81z^3+351z^2+246z-1)(-5-15z+36z^2)^2\)

Local exponents

≈\(-3.452681\)\(-1\) ≈\(-0.884694\)\(\frac{ 5}{ 24}-\frac{ 1}{ 24}\sqrt{ 105}\)\(0\) ≈\(0.004042\)\(\frac{ 5}{ 24}+\frac{ 1}{ 24}\sqrt{ 105}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(1\)
\(1\)\(1\)\(1\)\(1\)\(0\)\(1\)\(1\)\(1\)
\(1\)\(1\)\(1\)\(3\)\(0\)\(1\)\(3\)\(1\)
\(2\)\(2\)\(2\)\(4\)\(0\)\(2\)\(4\)\(1\)

Note:

This operator has a second MUM-point at infinity corresponding to operator 8.32

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