Summary

You searched for: sol=23760

Your search produced exactly one match

1

New Number: 8.83 |  AESZ:  |  Superseeker: 208 642704  |  Hash: 7314ca8e48f991223dc4e1c8b4893b95  

Degree: 8

\(\theta^4-2^{4} x\left(116\theta^4+160\theta^3+119\theta^2+39\theta+5\right)+2^{9} x^{2}\left(2096\theta^4+5600\theta^3+5694\theta^2+2366\theta+355\right)-2^{15} x^{3}\left(4232\theta^4+22416\theta^3+28566\theta^2+11646\theta+1745\right)-2^{21} x^{4}\left(20616\theta^4+8496\theta^3-69074\theta^2-48074\theta-9335\right)+2^{27} x^{5}\left(49408\theta^4+114208\theta^3-29684\theta^2-42372\theta-9585\right)+2^{34} x^{6}\left(46496\theta^4-21984\theta^3-28956\theta^2-5580\theta+375\right)-2^{41} 5 x^{7}(2\theta+1)^2(344\theta^2+416\theta+163)-2^{48} 5^{2} x^{8}(2\theta+1)^2(2\theta+3)^2\)

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Coefficients of the holomorphic solution: 1, 80, 23760, 9900800, 4805155600, ...
--> OEIS
Normalized instanton numbers (n0=1): 208, 3154, 642704, -4424361, 3864242160, ... ; Common denominator:...

Discriminant

\(-(16384z^2-768z+1)(4096z^2+704z-1)(128z+1)^2(320z-1)^2\)

Local exponents

\(-\frac{ 11}{ 128}-\frac{ 5}{ 128}\sqrt{ 5}\)\(-\frac{ 1}{ 128}\)\(0\)\(\frac{ 3}{ 128}-\frac{ 1}{ 64}\sqrt{ 2}\)\(-\frac{ 11}{ 128}+\frac{ 5}{ 128}\sqrt{ 5}\)\(\frac{ 1}{ 320}\)\(\frac{ 3}{ 128}+\frac{ 1}{ 64}\sqrt{ 2}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(\frac{ 1}{ 2}\)
\(1\)\(1\)\(0\)\(1\)\(1\)\(1\)\(1\)\(\frac{ 1}{ 2}\)
\(1\)\(3\)\(0\)\(1\)\(1\)\(3\)\(1\)\(\frac{ 3}{ 2}\)
\(2\)\(4\)\(0\)\(2\)\(2\)\(4\)\(2\)\(\frac{ 3}{ 2}\)

Note:

This is operator "8.83" from ...

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