Summary

You searched for: superseeker=361/21,120472/21

Your search produced exactly one match

1

New Number: 5.70 |  AESZ: 287  |  Superseeker: 361/21 120472/21  |  Hash: 97932196c46a8712f6dcb11165d698be  

Degree: 5

\(3^{2} 7^{2} \theta^4-3 7 x\left(3289\theta^4+6098\theta^3+4645\theta^2+1596\theta+210\right)+2^{2} 5 x^{2}\left(7712\theta^4-46168\theta^3-106885\theta^2-67410\theta-13629\right)+2^{4} x^{3}\left(106636\theta^4+493416\theta^3+420211\theta^2+116361\theta+6090\right)-2^{8} 5 x^{4}(2\theta+1)(1916\theta^3+2622\theta^2+1077\theta+91)-2^{12} 5^{2} x^{5}(2\theta+1)(\theta+1)^2(2\theta+3)\)

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Coefficients of the holomorphic solution: 1, 10, 510, 38260, 3473470, ...
--> OEIS
Normalized instanton numbers (n0=1): 361/21, 4780/21, 120472/21, 1537864/7, 216261320/21, ... ; Common denominator:...

Discriminant

\(-(64z^3+800z^2+149z-1)(-21+80z)^2\)

Local exponents

≈\(-12.310784\) ≈\(-0.195701\)\(0\) ≈\(0.006485\)\(\frac{ 21}{ 80}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(\frac{ 1}{ 2}\)
\(1\)\(1\)\(0\)\(1\)\(1\)\(1\)
\(1\)\(1\)\(0\)\(1\)\(3\)\(1\)
\(2\)\(2\)\(0\)\(2\)\(4\)\(\frac{ 3}{ 2}\)

Note:

This is operator "5.70" from ...

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