Summary

You searched for: superseeker=10/7,295/7

Your search produced exactly one match

1

New Number: 5.5 |  AESZ: 22  |  Superseeker: 10/7 295/7  |  Hash: 5b96eae0872756be1130d4b12ffe60a6  

Degree: 5

\(7^{2} \theta^4-7 x\left(155\theta^4+286\theta^3+234\theta^2+91\theta+14\right)-x^{2}\left(16105\theta^4+68044\theta^3+102261\theta^2+66094\theta+15736\right)+2^{3} x^{3}\left(2625\theta^4+8589\theta^3+9071\theta^2+3759\theta+476\right)-2^{4} x^{4}\left(465\theta^4+1266\theta^3+1439\theta^2+806\theta+184\right)+2^{9} x^{5}\left((\theta+1)^4\right)\)

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Coefficients of the holomorphic solution: 1, 2, 34, 488, 9826, ...
--> OEIS
Normalized instanton numbers (n0=1): 10/7, 65/7, 295/7, 3065/7, 4245, ... ; Common denominator:...

Discriminant

\((32z-1)(z^2-11z-1)(4z-7)^2\)

Local exponents

\(\frac{ 11}{ 2}-\frac{ 5}{ 2}\sqrt{ 5}\)\(0\)\(\frac{ 1}{ 32}\)\(\frac{ 7}{ 4}\)\(\frac{ 11}{ 2}+\frac{ 5}{ 2}\sqrt{ 5}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(1\)
\(1\)\(0\)\(1\)\(1\)\(1\)\(1\)
\(1\)\(0\)\(1\)\(3\)\(1\)\(1\)
\(2\)\(0\)\(2\)\(4\)\(2\)\(1\)

Note:

There is a second MUM-point at infinity, corresponding to Operator AESZ 118/5.16
A-Incarnation: five (1,1) sections in ${\bf P}^4 \times {\bf P}^4$.Quotient by ${\bf Z}/2$ of this:
the Reye congruence Calabi-Yau threefold.

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