Summary

You searched for: sol=19207440

Your search produced exactly one match

1

New Number: 6.15 |  AESZ:  |  Superseeker: 64 76608  |  Hash: 0130ee676bad42a2e117bca3367f8cf0  

Degree: 6

\(\theta^4+2^{4} x\left(56\theta^4+16\theta^3+22\theta^2+14\theta+3\right)+2^{10} x^{2}\left(308\theta^4+272\theta^3+347\theta^2+174\theta+35\right)+2^{18} x^{3}\left(212\theta^4+384\theta^3+473\theta^2+282\theta+69\right)+2^{26} x^{4}\left(77\theta^4+232\theta^3+327\theta^2+226\theta+62\right)+2^{35} x^{5}(7\theta^2+17\theta+13)(\theta+1)^2+2^{42} x^{6}(\theta+1)^2(\theta+2)^2\)

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Coefficients of the holomorphic solution: 1, -48, 3088, -231168, 19207440, ...
--> OEIS
Normalized instanton numbers (n0=1): 64, -1732, 76608, -4429212, 296488640, ... ; Common denominator:...

Discriminant

\((64z+1)^2(128z+1)^2(256z+1)^2\)

Local exponents

\(-\frac{ 1}{ 64}\)\(-\frac{ 1}{ 128}\)\(-\frac{ 1}{ 256}\)\(0\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(1\)
\(\frac{ 1}{ 2}\)\(\frac{ 1}{ 2}\)\(1\)\(0\)\(1\)
\(\frac{ 1}{ 2}\)\(\frac{ 1}{ 2}\)\(3\)\(0\)\(2\)
\(1\)\(1\)\(4\)\(0\)\(2\)

Note:

This is operator "6.15" from ...

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