Summary

You searched for: inst=22/5

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1

New Number: 5.47 |  AESZ: 246  |  Superseeker: -4/5 -108/5  |  Hash: f51a0c39f9179dc6a561b9afb6f9d85f  

Degree: 5

\(5^{2} \theta^4-2^{2} 5 x\left(12\theta^4+48\theta^3+49\theta^2+25\theta+5\right)-2^{4} x^{2}\left(544\theta^4+1792\theta^3+2444\theta^2+1580\theta+405\right)+2^{9} x^{3}\left(112\theta^4+960\theta^3+2306\theta^2+2130\theta+685\right)+2^{12} x^{4}\left(144\theta^4+768\theta^3+1308\theta^2+924\theta+235\right)+2^{20} x^{5}\left((\theta+1)^4\right)\)

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Coefficients of the holomorphic solution: 1, 4, 44, 400, 5356, ...
--> OEIS
Normalized instanton numbers (n0=1): -4/5, 22/5, -108/5, 694/5, -1040, ... ; Common denominator:...

Discriminant

\((1+16z)(16z+5)^2(16z-1)^2\)

Local exponents

\(-\frac{ 5}{ 16}\)\(-\frac{ 1}{ 16}\)\(0\)\(\frac{ 1}{ 16}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(1\)
\(1\)\(1\)\(0\)\(\frac{ 1}{ 2}\)\(1\)
\(3\)\(1\)\(0\)\(\frac{ 1}{ 2}\)\(1\)
\(4\)\(2\)\(0\)\(1\)\(1\)

Note:

There is a second MUM-point at infinity,
corresponding to Operator AESZ 247/5.48

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2

New Number: 7.16 |  AESZ:  |  Superseeker: 22/5 68  |  Hash: 660211ce6175f36772066594bfc33cbb  

Degree: 7

\(5^{2} \theta^4-2 5 x\theta(15+71\theta+112\theta^2+38\theta^3)-2^{2} x^{2}\left(4364\theta^4+15872\theta^3+24679\theta^2+19360\theta+6000\right)-2^{4} 3^{2} 5 x^{3}\left(92\theta^4+224\theta^3+103\theta^2-176\theta-165\right)+2^{6} 3^{2} x^{4}\left(1228\theta^4+10496\theta^3+30154\theta^2+35736\theta+14715\right)+2^{9} 3^{4} x^{5}(\theta+1)(38\theta^3+74\theta^2-304\theta-495)-2^{10} 3^{4} x^{6}(2\theta+13)(2\theta+3)(17\theta+39)(\theta+1)-2^{12} 3^{6} x^{7}(\theta+1)(2\theta+5)(2\theta+3)(\theta+3)\)

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Coefficients of the holomorphic solution: 1, 0, 60, 480, 16524, ...
--> OEIS
Normalized instanton numbers (n0=1): 22/5, 8, 68, 3292/5, 38826/5, ... ; Common denominator:...

Discriminant

\(-(-1+36z)(12z+5)^2(12z+1)^2(4z-1)^2\)

Local exponents

\(-\frac{ 5}{ 12}\)\(-\frac{ 1}{ 12}\)\(0\)\(\frac{ 1}{ 36}\)\(\frac{ 1}{ 4}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(1\)
\(1\)\(0\)\(0\)\(1\)\(\frac{ 1}{ 2}\)\(\frac{ 3}{ 2}\)
\(3\)\(1\)\(0\)\(1\)\(\frac{ 1}{ 2}\)\(\frac{ 5}{ 2}\)
\(4\)\(1\)\(0\)\(2\)\(1\)\(3\)

Note:

This is operator "7.16" from ...

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3

New Number: 8.79 |  AESZ:  |  Superseeker: 22/5 68  |  Hash: 064e5b590dd8b6a4daa1e905fbe693c2  

Degree: 8

\(5^{2} \theta^4-2 5 x\left(338\theta^4+412\theta^3+371\theta^2+165\theta+30\right)+2^{2} x^{2}\left(46396\theta^4+103408\theta^3+125291\theta^2+76370\theta+19080\right)-2^{4} 3 x^{3}\left(115508\theta^4+357896\theta^3+524149\theta^2+375205\theta+106530\right)+2^{6} 3^{2} x^{4}\left(173456\theta^4+669024\theta^3+1118292\theta^2+883484\theta+269049\right)-2^{11} 3^{3} x^{5}\left(20272\theta^4+91616\theta^3+168594\theta^2+142006\theta+45053\right)+2^{14} 3^{4} x^{6}\left(5792\theta^4+29504\theta^3+58300\theta^2+51220\theta+16641\right)-2^{21} 3^{5} x^{7}(\theta+1)^2(58\theta^2+208\theta+201)+2^{26} 3^{6} x^{8}(\theta+1)^2(\theta+2)^2\)

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Coefficients of the holomorphic solution: 1, 12, 204, 4368, 112140, ...
--> OEIS
Normalized instanton numbers (n0=1): 22/5, 8, 68, 3292/5, 38826/5, ... ; Common denominator:...

Discriminant

\((-1+48z)(16z-1)^2(48z-5)^2(12z-1)^3\)

Local exponents

\(0\)\(\frac{ 1}{ 48}\)\(\frac{ 1}{ 16}\)\(\frac{ 1}{ 12}\)\(\frac{ 5}{ 48}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(1\)
\(0\)\(1\)\(\frac{ 1}{ 2}\)\(\frac{ 1}{ 2}\)\(1\)\(1\)
\(0\)\(1\)\(\frac{ 1}{ 2}\)\(\frac{ 3}{ 2}\)\(3\)\(2\)
\(0\)\(2\)\(1\)\(2\)\(4\)\(2\)

Note:

This is operator "8.79" from ...

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