Summary

You searched for: Spectrum0=1/4,1/2,1/2,3/4

Your search produced 5 matches

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1

New Number: 4.61 |  AESZ: 289  |  Superseeker: 8224 15542388128  |  Hash: 673413653f5554d4f0cc1a8af33e8bbe  

Degree: 4

\(\theta^4-2^{4} x\left(400\theta^4+2720\theta^3+1752\theta^2+392\theta+33\right)-2^{15} x^{2}\left(4272\theta^4+6288\theta^3-3184\theta^2-1484\theta-177\right)-2^{24} 5 x^{3}\left(4688\theta^4-1536\theta^3-1384\theta^2-336\theta-27\right)+2^{36} 5^{2} x^{4}(4\theta+1)(2\theta+1)^2(4\theta+3)\)

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Coefficients of the holomorphic solution: 1, 528, 2434320, 18496262400, 174225386134800, ...
--> OEIS
Normalized instanton numbers (n0=1): 8224, 3407456, 15542388128, 54609260446560, 282477571639256928, ... ; Common denominator:...

Discriminant

\((16384z-1)(256z-1)(1+5120z)^2\)

Local exponents

\(-\frac{ 1}{ 5120}\)\(0\)\(\frac{ 1}{ 16384}\)\(\frac{ 1}{ 256}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(\frac{ 1}{ 4}\)
\(1\)\(0\)\(1\)\(1\)\(\frac{ 1}{ 2}\)
\(3\)\(0\)\(1\)\(1\)\(\frac{ 1}{ 2}\)
\(4\)\(0\)\(2\)\(2\)\(\frac{ 3}{ 4}\)

Note:

Sporadic Operator.

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2

New Number: 4.62 |  AESZ: 292  |  Superseeker: 4300/3 1701817028/3  |  Hash: bce26f214ee56f65c7a275cd8fdcc0c7  

Degree: 4

\(3^{2} \theta^4-2^{2} 3 x\left(4636\theta^4+7928\theta^3+5347\theta^2+1383\theta+126\right)+2^{9} x^{2}\left(59048\theta^4+50888\theta^3-26248\theta^2-16827\theta-2205\right)-2^{16} 7 x^{3}\left(9004\theta^4-2304\theta^3-2511\theta^2-504\theta-27\right)-2^{24} 7^{2} x^{4}(4\theta+1)(2\theta+1)^2(4\theta+3)\)

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Coefficients of the holomorphic solution: 1, 168, 279720, 737721600, 2391487698600, ...
--> OEIS
Normalized instanton numbers (n0=1): 4300/3, 1768292/3, 1701817028/3, 2484553593752/3, 1500880129466144, ... ; Common denominator:...

Discriminant

\(-(65536z^2+5584z-1)(-3+896z)^2\)

Local exponents

\(-\frac{ 349}{ 8192}-\frac{ 85}{ 8192}\sqrt{ 17}\)\(0\)\(s_1\)\(s_2\)\(-\frac{ 349}{ 8192}+\frac{ 85}{ 8192}\sqrt{ 17}\)\(\frac{ 3}{ 896}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(\frac{ 1}{ 4}\)
\(1\)\(0\)\(1\)\(1\)\(1\)\(1\)\(\frac{ 1}{ 2}\)
\(1\)\(0\)\(1\)\(1\)\(1\)\(3\)\(\frac{ 1}{ 2}\)
\(2\)\(0\)\(2\)\(2\)\(2\)\(4\)\(\frac{ 3}{ 4}\)

Note:

Sporadic Operator.

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3

New Number: 4.63 |  AESZ: 294  |  Superseeker: -80416 -15561562691488  |  Hash: f6b5a258285779facb6702e1a2a891bb  

Degree: 4

\(\theta^4-2^{4} x\left(18800\theta^4-14624\theta^3-8184\theta^2-872\theta-33\right)+2^{18} x^{2}\left(101744\theta^4-107920\theta^3+74968\theta^2+15100\theta+1191\right)-2^{30} 17 x^{3}\left(40048\theta^4+49152\theta^3+35848\theta^2+10752\theta+1143\right)+2^{50} 17^{2} x^{4}(4\theta+1)(2\theta+1)^2(4\theta+3)\)

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Coefficients of the holomorphic solution: 1, -528, -16919280, -95988076800, 3707387171888400, ...
--> OEIS
Normalized instanton numbers (n0=1): -80416, -872844376, -15561562691488, -341695175348542432, -8482586861983707669856, ... ; Common denominator:...

Discriminant

\((1073741824z^2-22272z+1)(-1+139264z)^2\)

Local exponents

\(0\)\(s_1\)\(s_2\)\(\frac{ 1}{ 139264}\)\(\frac{ 87}{ 8388608}-\frac{ 91}{ 8388608}\sqrt{ 7}I\)\(\frac{ 87}{ 8388608}+\frac{ 91}{ 8388608}\sqrt{ 7}I\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(\frac{ 1}{ 4}\)
\(0\)\(1\)\(1\)\(1\)\(1\)\(1\)\(\frac{ 1}{ 2}\)
\(0\)\(1\)\(1\)\(3\)\(1\)\(1\)\(\frac{ 1}{ 2}\)
\(0\)\(2\)\(2\)\(4\)\(2\)\(2\)\(\frac{ 3}{ 4}\)

Note:

Sporadic Operator.

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4

New Number: 4.74 |  AESZ: 363  |  Superseeker: -207 621972  |  Hash: 15f3be0c25c6a6ea1d78414f1cb31713  

Degree: 4

\(\theta^4+3^{2} x\left(231\theta^4+318\theta^3+231\theta^2+72\theta+8\right)+2^{3} 3^{5} x^{2}\left(774\theta^4+1854\theta^3+1869\theta^2+768\theta+100\right)+2^{6} 3^{8} x^{3}\left(951\theta^4+2304\theta^3+1740\theta^2+504\theta+50\right)+2^{10} 3^{12} x^{4}(4\theta+1)(2\theta+1)^2(4\theta+3)\)

Maple   LaTex

Coefficients of the holomorphic solution: 1, -72, 22680, -9424800, 4199995800, ...
--> OEIS
Normalized instanton numbers (n0=1): -207, 5544, 621972, -241386048, 59946723846, ... ; Common denominator:...

Discriminant

\((746496z^2+1647z+1)(1+216z)^2\)

Local exponents

\(-\frac{ 1}{ 216}\)\(-\frac{ 61}{ 55296}-\frac{ 5}{ 55296}\sqrt{ 15}I\)\(-\frac{ 61}{ 55296}+\frac{ 5}{ 55296}\sqrt{ 15}I\)\(0\)\(s_1\)\(s_2\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(\frac{ 1}{ 4}\)
\(1\)\(1\)\(1\)\(0\)\(1\)\(1\)\(\frac{ 1}{ 2}\)
\(3\)\(1\)\(1\)\(0\)\(1\)\(1\)\(\frac{ 1}{ 2}\)
\(4\)\(2\)\(2\)\(0\)\(2\)\(2\)\(\frac{ 3}{ 4}\)

Note:

Sporadic Operator.

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5

New Number: 1.6 |  AESZ: 6  |  Superseeker: 160 1956896  |  Hash: 483b4ca5270ed3bfca9243827b62064e  

Degree: 1

\(\theta^4-2^{4} x(4\theta+1)(2\theta+1)^2(4\theta+3)\)

Maple   LaTex

Coefficients of the holomorphic solution: 1, 48, 15120, 7392000, 4414410000, ...
--> OEIS
Normalized instanton numbers (n0=1): 160, 11536, 1956896, 485487816, 148865410272, ... ; Common denominator:...

Discriminant

\(1-1024z\)

Local exponents

\(0\)\(\frac{ 1}{ 1024}\)\(\infty\)
\(0\)\(0\)\(\frac{ 1}{ 4}\)
\(0\)\(1\)\(\frac{ 1}{ 2}\)
\(0\)\(1\)\(\frac{ 1}{ 2}\)
\(0\)\(2\)\(\frac{ 3}{ 4}\)

Note:

A-incarnation of $X(2,4)$ in $P^5$.

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