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You searched for: superseeker=44/3,220588/81

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1

New Number: 14.9 |  AESZ:  |  Superseeker: 44/3 220588/81  |  Hash: 32527b63e2e6c7ca027dfb5cb9afac16  

Degree: 14

\(3^{2} \theta^4+2^{2} 3 x\left(8\theta^4-128\theta^3-105\theta^2-41\theta-7\right)-2^{4} x^{2}\left(2720\theta^4-64\theta^3-3536\theta^2-680\theta+429\right)+2^{10} x^{3}\left(336\theta^4+3984\theta^3-826\theta^2+468\theta+1051\right)+2^{12} x^{4}\left(16640\theta^4-7232\theta^3+43840\theta^2+45800\theta+15969\right)-2^{18} x^{5}\left(5720\theta^4+6944\theta^3+19273\theta^2+22267\theta+9043\right)-2^{21} x^{6}\left(10216\theta^4+38016\theta^3+103024\theta^2+135096\theta+80559\right)+2^{28} x^{7}\left(2848\theta^4+11072\theta^3+24505\theta^2+27600\theta+12752\right)+2^{29} 3 x^{8}\left(888\theta^4+13312\theta^3+65952\theta^2+133944\theta+103073\right)-2^{34} x^{9}\left(8760\theta^4+63456\theta^3+203405\theta^2+310785\theta+183393\right)+2^{36} x^{10}\left(1024\theta^4-20032\theta^3-232944\theta^2-750136\theta-801269\right)+2^{42} x^{11}\left(3248\theta^4+34160\theta^3+146646\theta^2+293996\theta+228285\right)+2^{44} x^{12}\left(416\theta^4+11328\theta^3+98816\theta^2+340680\theta+408025\right)-2^{50} 5 x^{13}\left(104\theta^4+1408\theta^3+7395\theta^2+17845\theta+16643\right)-2^{56} 5^{2} x^{14}\left((\theta+4)^4\right)\)

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Coefficients of the holomorphic solution: 1, 28/3, 260, 116240/27, 7153796/81, ...
--> OEIS
Normalized instanton numbers (n0=1): 44/3, -1421/9, 220588/81, -14752264/243, 1138508000/729, ... ; Common denominator:...

Discriminant

\(-(1+16z)(16z+3)^2(1280z^2-32z-1)^2(256z^2+16z-1)^2(16z-1)^3\)

Local exponents

\(-\frac{ 3}{ 16}\)\(-\frac{ 1}{ 32}-\frac{ 1}{ 32}\sqrt{ 5}\)\(-\frac{ 1}{ 16}\)\(\frac{ 1}{ 80}-\frac{ 1}{ 80}\sqrt{ 6}\)\(0\)\(-\frac{ 1}{ 32}+\frac{ 1}{ 32}\sqrt{ 5}\)\(\frac{ 1}{ 80}+\frac{ 1}{ 80}\sqrt{ 6}\)\(\frac{ 1}{ 16}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(4\)
\(1\)\(\frac{ 1}{ 2}\)\(1\)\(1\)\(0\)\(\frac{ 1}{ 2}\)\(1\)\(0\)\(4\)
\(-2\)\(\frac{ 1}{ 2}\)\(1\)\(3\)\(0\)\(\frac{ 1}{ 2}\)\(3\)\(0\)\(4\)
\(3\)\(1\)\(2\)\(4\)\(0\)\(1\)\(4\)\(0\)\(4\)

Note:

This is operator "14.9" from ...

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2

New Number: 15.3 |  AESZ:  |  Superseeker: 44/3 220588/81  |  Hash: ae51313cd958206bb1b7a3c8ae23e509  

Degree: 15

\(3^{3} \theta^4+2^{2} 3^{2} x\left(12\theta^4-160\theta^3-153\theta^2-73\theta-15\right)-2^{4} 3 x^{2}\left(2688\theta^4+704\theta^3-6380\theta^2-6164\theta-2343\right)+2^{8} x^{3}\left(1312\theta^4+69632\theta^3+26456\theta^2+3928\theta-4305\right)+2^{12} x^{4}\left(51264\theta^4-16512\theta^3-16360\theta^2-16088\theta-1785\right)-2^{16} x^{5}\left(52000\theta^4+223680\theta^3+316652\theta^2+308700\theta+133179\right)-2^{21} x^{6}\left(42088\theta^4+36416\theta^3+31682\theta^2-15530\theta-24313\right)+2^{25} x^{7}\left(58136\theta^4+309440\theta^3+666728\theta^2+761160\theta+351769\right)+2^{29} x^{8}\left(30776\theta^4+26112\theta^3-81496\theta^2-231912\theta-165231\right)-2^{33} 3 x^{9}\left(16632\theta^4+120704\theta^3+332890\theta^2+441546\theta+227145\right)-2^{36} x^{10}\left(31968\theta^4+33600\theta^3-297916\theta^2-852260\theta-648637\right)+2^{40} x^{11}\left(40000\theta^4+381696\theta^3+1258584\theta^2+1813272\theta+964287\right)+2^{44} x^{12}\left(14240\theta^4+66688\theta^3+44952\theta^2-163928\theta-198345\right)-2^{48} x^{13}\left(5824\theta^4+76480\theta^3+307828\theta^2+490020\theta+272659\right)-2^{54} 5 x^{14}\left(164\theta^4+1536\theta^3+5043\theta^2+7113\theta+3693\right)-2^{60} 5^{2} x^{15}\left((\theta+2)^4\right)\)

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Coefficients of the holomorphic solution: 1, 20, 388, 7344, 141636, ...
--> OEIS
Normalized instanton numbers (n0=1): 44/3, -1421/9, 220588/81, -14752264/243, 1138508000/729, ... ; Common denominator:...

Discriminant

\(-(1+16z)(1280z^2-32z-1)^2(256z^2+16z-1)^2(16z+3)^3(16z-1)^3\)

Local exponents

\(-\frac{ 3}{ 16}\)\(-\frac{ 1}{ 32}-\frac{ 1}{ 32}\sqrt{ 5}\)\(-\frac{ 1}{ 16}\)\(\frac{ 1}{ 80}-\frac{ 1}{ 80}\sqrt{ 6}\)\(0\)\(-\frac{ 1}{ 32}+\frac{ 1}{ 32}\sqrt{ 5}\)\(\frac{ 1}{ 80}+\frac{ 1}{ 80}\sqrt{ 6}\)\(\frac{ 1}{ 16}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(2\)
\(2\)\(\frac{ 1}{ 2}\)\(1\)\(1\)\(0\)\(\frac{ 1}{ 2}\)\(1\)\(0\)\(2\)
\(3\)\(\frac{ 1}{ 2}\)\(1\)\(3\)\(0\)\(\frac{ 1}{ 2}\)\(3\)\(0\)\(2\)
\(5\)\(1\)\(2\)\(4\)\(0\)\(1\)\(4\)\(0\)\(2\)

Note:

This is operator "15.3" from ...

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