Summary

You searched for: sol=1944

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1

New Number: 4.69 |  AESZ: 350  |  Superseeker: 49 173876/9  |  Hash: e6de16eb3758d2ed5687f4b2a2abf36b  

Degree: 4

\(\theta^4-x\left(24+184\theta+545\theta^2+722\theta^3+289\theta^4\right)+2^{3} 3 x^{2}\left(214\theta^4+2734\theta^3+4861\theta^2+2640\theta+468\right)+2^{6} 3^{2} x^{3}\left(1391\theta^4+5184\theta^3+4252\theta^2+1296\theta+126\right)+2^{10} 3^{6} x^{4}\left((2\theta+1)^4\right)\)

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Coefficients of the holomorphic solution: 1, 24, 1944, 232800, 34133400, ...
--> OEIS
Normalized instanton numbers (n0=1): 49, 136, 173876/9, 781152, 57087750, ... ; Common denominator:...

Discriminant

\((256z-1)(81z-1)(1+24z)^2\)

Local exponents

\(-\frac{ 1}{ 24}\)\(0\)\(\frac{ 1}{ 256}\)\(\frac{ 1}{ 81}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(\frac{ 1}{ 2}\)
\(1\)\(0\)\(1\)\(1\)\(\frac{ 1}{ 2}\)
\(3\)\(0\)\(1\)\(1\)\(\frac{ 1}{ 2}\)
\(4\)\(0\)\(2\)\(2\)\(\frac{ 1}{ 2}\)

Note:

Sporadic Operator. There is a second MUM-point hiding at infinity, corresponding to Operator AESZ 351/4.70

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2

New Number: 5.100 |  AESZ: 347  |  Superseeker: 15 27140/3  |  Hash: f00de20026c099e75b447c475ab287e4  

Degree: 5

\(\theta^4-3 x\left(213\theta^4+186\theta^3+149\theta^2+56\theta+8\right)+2^{3} 3^{3} x^{2}\left(702\theta^4+1078\theta^3+949\theta^2+392\theta+60\right)-2^{6} 3^{3} x^{3}\left(9277\theta^4+18432\theta^3+16008\theta^2+6000\theta+840\right)+2^{13} 3^{4} 5 x^{4}(2\theta+1)^2(51\theta^2+69\theta+32)-2^{14} 3^{6} 5^{2} x^{5}(2\theta+1)^2(2\theta+3)^2\)

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Coefficients of the holomorphic solution: 1, 24, 1944, 218400, 28488600, ...
--> OEIS
Normalized instanton numbers (n0=1): 15, 1329/4, 27140/3, 220680, 5952570, ... ; Common denominator:...

Discriminant

\(-(192z-1)(1728z^2-207z+1)(-1+120z)^2\)

Local exponents

\(0\)\(\frac{ 23}{ 384}-\frac{ 11}{ 1152}\sqrt{ 33}\)\(\frac{ 1}{ 192}\)\(\frac{ 1}{ 120}\)\(\frac{ 23}{ 384}+\frac{ 11}{ 1152}\sqrt{ 33}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(\frac{ 1}{ 2}\)
\(0\)\(1\)\(1\)\(1\)\(1\)\(\frac{ 1}{ 2}\)
\(0\)\(1\)\(1\)\(3\)\(1\)\(\frac{ 3}{ 2}\)
\(0\)\(2\)\(2\)\(4\)\(2\)\(\frac{ 3}{ 2}\)

Note:

This is operator "5.100" from ...

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3

New Number: 14.2 |  AESZ:  |  Superseeker: 27/5 1619/5  |  Hash: c0f6d85270164c8c5a63d1bb2deaba83  

Degree: 14

\(5^{2} \theta^4+3^{2} 5 x\theta(6\theta^3-36\theta^2-23\theta-5)-x^{2}\left(43856\theta^4+189068\theta^3+226691\theta^2+135510\theta+33600\right)-3^{2} x^{3}\left(224236\theta^4+916896\theta^3+1403247\theta^2+1048995\theta+313920\right)-x^{4}\left(44621090\theta^4+199900036\theta^3+357072757\theta^2+304636250\theta+101358144\right)-3^{2} x^{5}\left(69593744\theta^4+347076728\theta^3+696076003\theta^2+653370139\theta+234075456\right)-3^{2} x^{6}\left(681084088\theta^4+3766244020\theta^3+8299124637\theta^2+8400442322\theta+3184811840\right)-3^{3} x^{7}\left(1616263276\theta^4+9835107968\theta^3+23484467027\theta^2+25311872719\theta+10046134656\right)-3^{3} x^{8}\left(8527956293\theta^4+56671723156\theta^3+145225420081\theta^2+165230257706\theta+68152357440\right)-2 3^{4} x^{9}\left(5575274615\theta^4+40185448970\theta^3+109721715457\theta^2+130944512374\theta+55834822464\right)-2^{3} 3^{3} x^{10}\left(12062719219\theta^4+93737716664\theta^3+271167874625\theta^2+337796659588\theta+148305175248\right)-2^{5} 3^{5} x^{11}(\theta+1)(691573543\theta^3+5071601663\theta^2+12510902832\theta+10260936720)-2^{7} 3^{6} x^{12}(\theta+1)(\theta+2)(80620421\theta^2+475174733\theta+711172676)-2^{14} 3^{6} 5 x^{13}(\theta+3)(\theta+2)(\theta+1)(107069\theta+369433)-2^{19} 3^{8} 5^{2} 29 x^{14}(\theta+1)(\theta+2)(\theta+3)(\theta+4)\)

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Coefficients of the holomorphic solution: 1, 0, 84, 1944, 70476, ...
--> OEIS
Normalized instanton numbers (n0=1): 27/5, 158/5, 1619/5, 51193/10, 485082/5, ... ; Common denominator:...

Discriminant

\(-(9z+1)(6z+1)(348z^2+51z-1)(5z+1)^2(4z+1)^2(576z^3+357z^2+72z+5)^2\)

Local exponents

≈\(-0.298314\)\(-\frac{ 1}{ 4}\)\(-\frac{ 1}{ 5}\)\(-\frac{ 1}{ 6}\)\(-\frac{ 17}{ 232}-\frac{ 11}{ 696}\sqrt{ 33}\) ≈\(-0.160739-0.057112I\) ≈\(-0.160739+0.057112I\)\(-\frac{ 1}{ 9}\)\(0\)\(-\frac{ 17}{ 232}+\frac{ 11}{ 696}\sqrt{ 33}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(1\)
\(1\)\(\frac{ 1}{ 2}\)\(\frac{ 1}{ 2}\)\(1\)\(1\)\(1\)\(1\)\(1\)\(0\)\(1\)\(2\)
\(3\)\(\frac{ 1}{ 2}\)\(\frac{ 1}{ 2}\)\(1\)\(1\)\(3\)\(3\)\(1\)\(0\)\(1\)\(3\)
\(4\)\(1\)\(1\)\(2\)\(2\)\(4\)\(4\)\(2\)\(0\)\(2\)\(4\)

Note:

This is operator "14.2" from ...

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