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You searched for: superseeker=27/5,1619/5

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1

New Number: 12.15 |  AESZ:  |  Superseeker: 27/5 1619/5  |  Hash: f7297f2850190f8613d1cbc3a7363a23  

Degree: 12

\(5^{2} \theta^4-5 x\left(296\theta^4+574\theta^3+457\theta^2+170\theta+25\right)-x^{2}\left(4531\theta^4+24118\theta^3+37791\theta^2+23710\theta+5550\right)+2^{2} x^{3}\left(559\theta^4+9744\theta^3+19448\theta^2+14280\theta+4055\right)+x^{4}\left(1455\theta^4-636\theta^3+151398\theta^2+254100\theta+114136\right)+x^{5}\left(80304\theta^4+79818\theta^3-776517\theta^2-952026\theta-338569\right)-x^{6}\left(18597\theta^4-67050\theta^3-680097\theta^2-608202\theta-164470\right)-2 x^{7}\left(19086\theta^4+454818\theta^3+525507\theta^2-112266\theta-235189\right)-2^{2} x^{8}\left(52779\theta^4-252492\theta^3-39867\theta^2+316368\theta+192050\right)-2^{3} x^{9}\left(27325\theta^4+45630\theta^3-118827\theta^2-223839\theta-101599\right)+2^{2} 17 x^{10}\left(8047\theta^4+9182\theta^3-8905\theta^2-20876\theta-9476\right)+2^{5} 17^{2} x^{11}(\theta+1)(19\theta^3+129\theta^2+246\theta+145)-2^{4} 17^{3} x^{12}(\theta+2)(\theta+1)(2\theta+3)^2\)

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Coefficients of the holomorphic solution: 1, 5, 109, 3329, 122581, ...
--> OEIS
Normalized instanton numbers (n0=1): 27/5, 158/5, 1619/5, 51193/10, 485082/5, ... ; Common denominator:...

Discriminant

\(-(4z+1)(z+1)(68z^2+61z-1)(z-1)^2(34z^3-12z^2+3z-5)^2\)

Local exponents

\(-1\)\(-\frac{ 61}{ 136}-\frac{ 11}{ 136}\sqrt{ 33}\)\(-\frac{ 1}{ 4}\) ≈\(-0.126959-0.475615I\) ≈\(-0.126959+0.475615I\)\(0\)\(-\frac{ 61}{ 136}+\frac{ 11}{ 136}\sqrt{ 33}\) ≈\(0.606859\)\(1\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(0\)\(1\)
\(1\)\(1\)\(1\)\(1\)\(1\)\(0\)\(1\)\(1\)\(\frac{ 1}{ 2}\)\(\frac{ 3}{ 2}\)
\(1\)\(1\)\(1\)\(3\)\(3\)\(0\)\(1\)\(3\)\(\frac{ 1}{ 2}\)\(\frac{ 3}{ 2}\)
\(2\)\(2\)\(2\)\(4\)\(4\)\(0\)\(2\)\(4\)\(1\)\(2\)

Note:

This is operator "12.15" from ...

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2

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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