User:Vladimir Reshetnikov/Sandbox

Let $B\left(z;\ a,b\right)$ denote the incomplete beta function: $$B\left(z;\ a,b\right)=\int_0^zu^{a-1}(1-u)^{b-1}du.$$

Conjecture 1
Let $\alpha$ be the unique algebraic number satisfying $$\alpha^4-540\,\alpha^3+270\,\alpha^2-972\,\alpha+729=0\ \land\ \alpha<1.$$ It can be expressed in radicals as follows $$\alpha=3\left(45+26\,\sqrt3-2\,\sqrt{6\left(168+97\,\sqrt3\right)}\right).$$ Then, conjecturally, $$B\left(\alpha;\ \frac12,\frac13\right)\stackrel?=\frac{\sqrt\pi}2\cdot\frac{\Gamma\left(\frac13\right)}{\Gamma\left(\frac56\right)}.$$

Conjecture 2
Let $\alpha$ be the unique algebraic number satisfying $$\alpha^3-99\,\alpha^2+243\,\alpha-81=0\ \land\ \alpha<1.$$ It can be expressed in radicals as follows $$\alpha=33-\frac{42\left(1+\sqrt{-3}\right)}{\sqrt[3]{\frac12\left(37+\sqrt{-3}\right)}}-3\times2^{2/3}\left(1-\sqrt{-3}\right)\sqrt[3]{37+\sqrt{-3}}.$$ Then, conjecturally, $$B\left(\alpha;\ \frac12,\frac13\right)\stackrel?=\sqrt\pi\cdot\frac{\Gamma\left(\frac43\right)}{\Gamma\left(\frac56\right)}.$$

Conjecture 3
Let $\alpha$ be the unique algebraic number satisfying $$\alpha^9-6129\,\alpha^8-63180\,\alpha^7-1698084\,\alpha^6+5690574\,\alpha^5-21611934\,\alpha^4+28579716\,\alpha^3-10628820\,\alpha^2+4782969\,\alpha-4782969=0\ \land\ \alpha<1.$$ Then, conjecturally, $$B\left(\alpha;\ \frac12,\frac13\right)\stackrel?=2\,\sqrt\pi\cdot\frac{\Gamma\left(\frac43\right)}{\Gamma\left(\frac56\right)}.$$

Conjecture 4
Let $\alpha$ be the unique algebraic number satisfying $$\alpha^8+1224\,\alpha^7-67284\,\alpha^6+328536\,\alpha^5-1115370\,\alpha^4+367416\,\alpha^3+1338444\,\alpha^2-1417176\,\alpha+531441=0\ \land\ 0<\alpha<1.$$ It can be expressed in radicals as follows $$\alpha=\frac1{7810}\left(-1194930+749760\sqrt3+140580\sqrt{158-91\sqrt3}\\-12\sqrt{3905\left(5494335-3163050\sqrt3-650436\sqrt{158-91\sqrt3}+35145\left(158-91\sqrt3\right)+342\left(158-91\sqrt3\right)^{3/2}+403872\sqrt{3\left(158-91\sqrt3\right)}\right)}\right).$$ Then, conjecturally, $$B\left(\alpha;\ \frac12,\frac13\right)\stackrel?=\frac{3\,\sqrt\pi}4\cdot\frac{\Gamma\left(\frac13\right)}{\Gamma\left(\frac56\right)}.$$

Conjecture 5
Let $\alpha$ be the unique algebraic number satisfying $$\alpha^8-2088\,\alpha^7+64908\,\alpha^6+21384\,\alpha^5+1917270\,\alpha^4-5616216\,\alpha^3+7007148\,\alpha^2-4251528\,\alpha+531441=0\ \land\ \alpha<1.$$ It can be expressed in radicals as follows $$\alpha=\frac1{115522}\left(30151242+13516074\sqrt5-190164\sqrt3\left(4745-2122\sqrt5\right)^{3/2}\\+835472448\sqrt{3\left(4745-2122\sqrt5\right)}-433427202\sqrt{15\left(4745-2122\sqrt5\right)}-\sqrt\beta\right),$$ where $$\beta=\left(-30151242-13516074\sqrt5+190164\sqrt3\left(4745-2122\sqrt5\right)^{3/2}-835472448\sqrt{3\left(4745-2122\sqrt5\right)}+433427202\sqrt{15\left(4745-2122\sqrt5\right)}\right)^2\\-231044\left(35869581+15595470\sqrt5-230202\sqrt3\left(4745-2122\sqrt5\right)^{3/2}+1011779154\sqrt{3\left(4745-2122\sqrt5\right)}-524500002\sqrt{15\left(4745-2122\sqrt5\right)}\right).$$ Then, conjecturally, $$B\left(\alpha;\ \frac12,\frac13\right)\stackrel?=\frac{\sqrt\pi}5\cdot\frac{\Gamma\left(\frac13\right)}{\Gamma\left(\frac56\right)}.$$

Conjecture 6
Let $\alpha$ be the unique algebraic number satisfying $$5\,\alpha^4+360\,\alpha^3-1350\,\alpha^2+729=0\ \land\ 0<\alpha<1.$$ It can be expressed in radicals as follows $$\alpha=3\left(-6+3\sqrt5-\sqrt{6\left(13-\frac{29}{\sqrt5}\right)}\right).$$ Then, conjecturally, $$B\left(\alpha;\ \frac12,\frac13\right)\stackrel?=\frac{3\,\sqrt\pi}5\cdot\frac{\Gamma\left(\frac13\right)}{\Gamma\left(\frac56\right)}.$$