Draft:The Balkans Continued Fraction

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The Balkans Continued Fraction Conjecture consists in proving a closed formula found using machine investigation. The conjecture was formulated by David Naccache and Ofer Yifrach-Stav in 2023 .

In the following description, $G$ represents Catalan's constant, and $C_{k}$ denotes Catalan numbers.

The closed formula computes the exact value of the following continued fraction, known as the "Balkans Continued Fraction," for odd $j$:

$Q_{j,\kappa ,c}=j(2-j+2\kappa )+K_{n=1}^{\infty }\left({\frac {-2n(c+n)(j+n-1)(1-j+2\kappa +n)}{j(2-j+2\kappa )+(3+4\kappa )n+3n^{2}}}\right)$

1. If $j\geq 2\kappa +3$ (Trivial)

This case, mentioned here for the sake of completeness, is not part of the conjecture as $Q_{j,\kappa ,c}$ is computed by straightforward finite summation.

$Q_{j,\kappa ,c}=j(2-j+2\kappa )+K_{n=1}^{j-2\kappa -1}{\frac {-2n(c+n)(j+n-1)(1-j+2\kappa +n)}{j(2-j+2\kappa )+(3+4\kappa )n+3n^{2}}}$

2. If $3+\kappa \leq j\leq 2\kappa +1$ (Trivial if conjectures 1 and 2 hold true)

This case uses the symmetry relation:

$Q_{j,\kappa ,c}=Q_{2(\kappa +1)-j,\kappa ,c}$

Replace $j$ by $j'=2(\kappa +1)-j$ and compute $Q_{j',\kappa ,c}$ using the conjectured formulae given in the next subsections.

3. If $j=1$ (Conjecture 1)

Define:

$V_{\kappa ,c,x,y}={\begin{cases}x+yc&{\text{if }}c<2\\-2c(2c-1)(2(c-\kappa )-1)^{2}\cdot V_{\kappa ,c-2,x,y}\\+(8c^{2}+(2-8\kappa )c-2\kappa +1)\cdot V_{\kappa ,c-1,x,y},&{\text{if }}c\geq 2\end{cases}}$

$\Gamma _{\kappa ,c,x,y}=(2c-1)!!^{2}G+V_{\kappa ,c-1,x,y}\prod _{i=0}^{\kappa -1}(2(c-i)-1)$

$\delta ={\frac {4^{\kappa -1}}{(2\kappa -1)C_{\kappa -1}}}\quad and\quad \rho ={\frac {\delta \cdot (-1)^{\kappa }(1-2\kappa )}{(2\kappa )!(2\kappa -3)!!}}$

$\alpha =\rho \cdot V_{1,\kappa -1,1,-2}\quad and\quad \beta =-\rho \cdot (2\kappa -3)^{2}\cdot V_{2,\kappa -1,1,12}-\alpha$

And output $Q_{1,\kappa ,c}={\frac {\delta \cdot (2c)!}{\Gamma _{\kappa ,c,\alpha ,\beta }}}$

4. If $3\leq j\leq \kappa +2$ (Conjecture 2)

Proceed in three steps:

Step 1 (involves only $j$)

For $j=3$ or $j=5$, define:

$\{\tau _{0,3},\lambda _{0,3},\tau _{1,3},\lambda _{1,3}\}=\{-1,4,-{\frac {1}{3}},-{\frac {14}{3}}\}$ and $\{\tau _{0,5},\lambda _{0,5},\tau _{1,5},\lambda _{1,5}\}=\{19,234,-17,-8\}$

If $j\geq 7$, define:

$\zeta _{j}={\frac {2^{j+1}(j-1)!}{(j-2)(j-4)}}\quad {\text{and}}\quad a_{j}=(j-1)j(2j-5)\quad {\text{and}}\quad b_{j}={\frac {(j-6)\cdot a_{j-1}a_{j}}{j-1}}$

$p_{j}={\frac {(j-6)(j-4)(j^{2}-1)}{4}},$

$d_{j}=6j^{6}-15j^{5}-68j^{4}+74j^{3}+89j^{2}-44j-18\quad {\text{and}}\quad e_{j}=(3j+1)(j^{2}-7)+3.$

and iterate using the following formulae to compute $\{\tau _{0,j},\lambda _{0,j},\tau _{1,j},\lambda _{1,j}\}:$

$\tau _{0,j+2}={\frac {4\tau _{0,j}a_{j}(2j-3)-(3j-2)\zeta _{j}}{j^{2}-1}}\quad {\text{and}}\quad \lambda _{0,j+2}={\frac {4\lambda _{0,j}a_{j}(2j+1)-e_{j-2}\zeta _{j}}{(j-3)(j+1)}},$

$\tau _{1,j+2}={\frac {\tau _{1,j}b_{j}-3\left((j-3)(j-1)-1\right)\zeta _{j}}{p_{j}}}\quad {\text{and}}\quad \lambda _{1,j+2}={\frac {\lambda _{1,j}b_{j}\left(j(2j-3)-1\right)-d_{j-2}\zeta _{j}}{p_{j}\left((j-2)(2j-7)-1\right)}}.$

Step 2 (involves both 𝑗 and 𝜅):

Define (for 𝑛 ∈ {0, 1}):

$\eta _{n,j,\kappa }=(2\kappa +2j-9-2n)(2\kappa +j-8-2n)(-2\kappa +5-j)(2\kappa +j-6)$

$\phi _{n,j,\kappa }=8\kappa ^{2}+\kappa (10j-48-8n)+3j^{2}-(28+4n)j+68+18n$

$U_{n,j,\kappa }={\begin{cases}\tau _{n,j}+\lambda _{n,j}\kappa &{\text{if }}\kappa <2,\\\eta _{n,j,\kappa }\cdot U_{n,j,\kappa -2}+\phi _{n,j,\kappa }\cdot U_{n,j,\kappa -1}&{\text{if }}\kappa \geq 2,\end{cases}}$

$s_{n,j,\kappa }={\kappa !\cdot (-2)^{3\kappa -2}\cdot U_{n,j,\kappa -j+2}}\cdot \prod _{i=0}^{\frac {j-3}{2}}(\kappa -i)(2\kappa -2i-1)^{2}$

$\ell _{n,j,\kappa }={\frac {(j-2\kappa )(2\kappa -1)((2\kappa -j-2)(3-2\kappa ))^{n}\cdot s_{n,j,\kappa }}{(2\kappa )!^{2}}}$

Step 3 (involves 𝑗, 𝜅, 𝑐):

Define:

$\Delta _{j,\kappa ,c}={\begin{cases}\ell _{c,j,\kappa }&{\text{if }}c<2\\-2c(2c-j)(2c-2\kappa +j-2)(2c-2\kappa -1)\Delta _{j,\kappa ,c-2}&{\text{if }}c\geq 2\\+\left(8c^{2}+(2-8\kappa )c+(j-2)(2\kappa -j)\right)\Delta _{j,\kappa ,c-1}&\end{cases}}$

$f_{j,\kappa ,c}=C_{\frac {j-3}{2}}C_{\kappa -1}(j-2)(2\kappa -1)(2c-1)!!^{2}\prod _{i=1}^{\frac {j-1}{2}}(2c-2\kappa +2i-1)(\kappa -i+1)$

$g_{j,\kappa ,c}=(2c)!2^{\frac {j+4\kappa -7}{2}}\prod _{i=1}^{\frac {j-1}{2}}(2c-2i+1)(2\kappa -2i+1)$

$h_{j,\kappa ,c}=\Delta _{j,\kappa ,c-1}\cdot \prod _{i=0}^{\frac {j-3}{2}}(2c-2i-1)\prod _{i=0}^{\kappa -1}(2c-2i-1)$

Output:

$Q_{j,\kappa ,c}={\frac {g_{j,\kappa ,c}}{h_{j,\kappa ,c}+f_{j,\kappa ,c}\cdot G}}$

Double factorial-free and $C_{x}$-free expressions

Note that:

$\prod _{i=1}^{\frac {j-1}{2}}(2c-2\kappa +2i-1)\cdot (\kappa -i+1)={\frac {(2c-2\kappa +j-2)!!}{(2c-2\kappa -1)!!}}\cdot {\frac {\kappa !}{(\kappa -{\frac {j-1}{2}})!}}$

$\prod _{i=1}^{\frac {j-1}{2}}(2c-2i+1)\cdot (2\kappa -2i+1)={\frac {(2c-1)!!}{(2c-j)!!}}\cdot {\frac {(2\kappa -1)!!}{(2\kappa -j)!!}}$

$\prod _{i=0}^{\frac {j-3}{2}}(2c-2i-1)\prod _{i=0}^{\kappa -1}(2c-2i-1)={\frac {(2c-1)!!}{(2c-j)!!}}\cdot {\frac {(2c-1)!!}{(2c-2\kappa -1)!!}}$

$\prod _{i=0}^{\frac {j-3}{2}}(\kappa -i)(2\kappa -2i-1)^{2}={\frac {k!}{(\kappa -{\frac {j-1}{2}})!}}\cdot {\frac {(2\kappa -1)!!^{2}}{(2\kappa -j)!!^{2}}}$

And the well-known identities:

$(2x-1)!!={\frac {(2x)!}{2^{x}x!}}={\frac {(2x-1)!}{2^{x-1}(x-1)!}}$ and $C_{x}={\frac {(2x)!}{(x+1)!\,x!}}$

yield expressions that avoid double factorials. The first identity is always usable because $j$ is odd.