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 $j$:

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

1. If $j\ge 2\kappa +3$ (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 $3+\kappa \le j\le 2\kappa +1$ (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 $j^{\prime }=2(\kappa +1)-j$ and compute $Q_{j^{\prime },\kappa ,c}$ using the conjectured formulae given in the next subsections.

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

Define:

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

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

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

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

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

4. If $3\le j\le \kappa +2$ (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 $j\ge 7$, define:

${\zeta }_j=\frac{2^{j+1}(j-1)!}{(j-2)(j-4)}\text{and}{a}_j=(j-1)j(2j-5)\text{and}{b}_j=\frac{(j-6)\cdot a_{j-1}{a}_j}{j-1}$

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

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

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

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

${\tau }_{1,j+2}=\frac{{\tau }_{1,j}{b}_j-3((j-3)(j-1)-1){\zeta }_j}{p_j}\text{and}{\lambda }_{1,j+2}=\frac{{\lambda }_{1,j}{b}_j(j(2j-3)-1)-d_{j-2}{\zeta }_j}{p_j((j-2)(2j-7)-1)}.$

Step 2 (involves both 𝑗 and 𝜅):

Define (for 𝑛 ∈ {0, 1}):

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

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

$U_{n,j,\kappa }=\{\begin{array}{cc}{\tau }_{n,j}+{\lambda }_{n,j}\kappa & \text{if }\kappa <2,\\ {\eta }_{n,j,\kappa }\cdot U_{n,j,\kappa -2}+{\varphi }_{n,j,\kappa }\cdot U_{n,j,\kappa -1}& \text{if }\kappa \ge 2,\end{array}$

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

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

Step 3 (involves 𝑗, 𝜅, 𝑐):

Define:

${\Delta }_{j,\kappa ,c}=\{\begin{array}{cc}{\ell }_{c,j,\kappa }& \text{if }c<2\\ -2c(2c-j)(2c-2\kappa +j-2)(2c-2\kappa -1){\Delta }_{j,\kappa ,c-2}& \text{if }c\ge 2\\ +(8c^2+(2-8\kappa )c+(j-2)(2\kappa -j)){\Delta }_{j,\kappa ,c-1}& \end{array}$

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

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

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

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.

References

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