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I-a. Some functions

As these will be used in the continued fraction evaluations below, recall the Riemann zeta function $\zeta(s),$ and Dirichlet beta function $\beta(s),$ \begin{align} \zeta(s) &= \sum_{n=1}^\infty\frac{1}{n^s}\\ \beta(s) &= \sum_{n=1}^\infty\frac{(-1)^{n-1}}{(2n-1)^s} \end{align}

I-b. Zagier's 6 sporadic sequences

Inspired by Apery's result in proving the irrationality of $\zeta(3)$ using certain integer sequences, Zagier (via a computer) searched for sequences with recurrence relation and deg-$2$ coefficients in $n$ of form,

I-c. Continued fractions

where $F_k(n)$ are polynomials all of degree $m$. Define two polynomial functions using the rules,

\begin{align} p(n) &= F_2(n)\\ q(n) &= F_1(n)\, F_3(n+1) \end{align}

which implies $q(n)$ has degree twice that of $p(n)$. Define the continued fraction,

$$C=\cfrac{1}{p(0) + \cfrac{q(0)}{p(1) + \cfrac{q(1)}{p(2)+ \cfrac{q(2)}{p(3)+\ddots } }}}$$

then it seems $C$ may have a nice closed-form based on the properties of the recurrence relation. Examples below.

II. Degree 2

\begin{align} p(n) &= \color{blue}{an^2+an+b}\\ q(n) &= \color{blue}{(n+1)^2}\times \color{blue}{c(n+1)^2} = c(n+1)^4 \end{align}

$$C_2(a,b,c)=\cfrac{1}{p(0) + \cfrac{q(0)}{p(1) + \cfrac{q(1)}{p(2)+ \cfrac{q(2)}{p(3)+\ddots } }}}$$

Note: The first evaluation is valid since it was found by Apery, while the second (in $\color{green}{\text{green}}$) is courtesy of H. Cohen's answer (though it has slow convergence which is why I missed it).

III. Degree 3

\begin{align} r(n) &= -(2n+1)(an^2+an+a-2b)\\ s(n) &= (n+1)^3\times(-a^2-4c)(n+1)^3 = -(a^2+4c)(n+1)^6 \end{align}

$$C_3(a,b,c)=\cfrac{1}{r(0) + \cfrac{s(0)}{r(1) + \cfrac{s(1)}{r(2)+ \cfrac{s(2)}{r(3)+\ddots } }}}$$

IV. Degree 4

where $P_i$ are polynomials of deg-$4$. Why?

V. Degree 5

But Zudilin found a 3-term recurrence,

VI. Questions

I. Some functions

As these will be used in the continued fraction evaluations below, recall the Riemann zeta function $\zeta(s),$ and Dirichlet beta function $\beta(s),$

$$\beta(s) = \sum_{n=1}^\infty\frac{(-1)^{n-1}}{(2n-1)^s}$$

II. Zagier's 6 sporadic sequences

Inspired by Apery's result in proving the irrationality of $\zeta(3)$ using certain integer sequences, Zagier (via a computer) searched for sequences with recurrence relation and deg-$2$ coefficients of form,

III. Continued fractions

where $F_i(n)$ are polynomials of degree $k$. Define two polynomial functions using the rules,

\begin{align} p(n) &= F_1(n-1)\, F_3(n)\\ q(n) &= F_2(n) \end{align}

which implies $p(n)$ has degree twice that of $q(n)$. Define the continued fraction,

$$C =\cfrac{1}{q(0) + \cfrac{p(1)}{q(1) + \cfrac{p(2)}{q(2)+ \cfrac{p(3)}{q(3)+\ddots } }}}$$

More compactly,

$$C(m) = \frac1{q(0) + \large{\underset{n=1}{\overset{m}{\mathrm K}} ~ \frac{p(n)}{q(n)}}}$$

or in Mathematica notation,

$$C(m) = \frac1{q(0) + \text{ContinuedFractionK}[p(n),\;q(n),\, \text{{n, 1, m}}]}$$

It seems $C$ may have a nice closed-form based on the properties of the recurrence relation. Examples below.

IV. Degree 2

\begin{align} p(n) &= \color{blue}{n^2}\times \color{blue}{cn^2} = cn^4\\ q(n) &= \color{blue}{an^2+an+b} \end{align}

$$C_2(a,b,c) = \frac1{q(0) + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{p(n)}{q(n)}}}$$

Note: The first evaluation is valid since it was found by Apery, while the second (in $\color{green}{\text{green}}$) is courtesy of H. Cohen's answer. (Update: May 22, 2023) It turns out $C_2(-9,-3,-27)$ has six limits, one of which is divergent. See this MO post.

V. Degree 3

\begin{align} r(n) &= n^3\times(-a^2-4c)n^3 = -(a^2+4c)n^6\\ s(n) &= -(2n+1)(an^2+an+a-2b) \end{align}

$$C_3(a,b,c) = \frac1{s(0) + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{r(n)}{s(n)}}}$$

VI. Degree 4 & 5

where $P_i$ are polynomials of deg-$4$. Why? But Zudilin found,

VII. Questions

\begin{align} K &=\operatorname{Cl}_2\left(\tfrac12\pi\right) = \beta(2) = \sum_{n=0}^\infty\frac{1}{(4n+1)^2}-\sum_{n=0}^\infty \frac{1}{(4n+3)^2} \\ \kappa &= \operatorname{Cl}_2\left(\tfrac13\pi\right) \,=\, \frac{3\sqrt{3}}{4} \left(\sum_{n=0}^\infty\frac{1}{(3n+1)^2}-\sum_{n=0}^\infty \frac{1}{(3n+2)^2} \right) \end{align}

$$(n+1)^2\,u_{n+1} = (\color{blue}{an^2+an+b})u_k+\color{blue}{cn^2}\,u_{n-1}$$

II. Degree 2

Recall Zagier's recurrence above and, using the coefficients on the RHS, define the polynomial functions,

\begin{align} p(n) &= \color{blue}{an^2+an+b}\\ q(n) &= \color{blue}{cn^2}\times n^2 = c\,n^4 \end{align}

where we affix the monomial $n^2$ so that $q(n)$ will have degree twice that of $p(n).$ Then define the continued fraction,

$$C_2(a,b,c)=\cfrac{1}{p(0) + \cfrac{q(1)}{p(1) + \cfrac{q(2)}{p(2)+ \cfrac{q(3)}{p(3)+\ddots } }}}$$

where we start with $q(1)$ to avoid $q(0)=0$.

Q: Is it true that,

$$(n+1)^3\,v_{n+1} = -(2n+1)(an^2+an+a-2b)v_n -(a^2+4c)n^3\,v_{n-1}$$

where Zagier's $(a,b,c)$ also apply. Again, using the coefficients on the RHS, define the polynomial functions,

\begin{align} r(n) &= -(2n+1)(an^2+an+a-2b)\\ s(n) &= -(a^2+4c)n^3\times n^3 = -(a^2+4c)\,n^6 \end{align}

where we affix the monomial $n^3$ so that $s(n)$ will have degree twice that of $r(n).$ Then define the continued fraction,

$$C_3(a,b,c)=\cfrac{1}{r(0) + \cfrac{s(1)}{r(1) + \cfrac{s(2)}{r(2)+ \cfrac{s(3)}{r(3)+\ddots } }}}$$

where again we start with $s(1)$ to avoid $s(0)=0$.

and $d = a^2+4c =125, 1, 64, 81, 16, -27,$ respectively (and all powers of the smallest primes $2,3,5$).

where $Q_i$ are polynomials of deg-$5$ and used it in an analogous continued fraction for $\zeta(4)$,

$$\frac{\zeta(4)}{13}=\cfrac{1}{u(0) + \cfrac{v(1)}{u(1) + \cfrac{v(2)}{u(2)+ \cfrac{v(3)}{u(3)+\ddots } }}}$$

with $u(n), v(n)$ as polynomial functions where, as usual, the latter has degree twice that of the former. (To be discussed in the next post.)

\begin{align} \operatorname{Cl}_2\left(\tfrac12\pi\right) &= K = \beta(2) \\ \operatorname{Cl}_2\left(\tfrac13\pi\right) &= \kappa \end{align}

$$(n+1)^2\,u_{n+1} = (an^2+an+b)u_k+ cn^2\,u_{n-1}$$

I-c. Continued fractions

Given a 3-term recurrence relation of form,

$$F_1(n)\,u_{n+1} = F_2(n)\,u_n + F_3(n)\,u_{n-1}$$

where $F_k(n)$ are polynomials all of degree $m$. Define two polynomial functions using the rules,

\begin{align} p(n) &= F_2(n)\\ q(n) &= F_1(n)\, F_3(n+1) \end{align}

which implies $q(n)$ has degree twice that of $p(n)$. Define the continued fraction,

$$C=\cfrac{1}{p(0) + \cfrac{q(0)}{p(1) + \cfrac{q(1)}{p(2)+ \cfrac{q(2)}{p(3)+\ddots } }}}$$

then it seems $C$ may have a nice closed-form based on the properties of the recurrence relation. Examples below.

II. Degree 2

Recall Zagier's recurrence,

$$\color{blue}{(n+1)^2}\,u_{n+1} = (\color{blue}{an^2+an+b})u_k+\color{blue}{cn^2}\,u_{n-1}$$

Define $p(n)$ and $q(n)$ according to the rules in the previous section,

\begin{align} p(n) &= \color{blue}{an^2+an+b}\\ q(n) &= \color{blue}{(n+1)^2}\times \color{blue}{c(n+1)^2} = c(n+1)^4 \end{align}

Then define the cfrac,

$$C_2(a,b,c)=\cfrac{1}{p(0) + \cfrac{q(0)}{p(1) + \cfrac{q(1)}{p(2)+ \cfrac{q(2)}{p(3)+\ddots } }}}$$

Q: Is it true that,

$$(n+1)^3\,v_{n+1} = -(2n+1)(an^2+an+a-2b)v_n +(-a^2-4c)n^3\,v_{n-1}$$

and Zagier's $(a,b,c).$ Using the same rules, let,

\begin{align} r(n) &= -(2n+1)(an^2+an+a-2b)\\ s(n) &= (n+1)^3\times(-a^2-4c)(n+1)^3 = -(a^2+4c)(n+1)^6 \end{align}

Define the cfrac,

$$C_3(a,b,c)=\cfrac{1}{r(0) + \cfrac{s(0)}{r(1) + \cfrac{s(1)}{r(2)+ \cfrac{s(2)}{r(3)+\ddots } }}}$$

where $d = a^2+4c =125, 1, 64, 81, 16, -27,$ respectively (and all powers of the smallest primes $2,3,5$).

where $Q_i$ are polynomials of deg-$5$ and used it in an analogous continued fraction for $\zeta(4).$ (To be discussed in the next post.)

As we will use these in the continued fraction evaluations below, recall the Riemann zeta function $\zeta(s),$ and Dirichlet beta function $\beta(s),$ \begin{align} \zeta(s) &= \sum_{n=1}^\infty\frac{1}{n^s}\\ \beta(s) &= \sum_{n=1}^\infty\frac{(-1)^{n-1}}{(2n-1)^s} \end{align}

$$(n+1)^2u_{n+1} = (\color{blue}{an^2+an+b})u_k+\color{blue}c\,n^2u_{n-1}$$

Define the polynomial function,

$$s_n = \color{blue}{an^2+an+b}$$

and the continued fraction,

$$C_2(a,b,c)=\cfrac{1}{s_0 + \cfrac{1^4\, \color{blue}c}{s_1 + \cfrac{2^4\, \color{blue}c}{s_2+ \cfrac{3^4\,\color{blue}c}{s_3+\ddots } }}}$$

where $K$ is Catalan's constant and $\kappa$ is Gieseking's constant, both of which not yet proven to be irrational. The first evaluation is valid since it was found by Apery, while the second (in green) is courtesy of H. Cohen's answer below (though it has slow convergence which is why I missed it).

$$(n+1)^3 v_{n+1} = \color{blue}{-(2n+1)(an^2+an+a-2b)}v_n \color{blue}{- (a^2+4c)}n^3v_{n-1}$$

where Zagier's $(a,b,c)$ also apply. Define the polynomial function,

$$t_n = \color{blue}{-(2n+1)(an^2+an+a-2b)}$$

and the continued fraction with constant $\color{blue}{d = -(a^2+4c)}$,

$$C_3(a,b,c)=\cfrac{1}{t_0 + \cfrac{1^6\, \color{blue}d}{t_1 + \cfrac{2^6\, \color{blue}d}{t_2+ \cfrac{3^6\,\color{blue}d}{t_3+\ddots } }}}$$

where $-d=125, 1, 64, 81, 16, -27,$ respectively (and all powers of the smallest primes $2,3,5$).  Note: The second closed-form is valid since it was also found by Apery which he used (together with other methods) to prove the irrationality of $\zeta(3)$.

$$C_4(a_1, a_2,\dots a_n) =\cfrac{1}{p_0 + \cfrac{1^8\, q_1}{p_1 + \cfrac{2^8\, q_2}{p_2+ \cfrac{3^8\,q_3}{p_3+\ddots } }}}$$

and where $p_i, q_i$ are polynomial functions. (To be discussed in the next post.)

As these will be used in the continued fraction evaluations below, recall the Riemann zeta function $\zeta(s),$ and Dirichlet beta function $\beta(s),$ \begin{align} \zeta(s) &= \sum_{n=1}^\infty\frac{1}{n^s}\\ \beta(s) &= \sum_{n=1}^\infty\frac{(-1)^{n-1}}{(2n-1)^s} \end{align}

$$(n+1)^2\,u_{n+1} = (\color{blue}{an^2+an+b})u_k+\color{blue}{cn^2}\,u_{n-1}$$

Recall Zagier's recurrence above and, using the coefficients on the RHS, define the polynomial functions,

\begin{align} p(n) &= \color{blue}{an^2+an+b}\\ q(n) &= \color{blue}{cn^2}\times n^2 = c\,n^4 \end{align}

where we affix the monomial $n^2$ so that $q(n)$ will have degree twice that of $p(n).$ Then define the continued fraction,

$$C_2(a,b,c)=\cfrac{1}{p(0) + \cfrac{q(1)}{p(1) + \cfrac{q(2)}{p(2)+ \cfrac{q(3)}{p(3)+\ddots } }}}$$

where we start with $q(1)$ to avoid $q(0)=0$.

where $K$ is Catalan's constant and $\kappa$ is Gieseking's constant, both of which not yet proven to be irrational.

Note: The first evaluation is valid since it was found by Apery, while the second (in $\color{green}{\text{green}}$) is courtesy of H. Cohen's answer (though it has slow convergence which is why I missed it).

$$(n+1)^3\,v_{n+1} = -(2n+1)(an^2+an+a-2b)v_n -(a^2+4c)n^3\,v_{n-1}$$

where Zagier's $(a,b,c)$ also apply. Again, using the coefficients on the RHS, define the polynomial functions,

\begin{align} r(n) &= -(2n+1)(an^2+an+a-2b)\\ s(n) &= -(a^2+4c)n^3\times n^3 = -(a^2+4c)\,n^6 \end{align}

where we affix the monomial $n^3$ so that $s(n)$ will have degree twice that of $r(n).$ Then define the continued fraction,

$$C_3(a,b,c)=\cfrac{1}{r(0) + \cfrac{s(1)}{r(1) + \cfrac{s(2)}{r(2)+ \cfrac{s(3)}{r(3)+\ddots } }}}$$

where again we start with $s(1)$ to avoid $s(0)=0$.

and $d = a^2+4c =125, 1, 64, 81, 16, -27,$ respectively (and all powers of the smallest primes $2,3,5$). 

Note: The second closed-form is valid since it was also found by Apery which he used (together with other methods) to prove the irrationality of $\zeta(3)$.

$$\frac{\zeta(4)}{13}=\cfrac{1}{u(0) + \cfrac{v(1)}{u(1) + \cfrac{v(2)}{u(2)+ \cfrac{v(3)}{u(3)+\ddots } }}}$$

with $u(n), v(n)$ as polynomial functions where, as usual, the latter has degree twice that of the former. (To be discussed in the next post.)