Continued Fractions with Quadratic Numerators via the Bauer–Muir Transform

Abstract

We study a class of continued fraction transformations where the partial numerators are quadratic polynomials and the denominators are linear or constant. Using the Bauer–Muir transform, we establish two theorems that yield structurally distinct but equivalent continued fractions—one with rational coefficients and another with alternating forms. These transformations provide a unified framework for evaluating and simplifying continued fractions, including classical identities such as one of Euler, a recent result by Campbell and Chen, and several conjectures from the Ramanujan Machine involving $\pi$ and $\log 2.$ We conclude by discussing the potential extension of our methods to more general polynomial cases.

1. Introduction

Given sequences ${\left(a_n\right)}_{n\in \mathbb{N}}$ and ${\left(b_n\right)}_{n\in \mathbb{N}}$, the continued fraction

$${\displaystyle \frac{a_1}{b_1+\frac{a_2}{b_2+\frac{a_3}{b_3+\ddots }}}}$$

(1)

is typically denoted by

$$\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{a_n}{b_n}}={\displaystyle \frac{a_1}{b_1}}{\displaystyle \genfrac{}{}{0.0pt}{}{}{+}}{\displaystyle \frac{a_2}{b_2}}{\displaystyle \genfrac{}{}{0.0pt}{}{}{+}}{\displaystyle \frac{a_3}{b_3}}{\displaystyle \genfrac{}{}{0.0pt}{}{}{+}}{\displaystyle \genfrac{}{}{0.0pt}{}{}{\cdots }}.$$

Continued fractions have a long and rich history, with deep connections to number theory, approximation theory, orthogonal polynomials, and special functions.

Among the many tools developed for their analysis, the Bauer–Muir transform stands out as a classical yet powerful method. It produces equivalent continued fractions by introducing a sequence of linear modifications to the numerators and denominators while preserving the overall value when convergence holds.

Originally introduced by M. E. Bauer in 1872 and generalized by T. Muir in 1877 [1], the transformation was later recast in terms of linear fractional transformations. It also appears in modern contexts, such as convergence acceleration algorithms and analytic continuation techniques.

Given a continued fraction of the form ${\mathbf{K}}_{n=1}^{\infty }{\displaystyle \frac{a_n}{b_n}}$, its Bauer–Muir transform with respect to a sequence ${\left(x_n\right)}_{n\in {\mathbb{N}}_0}$ is defined by

$$x_0+{\displaystyle \frac{{\varphi }_1}{b_1+x_1+{\mathbf{K}}_{n=1}^{\infty }\frac{a_n{\varphi }_{n+1}/{\varphi }_n}{b_{n+1}+x_{n+1}-x_{n-1}{\varphi }_{n+1}/{\varphi }_n}}},$$

(2)

where

$${\varphi }_n=a_n-x_{n-1}(b_n+x_n),\mathrm{for}n\in \mathbb{N},$$

(3)

and all ${\varphi }_n\ne 0$.

Proposition 1

([1]). Let ${\mathbf{K}}_{n=1}^{\infty }{\displaystyle \frac{a_n}{b_n}}$ be a continued fraction with positive elements that converges. If $x_n\ge 0$ for sufficiently large n, then its Bauer–Muir transform with respect to $\left(x_n\right)$ also converges to the same limit.

This result was originally established by Perron, who proved that under the stated conditions, the Bauer–Muir transform not only converges but also preserves the value of the continued fraction.

Particular choices of the auxiliary sequence $\left(x_n\right)$ yield elegant closed-form expressions. For instance, Belaghi, Khrushchev, and Bashirov [2] considered the case

$$a_n=an+b,b_n=n+c,\mathrm{with}0\le a\le 1,a+b\ge 1,c\ge 0,$$

and showed that the transformed fraction simplifies to the rational form

$$a+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{-an+a-a^2+b-ac}{n+2a+c}}.$$

Recently, the Ramanujan Machine project [3] has introduced a novel paradigm for discovering mathematical identities—especially continued fraction representations of constants such as $\pi$, $\log 2$, and values of the Riemann zeta function. These conjectures are not derived symbolically but through numerical experiments, employing algorithms such as meet-in-the-middle searches and optimization methods. Notably, several of the Ramanujan Machine’s conjectures for $\pi$ exhibit a recurring pattern: continued fractions with quadratic numerators and linear denominators.

This structural form is not limited to conjectures from the Ramanujan Machine. In fact, many classical continued fraction identities also feature quadratic numerators and linear or constant denominators. For example, Ramanujan, in his first letter to Hardy (see [4], p. 145), wrote

$$s+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{{(2n-1)}^2}{2s}}=4{\left({\displaystyle \frac{\Gamma \left(\frac{s+3}{4}\right)}{\Gamma \left(\frac{s+1}{4}\right)}}\right)}^2,\mathrm{for}s>0,$$

and Brouncker’s celebrated formula for $\pi$ (see [1], (3.19)),

$${\displaystyle \frac{4}{\pi }}=1+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{{(2n+1)}^2}{2}},$$

also takes this form.

Motivated by these observations, we aim to systematically investigate transformations of continued fractions of the form

$$\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{(n+a)(n+b)}{\alpha n+c}},$$

and show that they can be re-expressed in new forms—sometimes simpler or more regular—using Bauer–Muir-type transformations.

Polynomial continued fractions—those whose numerators and denominators are polynomial functions of the index—have been studied in various structural settings. In particular, Bowman and Mc Laughlin [5] examined numerous infinite classes of continued fractions where the degrees of the numerator and denominator polynomials are equal or nearly equal. Their work focuses on balanced-degree cases.

In a different direction, Cao, Tanigawa, and Zhai [6] demonstrated the utility of the Bauer–Muir transformation in proving continued fraction identities involving ratios of gamma functions. Notably, several of their identities require two successive applications of the transformation, each with nonlinear modifying sequences. This underscores the flexibility and depth of the method and parallels our approach in utilizing such transformations to obtain elegant continued fraction expansions for mathematical constants.

The recent Ramanujan Machine project [3] provides an algorithmic counterpart to this intuition-driven discovery process, generating candidate identities through large-scale numerical search and symbolic reconstruction.

As an illustrative example, consider the following identity conjectured by the Ramanujan Machine:

$${\displaystyle \frac{\pi }{2}}+1=3-{\displaystyle \frac{2}{6-\frac{9}{9-\frac{20}{12-\frac{35}{15-\ddots }}}}}=3+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{-(n+1)(2n-1)}{3n+3}}.$$

In this paper, we show that this continued fraction is closely related, via our main result (see Theorem 1), to the more regular form

$$\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{(n+1)(n+2)}{1}}={\displaystyle \frac{2·3}{1+\frac{3·4}{1+\frac{4·5}{1+\frac{5·6}{1+\ddots }}}}}.$$

This transformation reveals a hidden structure behind seemingly irregular continued fractions and provides a unified framework for proving many of the Ramanujan Machine’s conjectures.

Main contributions. The principal contributions of this paper are as follows:

• We establish two transformation theorems (Theorems 1 and 2) for continued fractions whose partial numerators are quadratic polynomials and whose denominators are linear or constant.
• The transformed continued fractions typically converge more rapidly than their original forms, offering computational advantages.
• Several continued fraction conjectures related to $\pi$ and $\log 2$ proposed by the Ramanujan Machine are verified by converting them into known classical continued fractions using our transformation formulas.

The remainder of the paper is organized as follows.

Section 2 reviews the general theory of continued fractions relevant to our discussion, including Euler’s continued fraction and an integral representation that will be used in later sections.

Section 3 presents and proves our two main transformation theorems, along with several corollaries.

In Section 4, we compare our results with a classical contraction identity, highlighting structural differences and computational advantages.

Section 5 provides applications of our theorems. We begin with a transformation of a classical Euler-type continued fraction ([1], p. 178), followed by an analysis of a recently discovered identity by Campbell and Chen [7], and conclude by verifying several continued fraction conjectures involving $\pi$ and $\log 2$ proposed by the Ramanujan Machine [3].

Finally, Section 6 concludes the paper with a summary of our contributions and a discussion of possible future directions.

2. Preliminaries

In this section, we review several classical results concerning continued fractions and their transformations. These foundational tools will be frequently invoked in our subsequent derivations and applications.

An ordered pair of sequnces $({\left\{a_n\right\}}_{n\in \mathbb{N}},{\left\{b_n\right\}}_{n\in {\mathbb{N}}_0})$ of complex numbers, with $a_n\ne 0$ for $n\ge 1$, gives rise to sequences ${\left\{s_n\left(w\right)\right\}}_{n\in {\mathbb{N}}_0}$ and ${\left\{S_n\left(w\right)\right\}}_{n\in {\mathbb{N}}_0}$ of linear fractional transformations

$$s_0\left(w\right)=S_0\left(w\right)=b_0+w,s_n\left(w\right)={\displaystyle \frac{a_n}{b_n+w}},S_n\left(w\right)=S_{n-1}\left(s_n\left(w\right)\right),$$

(4)

for $n\in \mathbb{N}$, and a sequence $\left\{f_n\right\}$ with $f_n=S_n\left(0\right)$. It is noted that

$$S_n\left(w\right)=b_0+{\displaystyle \frac{a_1}{b_1}}{\displaystyle \genfrac{}{}{0.0pt}{}{}{+}}{\displaystyle \frac{a_2}{b_2}}{\displaystyle \genfrac{}{}{0.0pt}{}{}{+}}\cdots {\displaystyle \genfrac{}{}{0.0pt}{}{}{+}}{\displaystyle \frac{a_{n-1}}{b_{n-1}}}{\displaystyle \genfrac{}{}{0.0pt}{}{}{+}}{\displaystyle \frac{a_n}{b_n+w}}.$$

(5)

The numbers $a_n$ and $b_n$ are called the n-th partial numerator and partial denominator, respectively, of the continued fraction. The value $f_n$ is called the n-th approximant or convergent. When we write $f_n={\displaystyle \frac{P_n}{Q_n}}$ and let $P_{-1}=1$, $P_0=b_0$, $Q_{-1}=0$, and $Q_0=1$. Then, ${\left\{P_n\right\}}_{n\in {\mathbb{N}}_0}$ and ${\left\{Q_n\right\}}_{n\in {\mathbb{N}}_0}$ satisfy the Euler–Wallis formulas [8]:

$$\begin{array}{cc}\hfill P_n& =b_n{P}_{n-1}+a_n{P}_{n-2},\hfill \end{array}$$

(6)

$$\begin{array}{cc}\hfill Q_n& =b_n{Q}_{n-1}+a_n{Q}_{n-2}.\hfill \end{array}$$

(7)

Assume that ${lim}_{n\to \infty }{f}_n=\xi$, i.e.,

$$\xi =b_0+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{a_n}{b_n}}=b_0+{\displaystyle \frac{a_1}{b_1}}{\displaystyle \genfrac{}{}{0.0pt}{}{}{+}}{\displaystyle \frac{a_2}{b_2}}{\displaystyle \genfrac{}{}{0.0pt}{}{}{+}}\cdots {\displaystyle \genfrac{}{}{0.0pt}{}{}{+}}{\displaystyle \frac{a_n}{b_n}}{\displaystyle \genfrac{}{}{0.0pt}{}{}{+}}{\displaystyle \frac{a_{n+1}}{{\xi }_{n+1}}}.$$

Let $P_n/Q_n$ be its n-th approximant. Then, we have [1]

$$\xi ={\displaystyle \frac{{\xi }_{n+1}{P}_n+a_{n+1}{P}_{n-1}}{{\xi }_{n+1}{Q}_n+a_{n+1}{Q}_{n-1}}}.$$

(8)

If the continued fraction has positive terms, then ([1], Thereom 1.7)

$${\displaystyle \frac{P_0}{Q_0}}<\cdots <{\displaystyle \frac{P_{2n}}{Q_{2n}}}<\cdots <{\displaystyle \frac{P_{2n+1}}{Q_{2n+1}}}<\cdots <{\displaystyle \frac{P_1}{Q_1}}<{\displaystyle \frac{P_{-1}}{Q_{-1}}}=\infty .$$

(9)

Lemma 1

(Euler, 1744 ). Let ${\left(c_k\right)}_{k\in {\mathbb{N}}_0}$ be a sequence of nonzero numbers. Then

$$\sum _{k=0}^n{c}_k={\displaystyle \frac{c_0}{1+{\mathbf{K}}_{k=1}^n{\displaystyle \frac{-c_k/c_{k-1}}{1+c_k/c_{k-1}}}}}.$$

(10)

This classical identity expresses a finite sum in terms of a finite continued fraction. Originally due to Euler, it plays a pivotal role in the development of continued fraction expansions for various series and generating functions.

A particularly useful special case arises when each term $c_k$ can be written as a product of the form $c_k={\prod }_{j=0}^k{r}_j$, where ${\left(r_k\right)}_{k\in {\mathbb{N}}_0}$ is a sequence of nonzero numbers. This multiplicative representation yields the well-known continued fraction identity; specifically, in this setting, the formula reduces to the Euler continued fraction:

$$1+\sum _{k=1}^n{r}_1{r}_2\cdots r_k={\displaystyle \frac{1}{1+{\mathbf{K}}_{k=1}^n{\displaystyle \frac{-r_k}{1+r_k}}}}.$$

(11)

This formulation often arises in the analysis of generating functions and forms a bridge between summation and continued fraction representations.

We also recall a useful rescaling identity that allows for the normalization of a continued fraction through an auxiliary sequence $\left(c_n\right)$. If ${\left(c_n\right)}_{n\in \mathbb{N}}$ is a sequence of nonzero numbers, then:

$$\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{a_n}{b_n}}={\displaystyle \frac{1}{c_1}}\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{a_n{c}_n{c}_{n+1}}{b_n{c}_{n+1}}}.$$

(12)

This transformation is particularly useful when simplifying or aligning the partial numerators and denominators of a continued fraction without altering its value.

In our analysis, we also make use of the following elegant integral identity involving the Laplace transform of the reciprocal of a hyperbolic cosine power (cf. [1], p. 198):

$${\int }_0^{\infty }{\displaystyle \frac{e^{-su}}{{cosh}^{m}u}}du={\displaystyle \frac{1}{s+{\mathbf{K}}_{n=1}^{\infty }\frac{n(n+m-1)}{s}}},(m\in \mathbb{R},s>0).$$

(13)

This identity illustrates a deep connection between special functions and continued fractions. In particular, it highlights the appearance of structured quadratic numerators in a naturally arising continued fraction.

In this paper, we focus on continued fractions with quadratic numerators of the form

$$a_n=\alpha n^2+\beta n+\gamma ,b_n=\delta n+ϵ,$$

and derive explicit transformations for such expressions under the Bauer–Muir framework.

By the Fundamental Theorem of Algebra, every quadratic polynomial with complex coefficients can be factored into linear terms. Moreover, thanks to the scaling property in Equation (12), we may, without loss of generality, normalize the leading coefficient to 1. Hence, it suffices to consider quadratic numerators of the form

$$a_n=(n+a)(n+b),$$

where $a,b\in \mathbb{C}$, which provides a canonical setting for the transformations developed in the next section.

3. Main Results

In this section, we establish a transformation formula for continued fractions whose numerators are quadratic polynomials and whose denominators are linear or constant. Specifically, we consider continued fractions of the form

$$\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{(n+a)(n+b)}{\alpha n+c}},$$

which are often amenable to classical analytic techniques and, in some cases, allow explicit evaluation.

However, many of the recently proposed continued fraction conjectures—such as those discovered by the Ramanujan Machine—exhibit more intricate structures that are not easily identified as classical forms. In such cases, our transformation theorem provides a new perspective by linking these expressions to more familiar continued fractions via analytic equivalence.

In particular, Theorem 1 transforms a continued fraction of the above classical type into a different form whose structure often resembles the conjectures, thereby offering a unified framework for recognizing and verifying these identities. Moreover, the result provides a general method for producing new continued fraction identities that, while differing significantly in form, converge to the same value.

Theorem 1.

Let $a,b,c,\alpha \in \mathbb{R}$, and define $\beta ={\displaystyle \frac{-\alpha +\sqrt{{\alpha }^2+4}}{2}}$. Assume that the continued fraction

$$\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{(n+a)(n+b)}{\alpha n+c}}$$

has positive elements and converges. Then its value is equal to

$$\beta (a+1)+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{-\beta (n+a)\left[\beta (n+a)+c+(n-b-1)(\alpha +\beta )\right]}{\alpha +c+2\beta (n+a+1)+(n-1)(\alpha +\beta )}},$$

provided that the transformed continued fraction also converges.

Proof. 

For any positive integer n, we define

$$\begin{array}{cc}\hfill {\theta }_n& =-\beta (n+a)\left[\beta \right(n+a)+c+(n-b-1\left)\right(\alpha +\beta \left)\right],\hfill \\ \hfill {\delta }_n& =\alpha +c+2\beta (n+a+1)+(n-1)(\alpha +\beta ),\hfill \\ \hfill {\mu }_n& =\underset{k=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{(n+k+a)(k+b)}{\alpha k+c+n(\alpha +\beta )}}.\hfill \end{array}$$

Since both the continued fractions ${\mu }_0$ and $\beta (a+1)+{\mathbf{K}}_{n=1}^{\infty }{\displaystyle \frac{{\theta }_n}{{\delta }_n}}$ converge, there exist two real numbers f and g such that

$${\mu }_0=f,\mathrm{and}\beta (a+1)+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{{\theta }_n}{{\delta }_n}}=g.$$

Let $({\left\{{\theta }_n\right\}}_{n\in \mathbb{N}},{\left\{{\delta }_n\right\}}_{n\in {\mathbb{N}}_0},{\{S_n\left(0\right)=f_n\}}_{n\in \mathbb{N}})$ be the continued fraction $\beta (a+1)+{\mathbf{K}}_{n=1}^{\infty }{\displaystyle \frac{{\theta }_n}{{\delta }_n}}$.

Let $a_n=(n+a)(n+b)$, $b_n=\alpha n+c$, $x_n=\beta (n+a+1)$. Then, by Proposition 1, its Bauer–Muir transform with respect to ${\left(x_n\right)}_{n\in {\mathbb{N}}_0}$ converges to the same value. Then, from the formula of ${\varphi }_n$ (see Equation (3)), we know that

$${\varphi }_n=\left(1-(\alpha +\beta )\beta \right)(n+a)(n+b)-\beta (n+a)(\beta +a\beta +c-\alpha b-\beta b).$$

Since $\beta =(-\alpha +\sqrt{{\alpha }^2+4})/2$, we have $\beta (\alpha +\beta )=1$. This implies that

$${\varphi }_n=-\beta (n+a)(\beta +a\beta +c-\alpha b-\beta b).$$

By Proposition 1, we have

$$\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{(n+a)(n+b)}{\alpha n+c}}=\beta (a+1)+{\displaystyle \frac{-\beta (1+a)(\beta +a\beta +c-\alpha b-\beta b)}{\alpha +c+\beta (a+2)+{\mathbf{K}}_{n=1}^{\infty }\frac{(n+a+1)(n+b)}{\alpha n+\alpha +c+\beta }}}.$$

(14)

Thus, ${\mu }_1$ converges. Using a similar method to treat the continued fraction ${\mu }_1$, we choose $x_n=\beta (n+a+2)$, and then we have

$$\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{(n+a+1)(n+b)}{\alpha n+c+\alpha +\beta }}=\beta (a+2)+{\displaystyle \frac{-\beta (2+a)\left(\beta (2+a)+c+(1-b)(\alpha +\beta )\right)}{\alpha +c+\alpha +\beta +\beta (a+3)+{\mathbf{K}}_{n=1}^{\infty }\frac{(n+a+2)(n+b)}{\alpha n+c+2(\alpha +\beta )}}}.$$

Therefore, we know that ${\mu }_2$ also converges. Moreover, the original continued fraction ${\mu }_0$ becomes

$$\beta (a+1)+{\displaystyle \frac{-\beta (1+a)(\beta +a\beta +c-\alpha b-\beta b)}{\alpha +c+2\beta (a+2)+\frac{-\beta (2+a)\left(\beta (2+a)+c+(1-b)(\alpha +\beta )\right)}{\alpha +c+\alpha +\beta +\beta (a+3)+{\mathbf{K}}_{n=1}^{\infty }\frac{(n+a+2)(n+b)}{\alpha n+c+2(\alpha +\beta )}}}}.$$

Continuing this process m times, we have that if the continued fraction ${\mu }_{m-1}$ converges, then it is equal to

$$\beta (a+m)+{\displaystyle \frac{-\beta (m+a)\left(\beta (m+a)+c+(m-1-b)(\alpha +\beta )\right)}{\alpha +c+(m-1)(\alpha +\beta )+\beta (a+m+1)+{\mathbf{K}}_{n=1}^{\infty }\frac{(n+a+m)(n+b)}{\alpha n+c+m(\alpha +\beta )}}}.$$

This implies that ${\mu }_m$ also converges, and we have the following identity for any positive integer m:

$$\begin{array}{cc}\hfill {\mu }_0& =S_m({\mu }_m-\beta (m+a+1))\hfill \end{array}$$

(15)

$$\begin{array}{cc}\hfill & =\beta (a+1)+{\displaystyle \frac{{\theta }_1}{{\delta }_1}}{\displaystyle \genfrac{}{}{0.0pt}{}{}{+}}{\displaystyle \frac{{\theta }_2}{{\delta }_2}}{\displaystyle \genfrac{}{}{0.0pt}{}{}{+}}{\displaystyle \genfrac{}{}{0.0pt}{}{}{\cdots }}{\displaystyle \genfrac{}{}{0.0pt}{}{}{+}}{\displaystyle \frac{{\theta }_{m-1}}{{\delta }_{m-1}}}{\displaystyle \genfrac{}{}{0.0pt}{}{}{+}}{\displaystyle \frac{{\theta }_m}{{\delta }_m}}{\displaystyle \genfrac{}{}{0.0pt}{}{}{+}}{\displaystyle \frac{{\mu }_m-\beta (m+a+1)}{1}}.\hfill \end{array}$$

(16)

That is to say,

$$S_n({\mu }_n-\beta (n+a+1))=f,\mathrm{for} n\in \mathbb{N}.$$

Let ${\displaystyle \frac{P_n}{Q_n}}=f_n=S_n\left(0\right)$ be the n-th approximant. We have $Q_{-1}=0$, $Q_0=1$, and

$$Q_n={\delta }_n{Q}_{n-1}+{\theta }_n{Q}_{n-2}.$$

This gives that

$$Q_1={\delta }_1,Q_2={\delta }_1{\delta }_2+{\theta }_2,Q_3={\delta }_1{\delta }_2{\delta }_3+{\delta }_3{\theta }_2+{\delta }_1{\theta }_3.$$

It is known that $|{\theta }_n|=\Theta \left(n^2\right)$ and ${\delta }_n=\Theta \left(n\right)$. Therefore, we have ${\displaystyle \frac{Q_n}{Q_{n-1}}}=\Theta \left(n\right)$.

Since ${\mu }_m$ has positive terms, we use Brouncker’s theorem (see Equation (9)):

$${\displaystyle \frac{(a+m+1)(b+1)}{\alpha +c+m(\alpha +\beta )+\frac{(a+m+2)(b+2)}{2\alpha +c+m(\alpha +\beta )}}}<{\mu }_m<{\displaystyle \frac{(a+m+1)(b+1)}{\alpha +c+m(\alpha +\beta )}}.$$

Taking $m\to \infty$, we have ${\mu }_m\to {\displaystyle \frac{b+1}{\alpha +\beta }}$. Applying Equations (8) and (16), we have

$$f={\displaystyle \frac{P_n+({\mu }_n-\beta (n+a+1))}{Q_n+({\mu }_n-\beta (n+a+1))}}.$$

Therefore,

$$\left|{\displaystyle \frac{P_n}{Q_n}}-f\right|=\left|{\displaystyle \frac{P_n}{Q_n}}-{\displaystyle \frac{P_n+({\mu }_n-\beta (n+a+1))}{Q_n+({\mu }_n-\beta (n+a+1))}}\right|={\displaystyle \frac{\left|{\mu }_n-\beta (n+a+1)\right|·\left|\frac{P_{n-1}}{Q_{n-1}}-\frac{P_n}{Q_n}\right|}{\left|\frac{Q_n}{Q_{n-1}}+{\mu }_n-\beta (n+a+1)\right|}}.$$

Since $Q_n/Q_{n-1}=\Theta \left(n\right)$,

$$\underset{n\to \infty }{lim}{\displaystyle \frac{\left|{\mu }_n-\beta (n+a+1)\right|}{\left|\frac{Q_n}{Q_{n-1}}+{\mu }_n-\beta (n+a+1)\right|}}$$

exists, and its value is a positive constant. Since ${lim}_{n\to \infty }{f}_n=g$, we have

$$\left|{\displaystyle \frac{P_{n-1}}{Q_{n-1}}}-{\displaystyle \frac{P_n}{Q_n}}\right|\to 0\mathrm{as}n\to \infty .$$

Therefore, $g={lim}_{n\to \infty }{f}_n=f$. We have the desired result. □

The preceding theorem admits an equivalent formulation by reparameterizing the expression in terms of the variable d, defined via $d=\beta$, so that $\alpha ={\displaystyle \frac{1-d^2}{d}}$. By applying the transformation in Equation (12), we obtain the following corollary.

Corollary 1.

Let $a,b,c,d\in \mathbb{R}$, and assume that the continued fraction on the left-hand side of the identity below has positive elements. Then

$$\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{d^2(n+a)(n+b)}{(1-d^2)n+cd}}=d^2(a+1)+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{-d^2(n+a)\left[(1+d^2)n+a{d}^2+cd-b-1\right]}{(2d^2+1)n+d^2+2a{d}^2+cd}},$$

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provided that both continued fractions converge.

While Theorem 1 provides a continued fraction transformation in which the partial numerators follow a unified expression, it is sometimes beneficial to express the continued fraction in a form where the numerators are defined separately for odd and even indices.

The next result presents such a refinement: the sequence $a_n$ is defined by two distinct formulas—one for odd n, and another for even n. Although this parity-based representation yields the same value as that in Theorem 1, it often reveals more transparent arithmetic structure and can be more amenable to symbolic analysis or application-specific adaptation.

We now state this alternative formulation as Theorem 2.

Theorem 2.

Let $a,b,c,\alpha \in \mathbb{R}$, and define $\beta ={\displaystyle \frac{-\alpha +\sqrt{{\alpha }^2+4}}{2}}$. Assume that the continued fraction

$$\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{(n+a)(n+b)}{\alpha n+c}}$$

has positive elements and converges. Then it has the same value as the continued fraction

$$\beta (a+1)+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{a_n}{\alpha +c+\beta (n+a+b+2)+(n-1)(\alpha +\beta )}},$$

where

$$\begin{array}{cc}\hfill a_{2n-1}& =-\beta (n+a)\left[\beta (n+a)+c+(\alpha +\beta )(n-b-1)\right],\hfill \\ \hfill a_{2n}& =-\beta (n+b)\left[\beta (n+b)+c+(\alpha +\beta )(n-a-1)\right],\mathit{for}n\ge 1,\hfill \end{array}$$

provided that this transformed continued fraction also converges.

Proof. 

Following the same approach as in the proof of Theorem 1, we begin with Equation (14). Instead of using the same $x_n$, we now choose a different $x_n=\beta (n+b+1)$. Applying the Bauer–Muir transform with this choice, we have

$$\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{(n+a+1)(n+b)}{\alpha n+c+\alpha +\beta }}=\beta (b+1)+{\displaystyle \frac{-\beta (1+b)\left[\beta (1+b)+c+(0-a)(\alpha +\beta )\right]}{\alpha +c+\alpha +\beta +\beta (b+2)+{\mathbf{K}}_{n=1}^{\infty }\frac{(n+a+1)(n+b+1)}{\alpha n+c+2(\alpha +\beta )}}}.$$

Substituting this expression into Equation (14), we obtain a nested form for

$$\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{(n+a)(n+b)}{\alpha n+c}},$$

expressed in terms of alternating shifts of a and b, along with recursive applications of the transformation:

$$\beta (a+1)+{\displaystyle \frac{-\beta (1+a)(\beta +a\beta +c-\alpha b-\beta b)}{\alpha +c+\beta (a+b+3)+\frac{-\beta (1+b)\left(\beta (1+b)+c+(0-a)(\alpha +\beta )\right)}{\alpha +c+\alpha +\beta +\beta (b+2)+{\mathbf{K}}_{n=1}^{\infty }\frac{(n+a+1)(n+b+1)}{\alpha n+c+2(\alpha +\beta )}}}}.$$

By iterating this process—alternating the roles of a and b and incrementing the index—we arrive at the continued fraction with partial numerators defined by parity as stated in Theorem 2. The proof that both converge to the same value parallels Theorem 1’s reasoning, but with significantly increased notational complexity; we therefore omit the details. □

The preceding theorem admits an equivalent formulation by reparameterizing the expression in terms of the variable d, defined by $d=\beta$, so that $\alpha ={\displaystyle \frac{1-d^2}{d}}$. We also apply the transformation identity (12) to simplify coefficients. This leads to the following corollary.

Corollary 2.

Let $a,b,c,d\in \mathbb{R}$, and assume that the continued fraction

$$\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{d^2(n+a)(n+b)}{(1-d^2)n+cd}}$$

has positive elements and converges. Then it is equal in value to

$$d^2(a+1)+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{a_n}{(d^2+1)n+d^2(a+b+1)+cd}},$$

where the partial numerators $a_n$ are defined by

$$\begin{array}{cc}\hfill a_{2n-1}& =-d^2(n+a)\left[d^2(n+a)+cd+(n-b-1)\right],\hfill \\ \hfill a_{2n}& =-d^2(n+b)\left[d^2(n+b)+cd+(n-a-1)\right],\mathit{for}n\ge 1,\hfill \end{array}$$

provided that the transformed continued fraction also converges.

Together, Theorems 1 and 2 provide two complementary transformation formulas for infinite continued fractions with quadratic numerators and linear denominators. The first presents a unified expression, while the second distinguishes between odd- and even-indexed terms, revealing finer structural features in certain cases.

To better position our main results within the broader theory of continued fractions, the next section provides a comparison between our transformation theorems and certain classical identities, such as contraction formulas. This comparative discussion highlights the structural differences and advantages of our approach.

We then proceed to demonstrate the utility of our results in Section 5, where we apply them to a variety of examples. These examples comprise a classical Euler-type continued fraction identity, a recently obtained formula of Campbell and Chen, and several conjectures proposed by the Ramanujan Machine. In particular, we show how our theorems can be used to verify, reinterpret, or simplify these identities by linking them to known evaluations or rewriting them in more tractable forms. The applications are organized into four thematic subsections, each highlighting a different context in which our transformation formulas prove effective.

4. Comparison with Classical Continued Fraction Transformations

Before proceeding to applications, we briefly compare our transformation theorems with classical identities in the theory of continued fractions. One well-known result expresses a continued fraction with unit denominators in terms of a structurally modified form involving products of consecutive partial numerators. Specifically, for a sequence $\left\{a_n\right\}$, one has (see [9], Lemma 1; [1], (5.38)):

$$\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{a_n}{1}}=a_1+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{-a_{2n-1}{a}_{2n}}{1+a_{2n}+a_{2n+1}}},$$

(18)

provided that both sides converge. This identity, a special case of the general contraction principle, preserves the value of the original continued fraction while reorganizing its structure via grouped numerator products.

Recently, Dougherty-Bliss and Zeilberger [10], as well as Laohakosol, Meesa, and Sutthimat [11], highlighted an elegant continued fraction expansion for the reciprocal of the Riemann zeta function $\zeta \left(s\right)={\sum }_{n=1}^{\infty }{n}^{-s}$, valid for $\Re \left(s\right)>1$:

$${\displaystyle \frac{1}{\zeta \left(s\right)}}=1+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{-n^{2s}}{{(n+1)}^s+n^s}}.$$

To further illustrate the differences between classical and new transformations, we consider the generalized Dirichlet beta function:

$${\beta }_{a,b}\left(s\right):=\sum _{n=0}^{\infty }{\displaystyle \frac{{(-1)}^n}{{(an+b)}^s}},a>0,b>0,\Re \left(s\right)>0.$$

This function interpolates between classical examples:

• $\beta \left(s\right)={\beta }_{2,1}\left(s\right)$, the Dirichlet beta function;
• $\eta \left(s\right)={\beta }_{1,1}\left(s\right)$, the alternating Riemann zeta function (Dirichlet eta function).

Using Euler’s formula (cf. Lemma 1), we express ${\beta }_{a,b}\left(s\right)$ as a continued fraction:

$${\beta }_{a,b}\left(s\right)={\displaystyle \frac{1}{b^s+{\mathbf{K}}_{n=1}^{\infty }\frac{{(an+b-a)}^{2s}}{{(an+b)}^s-{(an+b-a)}^s}}}.$$

(19)

Alternatively, using the Mellin transform, we have the integral representation:

$${\beta }_{a,b}\left(s\right)={\displaystyle \frac{1}{\Gamma \left(s\right)}}{\int }_0^{\infty }{\displaystyle \frac{x^{s-1}{e}^{-bx}}{1+e^{-ax}}}dx.$$

(20)

By substituting specific parameter values into (19), we recover two well-known classical identities. When $(a,b)=(2,1)$ (the Dirichlet beta case) and $(a,b)=(1,1)$ (the Dirichlet eta case), we obtain

$${\displaystyle \frac{1}{\beta \left(s\right)}}=1+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{{(2n-1)}^{2s}}{{(2n+1)}^s-{(2n-1)}^s}},{\displaystyle \frac{1}{\eta \left(s\right)}}=1+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{n^{2s}}{{(n+1)}^s-n^s}}.$$

These formulas emerge directly as special cases of the general continued-fraction representation (19).

We now consider a concrete example with parameters $a=3$, $b=1$, and $s=1$ to illustrate the practical difference in convergence behavior under various transformation methods. In this case,

$$\sum _{n=0}^{\infty }{\displaystyle \frac{{(-1)}^n}{3n+1}}={\displaystyle \frac{\sqrt{3}\pi +3\log 2}{9}}={\displaystyle \frac{1}{1+{\mathbf{K}}_{n=1}^{\infty }\frac{{(3n-2)}^2}{3}}}.$$

Letting

$$M:=\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{{(3n-2)}^2}{3}},$$

we compute its exact value:

$$M={\displaystyle \frac{9}{\sqrt{3}\pi +3\log 2}}-1\approx 0.1966748977,$$

accurate to at least 10 decimal places.

Applying the classical transformation (18), we write

$$M=3\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{{(n-\frac{2}{3})}^2}{1}},$$

and set $a_n={(n-\frac{2}{3})}^2$. This gives:

$${\displaystyle \frac{1}{3}}+{\displaystyle \frac{2}{3}}\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{-{(3n-1)}^2{(6n-5)}^2}{36n^2-6n+7}}\approx 0.1966868184,$$

evaluated using 10,000 terms. The approximation is accurate only up to four decimal places.

In contrast, applying Theorem 1 yields:

$$1+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{-6n(3n-2)}{9n+2}}\approx 0.1966748977,$$

which matches the exact value to all 10 decimal places using the same number of terms.

In summary, when applicable, our transformation theorems significantly enhance both numerical convergence and computational efficiency. Although we do not pursue the computational aspect in detail here, the above example clearly illustrates the potential benefits.

More importantly, Theorems 1 and 2 provide structurally novel continued fractions—with variable denominators and tunable parameters—that frequently yield elegant evaluations involving $\pi$, $\log 2$, or gamma function quotients.

This highlights a key advantage of our method: classical transformations often rely on recursive or contraction-based patterns, while our approach offers a unifying, parameter-dependent framework that simplifies and explains various identities—including recent conjectures generated by the Ramanujan Machine.

5. Applications

In this section, we demonstrate the applicability of Theorems 1 and 2 through a series of concrete examples. These examples are organized into four subsections, beginning with a classical transformation originally due to Euler, followed by a recent evaluation by Campbell and Chen. We then turn to several conjectural continued fractions proposed by the Ramanujan Machine, including identities involving $\pi$ and $\log 2$, and show how our transformation framework allows these conjectures to be verified or reformulated.

5.1. A Continued Fraction Identity Derived from Euler’s Formula

We begin by transforming a classical result due to Euler, stated in 1750 (cf. [1], p. 178). For parameters $s,q>0$ and $r>q$, Euler established the identity

$$s+\underset{n=0}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{(rn+q)(rn+r-q)}{2s}}=(q+s)·{\displaystyle \frac{{\int }_0^1{x}^{r+s+q-1}{(1-x^{2r})}^{-1/2}dx}{{\int }_0^1{x}^{r+s-q-1}{(1-x^{2r})}^{-1/2}dx}}.$$

(21)

We first apply the transformation identity (12) and shift the index to begin at $n=1$, yielding

$$\underset{n=0}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{(rn+q)(rn+r-q)}{2s}}=r·\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{(n+\frac{q}{r}-1)(n-\frac{q}{r})}{2s/r}}.$$

We now apply Theorem 1 with parameter choices

$$a={\displaystyle \frac{q}{r}}-1,b=-{\displaystyle \frac{q}{r}},c={\displaystyle \frac{2s}{r}},\alpha =0,$$

which leads to $\beta =1$. Under these values, Theorem 1 yields the identity

$$\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{(n+\frac{q}{r}-1)(n-\frac{q}{r})}{2s/r}}={\displaystyle \frac{q}{r}}+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{-(n+\frac{q}{r}-1)\left(2n+\frac{2q}{r}-2+\frac{2s}{r}\right)}{3n+\frac{2q}{r}+\frac{2s}{r}-1}}.$$

Applying transformation (12) again, we conclude that Euler’s identity is equivalent to the continued fraction

$$s+q+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{-(rn-r+q)\left(2rn+2q+2s-2r\right)}{3rn+2q+2s-r}}.$$

(22)

As a special case, let $r=2$ and $q=1$. Then, based on our transformation framework, we obtain the following new continued fraction representation:

$$s+1+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{(4n+2s-2)(1-2n)}{6n+2s}}.$$

(23)

This expression is equivalent to Euler’s formula (21) under these parameter values.

On the other hand, the classical evaluation of the same expression is given by

$$s+\underset{n=0}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{{(2n+1)}^2}{2s}}=4{\left({\displaystyle \frac{\Gamma \left(\frac{s+3}{4}\right)}{\Gamma \left(\frac{s+1}{4}\right)}}\right)}^2,$$

(24)

which was originally proved by Bauer in 1872 and later independently rediscovered by Ramanujan (see Part II of Ramanujan’s Notebooks [4]).

Hence, the identity

$$s+1+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{(4n+2s-2)(1-2n)}{6n+2s}}=s+\underset{n=0}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{{(2n+1)}^2}{2s}}$$

serves as a new representation of the Bauer–Ramanujan formula, obtained via Theorem 1.

In particular, Brouncker’s classical formula arises by setting $s=1$, yielding

$${\displaystyle \frac{4}{\pi }}=2+2\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{n(1-2n)}{3n+1}}=1+\underset{n=0}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{{(2n+1)}^2}{2}}.$$

(25)

Here, the right-hand side corresponds to the well-known Brouncker’s formula for ${\displaystyle \frac{4}{\pi }}$, which has appeared in classical literature since the 17th century. In contrast, the left-hand side is one of the continued fraction identities recently conjectured by the Ramanujan Machine [3].

In Section 5.3, we will revisit this identity and present an independent proof using Theorem 1, thereby offering a rigorous justification for one of the machine-generated conjectures.

5.2. A Continued Fraction Evaluation by Campbell and Chen

For $s>0$, Campbell and Chen [7] derived the following continued fraction identity:

$${\displaystyle \frac{s}{2}}+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{n^2-n-1}{s}}=2·{\displaystyle \frac{\Gamma \left(\frac{s-\sqrt{5}+3}{4}\right)\Gamma \left(\frac{s+\sqrt{5}+3}{4}\right)}{\Gamma \left(\frac{s-\sqrt{5}+1}{4}\right)\Gamma \left(\frac{s+\sqrt{5}+1}{4}\right)}}.$$

(26)

Observe that the numerator satisfies the factorization

$$n^2-n-1=(n-\varphi )(n-1+\varphi ),$$

where $\varphi ={\displaystyle \frac{1+\sqrt{5}}{2}}$ is the golden ratio.

Applying Theorem 1 with parameters $a=-\varphi$, $b=\varphi -1$, $c=s$, and $\alpha =0$, we obtain $\beta =1$. Under these parameters, Theorem 1 transforms the left-hand side of Equation (26) into the equivalent expression:

$$1-\varphi +{\displaystyle \frac{s}{2}}+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{(n-\varphi )(2\varphi -s-2n)}{3n-2\varphi +s+1}}.$$

(27)

As an example, let $s=2\varphi$. Then we obtain the identity

$$\varphi +\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{n^2-n-1}{2\varphi }}=1+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{2n(\varphi -n)}{3n+1}}={\displaystyle \frac{2\Gamma \left(\varphi +\frac{1}{2}\right)}{\sqrt{\pi }\Gamma \left(\varphi \right)}}.$$

(28)

Remark 1.

Strictly speaking, the continued fraction in this example has a first partial numerator $a_1=-1$, which is negative and thus violates the positivity condition required by Theorem 1. Nevertheless, we are still justified in applying the Bauer–Muir transform in this case. In fact, the conditions under which the transform preserves convergence and value can be relaxed. Jacobsen [12] remarked that the Bauer–Muir transformation of a convergent continued fraction “almost always” converges to the same limit.

To apply the transformation in a rigorous manner, one may observe that the continued fraction satisfies the required positivity condition from $n\ge 2$ onward. We can therefore isolate the initial term and apply the transformation to the tail:

$${\displaystyle \frac{s}{2}}+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{n^2-n-1}{s}}={\displaystyle \frac{s}{2}}-{\displaystyle \frac{1}{s+{\mathbf{K}}_{n=1}^{\infty }\frac{n^2+n-1}{s}}}.$$

Applying Theorem 1, we have

$$\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{n^2+n-1}{s}}=1+\varphi +\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{(n+\varphi )(2-2\varphi -s-2n)}{3n+2\varphi +s+1}}.$$

Thus, the following identity holds:

$${\displaystyle \frac{s}{2}}+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{n^2-n-1}{s}}={\displaystyle \frac{s}{2}}-{\displaystyle \frac{1}{s+\varphi +1+{\mathbf{K}}_{n=1}^{\infty }\frac{(n+\varphi )(2-2\varphi -s-2n)}{3n+2\varphi +s+1}}}.$$

On the other hand, applying the same parameters to Theorem 2 yields another equivalent representation of the left-hand side of Equation (26):

$$1-\varphi +{\displaystyle \frac{s}{2}}+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{a_n}{2n+s}},$$

where the sequence $\left(a_n\right)$ is defined by

$$\begin{array}{cc}\hfill a_{2n-1}& =(2n-2\varphi +s)(\varphi -n),\hfill \\ \hfill a_{2n}& =(2n+2\varphi +s-2)(1-\varphi -n),\mathrm{for}n\ge 1.\hfill \end{array}$$

Since the coefficients $a_n$ alternate depending on parity, we do not pursue a specific numerical example here (such as setting $s=2\varphi$). Instead, we note that Theorem 2 offers an alternative continued fraction representation of a different structural form, illustrating the flexibility of the transformation.

The example considered in this section illustrates how newly established identities involving special functions can be reinterpreted and transformed into continued fractions with algebraic or even symmetric structure using our framework. In particular, the flexibility of Theorems 1 and 2 allows one to derive structurally distinct but equivalent continued fraction representations, which may offer additional insight into analytic properties or facilitate further symbolic or numerical evaluations. We believe such connections enrich the interplay between special functions, continued fractions, and algebraic parameterizations.

5.3. Ramanujan’s Machine Conjectures Involving $\pi$

In this subsection, we verify several continued fraction identities related to $\pi$ that were conjectured by the Ramanujan Machine [3]. Specifically, we confirm the following five formulas:

$$\begin{array}{cc}\hfill 2+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{n(1-2n)}{3n+2}}& ={\displaystyle \frac{2}{\pi -2}},\hfill \end{array}$$

(29)

$$\begin{array}{cc}\hfill 1+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{n(1-2n)}{3n+1}}& ={\displaystyle \frac{2}{\pi }},\hfill \end{array}$$

(30)

$$\begin{array}{cc}\hfill 1+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{n(3-2n)}{3n+1}}& ={\displaystyle \frac{4}{\pi }},\hfill \end{array}$$

(31)

$$\begin{array}{cc}\hfill 2+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{(n+1)(1-2n)}{3n+3}}& ={\displaystyle \frac{\pi }{2}},\hfill \end{array}$$

(32)

$$\begin{array}{cc}\hfill \underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{n(3-2n)}{3n-1}}& ={\displaystyle \frac{\pi +2}{\pi +4}}.\hfill \end{array}$$

(33)

To demonstrate the validity of these conjectures, we begin by applying Theorem 1 with the parameter choices $a=0$, $b=c=1$, and $\alpha =0$, which yields $\beta =1$. Under these values, Theorem 1 gives the identity

$$\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{n(n+1)}{1}}=1+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{n(1-2n)}{3n+2}}.$$

On the other hand, applying $s=1$ and $m=2$ in the integral identity (13), we obtain

$$1+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{n(n+1)}{1}}={\displaystyle \frac{1}{{\displaystyle {\int }_0^{\infty }\frac{e^{-u}}{{cosh}^{2}u}du}}}={\displaystyle \frac{1}{(\pi -2)/2}}.$$

(34)

Combining the two identities above, we confirm the first Ramanujan Machine conjecture:

$$2+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{n(1-2n)}{3n+2}}={\displaystyle \frac{2}{\pi -2}}.$$

To verify the second identity (30), we again apply Theorem 1, this time with parameter choices $a=c=1$, $b=0$, and $\alpha =0$, which again yields $\beta =1$. Under these values, Theorem 1 gives the continued fraction identity

$$\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{n(n+1)}{1}}=2+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{-(n+1)(2n+1)}{3n+4}}=2+\underset{n=2}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{n(1-2n)}{3n+1}}.$$

Combining this with the evaluation in (34), we deduce that

$$1+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{n(1-2n)}{3n+1}}={\displaystyle \frac{2}{\pi }},$$

thereby confirming the second conjecture.

To verify the third identity (31), we apply Theorem 1 with parameters $a=c=1$, $b=2$, and $\alpha =0$, yielding again $\beta =1$. In this case, we obtain the identity

$$\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{(n+1)(n+2)}{1}}=2+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{(n+1)(1-2n)}{3n+4}}=2+\underset{n=2}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{n(3-2n)}{3n+1}}.$$

On the other hand, we note that

$$\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{n(n+1)}{1}}={\displaystyle \frac{2}{1}}{\displaystyle \genfrac{}{}{0.0pt}{}{}{+}}\underset{n=2}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{n(n+1)}{1}}={\displaystyle \frac{2}{1}}{\displaystyle \genfrac{}{}{0.0pt}{}{}{+}}\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{(n+1)(n+2)}{1}}.$$

Using Equation (34), we thus deduce:

$$1+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{n(3-2n)}{3n+1}}={\displaystyle \frac{4}{\pi }},$$

which confirms the third conjecture.

To verify the fourth identity (32), we apply Theorem 1 with parameters $a=2$, $b=c=1$, and $\alpha =0$, again yielding $\beta =1$. Under this specialization, the resulting identity is

$$\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{(n+1)(n+2)}{1}}=3+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{-(n+2)(2n+1)}{3n+6}}=3+\underset{n=2}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{(n+1)(1-2n)}{3n+3}}.$$

From this, we obtain the value of the continued fraction

$$\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{(n+1)(1-2n)}{3n+3}},$$

and hence verify the fourth identity (32).

Finally, we turn to the fifth identity (33). Applying Theorem 1 with parameters $a=3$, $b=0$, $c=1$, and $\alpha =0$, we again obtain $\beta =1$. Under these values, the continued fraction identity becomes

$$\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{n(n+3)}{1}}=4+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{-(n+3)(2n+3)}{3n+8}}=4+\underset{n=4}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{n(3-2n)}{3n-1}}.$$

Next, we evaluate the left-hand side using Equation (13) with $s=1$ and $m=4$, which yields

$$1+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{n(n+3)}{1}}={\displaystyle \frac{1}{{\displaystyle {\int }_0^{\infty }\frac{e^{-u}}{{cosh}^{4}u}du}}}={\displaystyle \frac{12}{3\pi -4}}.$$

On the other hand, from the identity above, we also have

$$\underset{n=4}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{n(3-2n)}{3n-1}}=\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{n(n+3)}{1}}-4={\displaystyle \frac{12}{3\pi -4}}-5={\displaystyle \frac{32-15\pi }{3\pi -4}}.$$

Therefore, we solve for the tail continued fraction and conclude

$$\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{n(3-2n)}{3n-1}}={\displaystyle \frac{\pi +2}{\pi +4}},$$

which confirms the fifth and final conjecture.

The five continued fraction identities conjectured by the Ramanujan Machine and related to $\pi$ have now all been rigorously confirmed using Theorem 1 and the integral identity (13). We now turn to another remarkable continued fraction from the Ramanujan Machine, this time involving the constant $\log 2$.

5.4. A Ramanujan’s Machine Conjecture Involving log2

We now turn to a continued fraction identity involving $\log 2$, as conjectured by the Ramanujan Machine [3]:

$$4+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{-2n^2{(n+1)}^2}{(n+1)(3n+4)}}={\displaystyle \frac{1}{1-\log 2}}.$$

(35)

To verify this, we apply Theorem 1 with parameters $a=0$, $b=0$, $c=3$, and $\alpha =0$, which yields $\beta =1$. Under these values, the continued fraction becomes

$$\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{n^2}{3}}=1+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{-n(2n+2)}{3n+4}}.$$

Next, we evaluate the left-hand side using the integral identity (13) with $s=3$ and $m=1$, giving

$$3+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{n^2}{3}}={\displaystyle \frac{1}{{\displaystyle {\int }_0^{\infty }\frac{e^{-3u}}{coshu}du}}}={\displaystyle \frac{1}{1-\log 2}}.$$

On the other hand, applying the rescaling transformation (12), we find

$$\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{-2n^2{(n+1)}^2}{(n+1)(3n+4)}}=\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{-2n(n+1)}{3n+4}}.$$

Combining the above identities, we obtain

$$\begin{array}{cc}\hfill 4+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{-2n^2{(n+1)}^2}{(n+1)(3n+4)}}& =4+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{-2n(n+1)}{3n+4}}\hfill \\ \hfill & =3+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{n^2}{3}}={\displaystyle \frac{1}{1-\log 2}},\hfill \end{array}$$

which confirms the conjectured identity (35).

The examples presented in this section demonstrate the scope and flexibility of the transformation theorems developed earlier. From classical formulas to recently established evaluations, and from conjectural identities proposed by the Ramanujan Machine to explicit representations involving special constants like $\pi$ and $\log 2$, our results provide a coherent framework that connects seemingly diverse continued fractions through unified structural transformations.

Beyond confirming individual identities, the approach also offers deeper insight into how different continued fractions may arise from common algebraic templates, further revealing the hidden symmetries and relationships within their parameterizations.

We conclude the paper with a final remark highlighting the potential directions for further generalization and investigation.

6. Concluding Remark

We apply Theorem 2 with parameters $a=0$, $b=1$, $c=1$, and $\alpha =0$, which yields $\beta =1$. Under these values, we obtain the following continued fraction transformation:

$$\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{n(n+1)}{1}}=1+\underset{n=1}{\overset{\infty }{\mathbf{K}}}{\displaystyle \frac{a_n}{2n+3}}=1-{\displaystyle \frac{1}{5-\frac{6}{7-\frac{6}{9-\frac{15}{11-\frac{15}{13-\ddots }}}}}},$$

(36)

where the coefficients $a_n$ are given by

$$\begin{array}{cc}\hfill a_{2n-1}& =-n(2n-1),\hfill \\ \hfill a_{2n}& =-(n+1)(2n+1),\mathrm{for}n\ge 1.\hfill \end{array}$$

The numerical value of this continued fraction, as evaluated in Equation (34), is known to be

$${\displaystyle \frac{4-\pi }{\pi -2}}.$$

This example revisits the central continued fraction ${\mathbf{K}}_{n=1}^{\infty }{\displaystyle \frac{n(n+1)}{1}}$ that appeared repeatedly in verifying several conjectures proposed by the Ramanujan Machine. While Theorem 1 enabled us to transform such expressions into forms directly evaluable via known integral identities, Theorem 2 provides an alternative but structured representation. In particular, the resulting continued fraction exhibits a striking pattern: the partial numerators alternate in a predictable way, and the partial denominators increase along odd integers.

Our investigation reveals that seemingly irregular continued fractions conjectured through numerical methods often conceal highly structured and analyzable forms. The two transformation theorems developed in this work provide a unified and explicit mechanism to identify these underlying patterns, especially in the context of continued fractions with quadratic numerators and linear or constant denominators.

In addition to validating known identities and recent conjectures, our approach opens the door to the generation and systematic transformation of entire families of continued fractions. In particular, it provides a versatile tool for the following:

• Producing new formulas for special values such as $\pi$, $\log 2$, or gamma function quotients;
• Simplifying or reinterpreting identities from experimental discovery platforms like the Ramanujan Machine;
• Accelerating convergence in numerical evaluations of slowly converging continued fractions.

Furthermore, while the current formulation of our theorems imposes certain algebraic conditions on the parameters, it is worth noting that these constraints can be relaxed. As observed by Jacobsen [12], broader classes of Bauer–Muir type transformations remain valid under weaker hypotheses, particularly when the sign and positivity constraints are lifted or adjusted. This suggests promising avenues for generalization, which interested readers may explore further.

Future directions also include extending our framework to encompass continued fractions with higher-degree polynomial numerators, q-analogues, or multivariate settings. Moreover, the integration of our transformation technique into symbolic computation tools could facilitate the automatic analysis and proof of newly discovered identities.

While these generalizations are promising, handling higher-degree numerators poses added complexity: the application of the Bauer–Muir transform often leads to complicated expressions, and finding suitable sequences $x_n$ that yield manageable and simplifiable transformed continued fractions remains a significant challenge.

In conclusion, the transformation theorems presented here offer not only a new lens through which to interpret classical and modern continued fraction identities but also a foundation for further theoretical development and computational exploration.