1 Introduction

In recent work [5], we considered the integer sequence

$$\begin{aligned} 1,1,2,12,936,68408496,342022190843338960032,\ldots \end{aligned}$$
(1.1)

(sequence A112373 in Sloane’s Online Encyclopedia of Integer Sequences), which is generated from the initial values \(x_0=x_1=1\) by the nonlinear recurrence relation

$$\begin{aligned} x_{n+2} \, x_n = x_{n+1}^2(x_{n+1}+1), \end{aligned}$$
(1.2)

and proved some observations of Hanna, namely that the sum

$$\begin{aligned} \sum _{j=1}^\infty \frac{1}{x_j} \end{aligned}$$
(1.3)

has the continued fraction expansion

$$\begin{aligned}{}[x_0; y_0,x_1,y_1,x_2,\ldots , y_{j-1},x_{j},\ldots ], \end{aligned}$$
(1.4)

where \(y_j=x_{j+1}/x_j\in {\mathbb N}\) and we use the notation

$$\begin{aligned}{}[a_0; a_1,a_2,a_3,\ldots ,a_n,\ldots ]=a_0+\frac{1}{a_1+\frac{1}{a_2+\frac{1}{a_3+\ldots \frac{1}{a_n+\ldots }}}} \end{aligned}$$

for continued fractions. Furthermore, we generalized this result by obtaining the explicit continued fraction expansion for the sum of reciprocals (1.3) in the case of a sequence \((x_n)\) generated by a nonlinear recurrence of the form

$$\begin{aligned} x_{n+1} \, x_{n-1} = x_{n}^2\, F(x_{n}), \end{aligned}$$
(1.5)

with \(F(x)\in {\mathbb Z}_{\ge 0}[x]\) and \(F(0)=1\); so (1.2) corresponds to the particular case \(F(x)=x+1\).

All of the recurrences (1.5) exhibit the Laurent phenomenon [4], and starting from \(x_0=x_1=1\) they generate a sequence of positive integers satisfying \(x_n|x_{n+1}\). The latter fact means that the sum (1.3) is an Engel series (see Theorem 2.3 in Duverney’s book [3], for instance).

The purpose of this note is to present a further generalization of the results in [5], by considering a sum

$$\begin{aligned} S= \frac{p}{q}+\sum _{j=2}^\infty \frac{1}{x_j}, \end{aligned}$$
(1.6)

with the terms \(x_n\) satisfying the recurrence

$$\begin{aligned} x_{n+1} \, x_{n-1} = x_{n}^2\, (z_nx_{n}+1), \end{aligned}$$
(1.7)

for \(n\ge 2\), where \((z_n)\) is a sequence of positive integers, \(x_1=q\), and \(x_2\) is specified suitably. Observe that, in contrast to (1.5), the recurrence (1.7) can be viewed as a non-autonomous dynamical system for \(x_n\), because the coefficient \(z_n\) can vary independently (unless it is taken to be \(G(x_n)\), for some function G). The same argument as used in [5], based on Roth’s theorem, shows the transcendence of any number S defined by a sum of the form (1.6) with such a sequence \((x_n)\).

2 The main result

We start with a rational number written in lowest terms as p / q, and suppose that the continued fraction of this number is given as

$$\begin{aligned} \frac{p}{q}= [a_0; a_1,a_2,a_3,\ldots ,a_{2k}] \end{aligned}$$
(2.1)

for some \(k\ge 0\). Note that, in accordance with a comment on p. 230 of [7], there is no loss of generality in assuming that the index of the final coefficient is even. For the convergents we denote numerators and denominators by \(p_n\) and \(q_n\), respectively, and use the correspondence between matrix products and continued fractions, which says that

$$\begin{aligned} \mathbf {M}_n:= \left( \begin{array}{cc} p_n &{}\quad p_{n-1} \\ q_n &{} q_{n-1} \end{array}\right) = \left( \begin{array}{cc} a_0 &{}\quad 1 \\ 1 &{} 0 \end{array}\right) \left( \begin{array}{cc} a_1 &{}\quad 1 \\ 1 &{} 0 \end{array}\right) \ldots \left( \begin{array}{cc} a_n &{}\quad 1 \\ 1 &{} 0 \end{array}\right) , \end{aligned}$$
(2.2)

yielding the determinantal identity

$$\begin{aligned} \det \mathbf {M}_n = p_nq_{n-1}-p_{n-1}q_n = (-1)^{n+1}. \end{aligned}$$
(2.3)

Now for a given sequence \((z_n)\) of positive integers, we define a new sequence \((x_n)\) by

$$\begin{aligned} x_1=q, \qquad x_{n+1}=x_ny_{n-1}(x_nz_n+1) \quad \mathrm {for} \quad n\ge 1, \end{aligned}$$
(2.4)

where

$$\begin{aligned} y_0 = q_{2k-1}+1, \qquad y_n=\frac{x_{n+1}}{x_n}\quad \mathrm {for} \quad n\ge 1. \end{aligned}$$
(2.5)

It is clear from (2.4) and (2.5) that \((x_n)\) is an increasing sequence of positive integers such that \(x_n|x_{n+1}\) for all \(n\ge 1\); \((y_n)\) also consists of positive integers, and is an increasing sequence as well. The recurrence (1.7) for \(n\ge 2\) follows immediately from (2.4) and (2.5).

Theorem 2.1

The partial sums of (1.6) are given by

$$\begin{aligned} S_n:= \frac{p}{q}+\sum _{j=2}^n \frac{1}{x_j} =[a_0;a_1,\ldots ,a_{2(k+n-1)}] \end{aligned}$$

for all \(n\ge 1\), where the coefficients appearing after \(a_{2k}\) are

$$\begin{aligned} a_{2k+2j-1} = y_{j-1}z_{j}, \qquad a_{2k+2j}=x_j \quad for \quad j\ge 1. \end{aligned}$$

Proof

For \(n=1\), \(S_1\) is just (2.1), and we note that \(q_{2k-1}=y_0-1\) and \(q_{2k}=q=x_1\). Proceeding by induction, we suppose that \(q_{2k+2n-3}=y_{n-1}-1\) and \(q_{2k+2n-2}=x_n\), and calculate the product

$$\begin{aligned} \begin{array}{l} \mathbf {M}_{2k+2n}=\mathbf {M}_{2k+2n-2} \left( \begin{array}{cc} a_{2k+2n-1} &{}\quad 1 \\ 1 &{}\quad 0 \end{array}\right) \left( \begin{array}{cc} a_{2k+2n} &{}\quad 1 \\ 1 &{}\quad 0 \end{array}\right) \\ \qquad \qquad \, = \mathbf {M}_{2k+2n-2} \left( \begin{array}{cc} y_{n-1}z_{n} &{}\quad 1 \\ 1 &{}\quad 0 \end{array}\right) \left( \begin{array}{cc} x_n &{}\quad 1 \\ 1 &{}\quad 0 \end{array}\right) \\ \qquad \qquad \, = \left( \begin{array}{cc} p_{2k+2n-2}&{}\quad p_{2k+2n-3} \\ q_{2k+2n-2} &{}\quad q_{2k+2n-3} \end{array}\right) \left( \begin{array}{cc} x_ny_{n-1}z_n &{}\quad y_{n-1}z_n \\ x_n &{}\quad 1 \end{array}\right) . \end{array} \end{aligned}$$

By making use of (2.4) and (2.5), this gives \(p_{2k+2n}=(x_ny_{n-1}z_n+1)p_{2k+2n-2}+x_n p_{2k+2n-3}\),

$$\begin{aligned} q_{2k+2n-1}= & {} y_{n-1}z_n\, q_{2k+2n-2}+ q_{2k+2n-3} = x_n y_{n-1}z_n +y_{n-1}-1 \\ \qquad \qquad \,= & {} \frac{x_{n+1}}{x_n} -1 = y_n-1, \end{aligned}$$

and

$$\begin{aligned} q_{2k+2n}= & {} (x_ny_{n-1}z_n+1)q_{2k+2n-2}+x_n q_{2k+2n-3}\\ \qquad \quad \,= & {} (x_ny_{n-1}z_n+1)x_n + x_n (y_{n-1}-1)=x_{n+1}, \end{aligned}$$

which are the required denominators for the \((2k+2n-1)\)th and \((2k+2n)\)th convergents. Thus we have

$$\begin{aligned} S_{n+1}=S_n + \frac{1}{x_{n+1}} =\frac{p_{2k+2n-2}}{q_{2k+2n-2}} +\frac{1}{q_{2k+2n}} = \frac{1}{q_{2k+2n}}\left( \frac{x_{n+1}}{x_n}p_{2k+2n-2}+1\right) . \end{aligned}$$

From (2.3) and (2.4), the bracketed expression above can be rewritten as

$$\begin{aligned} \begin{array}{l} \Big (y_{n-1}(x_nz_n+1) -q_{2n+2k-3}\Big )p_{2k+2n-2} +q_{2n+2k-2}p_{2k+2n-3} \\ \qquad = \Big (y_{n-1}(x_nz_n+1) -y_{n-1}+1\Big )p_{2k+2n-2} +x_np_{2k+2n-3}, \end{array} \end{aligned}$$

giving

$$\begin{aligned} S_{n+1} = \frac{1}{q_{2k+2n}}\Big ( (x_ny_{n-1}z_n+1)p_{2k+2n-2}+x_n p_{2k+2n-3}\Big )= \frac{p_{2k+2n}}{q_{2k+2n}}, \end{aligned}$$

which is the required result.\(\square \)

Upon taking the limit \(n\rightarrow \infty \) we obtain the infinite continued fraction expansion for the sum S, which is clearly irrational. To show that S is transcendental, we need the following growth estimate for \(x_n\):

Lemma 2.2

The terms of a sequence defined by (2.4) satisfy

$$\begin{aligned} x_{n+1}>x_n^{5/2} \end{aligned}$$

for all \(n\ge 3\).

Proof

Since \((x_n)\) is an increasing sequence, the recurrence relation (1.7) gives

$$\begin{aligned} x_{n+1}>\frac{x_n^3}{x_{n-1}}>x_n^2 \end{aligned}$$

for \(n\ge 2\). Hence \(x_{n-1}<x_n^{1/2}\) for \(n\ge 3\), and putting this back into the first inequality above yields \(x_{n+1}>x_n^3/x_n^{1/2}=x_n^{5/2}\), as required.\(\square \)

The preceding growth estimate for \(x_n\) means that S can be well approximated by rational numbers.

Theorem 2.3

The sum

$$\begin{aligned} S= \frac{p}{q}+\sum _{j=2}^\infty \frac{1}{x_j} =[a_0;a_1,\ldots ,a_{2k},y_0z_1,x_1,y_1z_2,\ldots , y_{j-1}z_j,x_j, \ldots ] \end{aligned}$$

is a transcendental number.

Proof

This is the same as the proof of Theorem 4 in [5], which we briefly outline here. Let \(P_n=p_{2k+2n-2}\) and \(Q_n=q_{2k+2n-2}\). Approximating the irrational number S by the partial sum \(S_n=P_n/Q_n\), then using Lemma 2.2 and a comparison with a geometric sum, gives the upper bound

$$\begin{aligned} \left| S-\frac{P_n}{Q_n} \right| =\sum _{j=n+1}^\infty \frac{1}{x_j} < \frac{1}{x_{n}^{5/2-\epsilon }}=\frac{1}{Q_n^{5/2-\epsilon }} \end{aligned}$$

for any \(\epsilon >0\), whenever n is sufficiently large. Roth’s theorem [6] (see also chapter VI in [1]) says that, for an arbitrary fixed \({\kappa }>2\), an irrational algebraic number \({\alpha }\) has only finitely many rational approximations P / Q for which \( \left| {\alpha }-\frac{P}{Q}\right| <\frac{1}{Q^{{\kappa }}}; \) so S is transcendental.\(\square \)

For other examples of transcendental numbers whose continued fraction expansion is explicitly known, see [2] and references therein.

3 Examples

The autonomous recurrences (1.5) considered in [5], where the polynomial F has positive integer coefficients and \(F(0)=1\), give an infinite family of examples. In that case, one has \(p=1\) and \(x_1=q=1\), so that \(k=0\), \(y_0=1\) and \(z_n=(F(x_n)-1)/x_n\). More generally, one could take \(z_n=G(x_n)\) for any non-vanishing arithmetical function G.

In general, it is sufficient to take the initial term in (1.6) lying in the range \(0<p/q\le 1\), since going outside this range only alters the value of \(a_0\). As a particular example, we take

$$\begin{aligned} \frac{p}{q}=\frac{2}{7}=[0;3,2], \qquad z_n=n \quad \mathrm {for} \quad n\ge 1, \end{aligned}$$

so that \(k=1\), and \(q_{1}=3\) which gives \(y_0=2\). Hence \(x_1=7\), \(x_2=112\), and the sequence \((x_n)\) continues with

$$\begin{aligned} 403200,1755760043520000,53695136666462381094317154204367872000000,\ldots . \end{aligned}$$

The sum S is the transcendental number

$$\begin{aligned} \frac{2}{7}+\frac{1}{112}+\frac{1}{403200}+ \frac{1}{1755760043520000}+\cdots \approx 0.2946453373015879, \end{aligned}$$

with continued fraction expansion

$$\begin{aligned}{}[0;3,2,2,7,32,112,10800,403200,17418254400,1755760043520000,\ldots ]. \end{aligned}$$