June 13, 2026
We study the concatenation of two \(k\)-generalized Pell numbers. More precisely, we determine all solutions of the equation \(P_n^{(k)} = P_m^{(k)} \cdot 10^{d} + P_p^{(k)}\), where \(d\) is the number of decimal digits of \(P_p^{(k)}\). We prove that for \(k \ge 3\) there are no solutions, while for \(k = 2\) the only solution is \(P_4 = 12 = 1\|2\).
For an integer \(k\ge 2\), the sequence of \(k\)-generalized Pell numbers (or \(k\)-Pell numbers) \((P_n^{(k)})_{n\ge 2-k}\) is defined by the linear recurrence \[P_n^{(k)} = 2P_{n-1}^{(k)} + P_{n-2}^{(k)} + \dots + P_{n-k}^{(k)}\qquad (n\ge 2),\] with initial conditions \(P_{-(k-2)}^{(k)} = \dots = P_0^{(k)} = 0\) and \(P_1^{(k)} = 1\). For \(k=2\) we recover the classical Pell sequence \(P_0=0\), \(P_1=1\), \(P_n=2P_{n-1}+P_{n-2}\).
The problem of concatenations of two terms of a sequence is classical. For Fibonacci numbers, Banks and Luca [1] proved that \(F_{10}=55 = 5\|5\) is the largest such number. For Lucas numbers, Erdüvan [2] showed that the only solutions are \(L_6=11 = 1\|1\) and \(L_9=47 = 4\|7\).
In [3], Altassan and Alan studied the analogous problem for \(k\)-generalized Fibonacci numbers and found the solutions \[F_7^{(3)} = 24 = \overline{F_3^{(3)}F_4^{(3)}},\quad F_8^{(3)} = 44 = \overline{F_4^{(3)}F_4^{(3)}},\quad F_{16}^{(8)} = 16128 = \overline{F_6^{(8)}F_9^{(8)}}.\] Recently, Tumwesigye, Ddamulira and Kaggwa [4] extended the result to \(k\)-generalized Lucas numbers.
In this paper we study the analogous problem for \(k\)-Pell numbers. Our main result is the following.
Theorem 1. Let \(k \ge 2\) and let \((P_n^{(k)})_{n \ge 2-k}\) be the \(k\)-generalized Pell sequence. Consider the Diophantine equation
\[\label{eq1} P_n^{(k)} = P_m^{(k)} \cdot 10^{d} + P_p^{(k)},\qquad{(1)}\]
where \(n > m, p \ge 0\) and \(d = \lfloor \log_{10} P_p^{(k)} \rfloor + 1\).
If \(k = 2\), the only solution of ?? is \((n, m, p) = (4, 1, 2)\), since \(P_4 = 12 = 1 \| 2\).
If \(k \ge 3\), equation ?? has no solutions.
In the next sections we develop the necessary tools and give the proof.
Let \(k \geqslant 2\) be an integer. We consider the linear recurrence sequence of order \(k\), \(G^{(k)}:=(G_n^{(k)})_{n\ge 2-k}\) defined as
\[G^{(k)}_n=rG^{(k)}_{n-1}+G^{(k)}_{n-2}+\cdots + G^{(k)}_{n-k}\quadfor all n\geq 2,\] with the initial conditions \[G^{(k)}_{-(k-2)}=G^{(k)}_{-(k-3)}=\cdots=G^{(k)}_{-1}=0, G^{(k)}_0=a\quad and\quad G^{(k)}_1=b.\] Observe that if \(a=0\) and \(b=1\), then \(G^{(k)}\) is nothing that just the \(k\)- generalized Fibonacci sequence or for simplicity, the \(k\)-Fibonacci sequence \(F^{(k)}:=(F ^{(k)}_n)_{n\geq 2-k}\). In this case, if we choose \(k=2\) we obtain the classical Fibonacci sequence \((F_n)_{n}\).
On the other hand, if \(r=2, a=0\) and \(b=1\) then \(G^{k}\) is known as the \(k\)-generalized Pell sequence \(P^{(k)}:=(P^{(k)}_n)_{n\geq 2-k}\). The terms of this sequence are called \(k\)-generalized Pell numbers.
The \(k\)-Pell numbers and their properties have been studied by many authors. For example, Kiliç [5] showed that the first \(k+1\) non-zero terms in \(P^(k)\) are the Fibonacci numbers with odd index, namely
\[\label{eq2} P_n^{(k)}=F_{2n-1} \quad for all 1\leq n\leq k+1.\tag{1}\] In addition, it was also proved in [5] that if \(k+2\leq n\leq 2k+2\), then
\[\label{eq3} P_n^{(k)}=F_{2n-1}-\sum_{j=1}^{n-k-1}F_{2j-1}F_{2(n-k-1)} .\tag{2}\]
It is known that the characteristic polynomial of the \(k\)–generalized Pell numbers \(P^{(k)}:=(P_m^{(k)})_{m\geq 2-k}\), namely \[\Psi_k(x) := x^k - 2x^{k-1} - \cdots - x - 1,\] is irreducible over \(\mathbb{Q}[x]\) and has just one root outside the unit circle. Let \(\alpha := \alpha(k)\) denote that single root. It was proved in [6] that \(\alpha(k)\) is located between \(\varphi ^2\left(1-\varphi^{-k} \right)\) and \(\varphi^2\) where \(\varphi\) denotes the golden section.This is called the dominant root of \(P^{(k)}\). To simplify notation, in our application we shall omit the dependence on \(k\) of \(\alpha\). We shall use \(\alpha^{(1)}, \dotso, \alpha^{(k)}\) for all roots of \(\Psi_k(x)\) with the convention that \(\alpha^{(1)} := \alpha\).
We now consider for an integer \(k\geq 2\), the function \[\begin{align} \label{eq4} g_{k}(z) = \dfrac{z-1}{(k+1)z^2-3kz+k-1}=\frac{z-1}{k(z^2-3z+1)+z^2-1} \qquad {\text{for}}\quad z \in \mathbb{C}. \end{align}\tag{3}\] In the following lemma, we give some properties of the sequence \(P^{(k)}\) which will be used in the proof of the equation ?? . The following lemmas was proved by Bravo and al in [6] and [7], respectively.
Lemma 1 (Bravo and al., [6]). Let \(k\geq 2\) be an integer. Then, \(\alpha\) be the dominant root of \(\{P^{(k)}_m\}_{m\ge 2-k}\). Then,
\(\alpha^{n-2}\leq P_n^{(k)}\leq \alpha^{n-1}\) for all \(n\geq 1;\)
\(P^{(k)}\) satisfies the following Binet formula \[\begin{align} \label{Binet} P_n^{(k)} = \sum_{i=1}^{k}g_k(\alpha_i)\alpha^n_i. \end{align}\qquad{(2)}\] where \(\alpha_1,\ldots, \alpha_k\) are the roots of the \(\Psi_k(x)\);
\[\label{eq5} \left| P_n^{(k)} - g_{k}(\alpha)\alpha^{n} \right| < \dfrac{1}{2} \quad holds~ for~ all~n \geqslant 2 - k.\qquad{(3)}\]
\(0.276< g_{k}(\alpha)<0.5.\)
Lemma 2 (Bravo and al., [7]). Let \(k\geq 2\), \(\alpha\) be the dominant root of \(\{P^{(k)}_m\}_{m\ge 2-k}\), and consider the function \(g_{k}(z)\) defined in 3 .
We begin by recalling some basic notions from algebraic number theory.
Let \(\eta\) be an algebraic number of degree \(d\) with minimal primitive polynomial over the integers \[a_0x^{d}+ a_1x^{d-1}+\cdots+a_d = a_0\prod_{i=1}^{d}(x-\eta^{(i)}),\] where the leading coefficient \(a_0\) is positive and the \(\eta^{(i)}\)’s are the conjugates of \(\eta\). Then the logarithmic height of \(\eta\) is given by \[h(\eta):=\dfrac{1}{d}\left( \log a_0 + \sum_{i=1}^{d}\log\left(\max\{|\eta^{(i)}|, 1\}\right)\right).\] In particular, if \(\eta=p/q\) is a rational number with \(\gcd (p,q)=1\) and \(q>0\), then \(h(\eta)=\log\max\{|p|, q\}\). The following are some of the properties of the logarithmic height function \(h(\cdot)\), which will be used without reference: \[\begin{align} h(\eta\pm \gamma) &\leq& h(\eta) +h(\gamma) +\log 2,\nonumber\\ h(\eta\gamma^{\pm 1})&\leq & h(\eta) + h(\gamma),\\ h(\eta^{s}) &=& |s|h(\eta) \qquad (s\in\mathbb{Z}). \nonumber \end{align}\] Using the above properties of the logarithmic height, Bravo and al. showed in [6] that the logarithmic height of \(g_k(\alpha)\) satisfies \[\label{height} h(g_{k}(\alpha))< 4\log k \quadfor k\geq 3,\tag{4}\] which will be used in the proof of the main problem.
We will also need the following Lemma.
Lemma 3 (Guzman–Luca, Lemma 7 in [8]). If \(s \ge 1\) and \(T > (4s^2)^s\), then the inequality \[\frac{y}{(\log y)^s} < T\] implies \[y < 2^s \, T \, (\log T)^s.\]
Lemma 4 (Cooper–Howard for \(k\)-Pell [9]). For \(k \ge 2\) and \(n \ge k+2\), we have \[P_n^{(k)} = 2^{\,n-1} + \sum_{j=1}^{\ell-1} C_{n,j} \, 2^{\,n-(k+1)j-1},\] where \[\ell = \left\lfloor \frac{n+k}{k+1} \right\rfloor,\] and the coefficients \(C_{n,j}\) are given by \[C_{n,j} = (-1)^j \left[ \binom{n - jk}{j+1} - \binom{n - jk - 2}{j-1} \right],\] with the convention \(\binom{a}{b}=0\) if \(a < b\) or if \(a\) or \(b\) is negative.
In order to solve our main equation ?? , we need to use a Baker–type lower bound for a nonzero linear form in logarithms of algebraic numbers. There are many such in the literature like that of Baker and Wüstholz. We use the following result by Matveev [10], which is one of our main tools in this project.
Theorem 2 (Matveev). Let \(\gamma_1,\ldots,\gamma_t\) be positive real algebraic numbers in a real algebraic number field \(\mathbb{K}\) of degree \(D\), \(b_1,\ldots,b_t\) be nonzero integers, and assume that \[\label{eq:Lambda} \Lambda:=\gamma_1^{b_1}\cdots\gamma_t^{b_t} - 1\qquad{(4)}\] is nonzero. Then \[\log |\Lambda| > -1.4\times 30^{t+3}\times t^{4.5}\times D^{2}(1+\log D)(1+\log B)A_1\cdots A_t,\] where \[B\geq\max\{|b_1|, \ldots, |b_t|\},\] and \[A _i \geq \max\{Dh(\gamma_i), |\log\gamma_i|, 0.16\},\qquad {\text{for all}}\qquad i=1,\ldots,t.\]
Finally, with the help of a computer program we can found all possible solutions.
The bounds on the variables obtained via Baker’s theorem are usually too large for any computational purposes. Therefore, to lower the upper bounds of the integer unknowns, one can either use a reduction method, usually called the Baker-Davenport reduction procedure or the LLL. reduction algorithm. In this project, we will use both.
For the Baker-Davenport, the variant we apply here is the one due to Dujella and Pethő ([11], Lemma 5a). For a real number \(r\), we denote by \(\parallel r \parallel\) the quantity \(\min \{|r - n| : n \in \mathbb{Z} \}\), the distance from \(r\) to the nearest integer.
Lemma 5 (Dujella, Pethő, [11]). Let \(\kappa \neq 0, A, B\) and \(\mu\) be real numbers such that \(A > 0\) and \(B > 1\). Let \(M > 1\) be a positive integer and suppose that \(\frac{p}{q}\) is a convergent of the continued fraction expansion of \(\kappa\) with \(q > 6M\). Let \[\begin{align} \varepsilon := \parallel \mu q \parallel - M \parallel \kappa q \parallel. \end{align}\] If \(\varepsilon > 0\), then there is no solution of the inequality \[\begin{align} 0 < |m \kappa - n + \mu| < AB^{-k} \end{align}\] in positive integers \(m,n,k\) with \[\begin{align} \frac{\log (Aq/\varepsilon)}{\log B} \leq k \quad \text{and} \quad m \leq M. \end{align}\]
Lemma 5 cannot be applied when \(\mu=0\) (since then \(\varepsilon<0\)). In this case, we use the following criterion due to Legendre, a well–known result from the theory of Diophantine approximation. For further details, we refer the reader to the books of Cohen [12], [13].
Lemma 6 (Legendre, [12], [13]). Let \(\kappa\) be real number and \(x,y\) integers such that \[\begin{align} \left|\kappa-\frac{x}{y}\right|<\frac{1}{2y^2}. \end{align}\] Then \(x/y=p_k/q_k\) is a convergent of \(\kappa\). Furthermore, let \(M\) and \(N\) be a nonnegative integers such that \(q_N> M\). Then putting \(a(M):=\max\{a_{i}: i=0, 1, 2, \ldots, N\}\), the inequality \[\begin{align} \left|\kappa-\frac{x}{y}\right|\ge \frac{1}{(a(M)+2)y^2}, \end{align}\] holds for all pairs \((x,y)\) of positive integers with \(0<y<M\).
We will also need the following lemma by Gúzman Sánchez and Luca ([14], Lemma 7):
Lemma 7 (Gúzman Sánchez, Luca, [14]). Let \(r \geq 1\) and \(H > 0\) be such that \(H > (4r^2)^r\) and \(H > L/(\log L)^r\). Then \[\begin{align} L < 2^r H (\log H)^r. \end{align}\]
Let \(k\) be a positive integer. A subset \(\mathcal{L}\) of the \(k\)–dimensional real vector space \({ \mathbb{R}^k}\) is called a lattice if there exists a basis \(\{b_1, b_2, \ldots, b_k \}\) of \(\mathbb{R}^k\) such that \[\begin{align} \mathcal{L} = \sum_{i=1}^{k} \mathbb{Z} b_i = \left\{ \sum_{i=1}^{k} r_i b_i \mid r_i \in \mathbb{Z} \right\}. \end{align}\] We say that \(b_1, b_2, \ldots, b_k\) form a basis for \(\mathcal{L}\), or that they span \(\mathcal{L}\). We call \(k\) the rank of \(\mathcal{L}\). The determinant \(\text{det}(\mathcal{L})\), of \(\mathcal{L}\) is defined by \[\begin{align} \text{det}(\mathcal{L}) = | \det(b_1, b_2, \ldots, b_k) |, \end{align}\] with the \(b_i\)’s being written as column vectors. This is a positive real number that does not depend on the choice of the basis (see [15], Section 1.2).
Given linearly independent vectors \(b_1, b_2, \ldots, b_k\) in \(\mathbb{R}^k\), we refer back to the Gram–Schmidt orthogonalization technique. This method allows us to inductively define vectors \(b^*_i\) (with \(1 \leq i \leq k\)) and real coefficients \(\mu_{i,j}\) (for \(1 \leq j \leq i \leq k\)). Specifically, \[\begin{align} b^*_i &= b_i - \sum_{j=1}^{i-1} \mu_{i,j} b^*_j,~~~ \mu_{i,j} = \dfrac{\langle b_i, b^*_j\rangle }{\langle b^*_j, b^*_j\rangle}, \end{align}\] where \(\langle \cdot , \cdot \rangle\) denotes the ordinary inner product on \(\mathbb{R}^k\). Notice that \(b^*_i\) is the orthogonal projection of \(b_i\) on the orthogonal complement of the span of \(b_1, \ldots, b_{i-1}\), and that \(\mathbb{R}b_i\) is orthogonal to the span of \(b^*_1, \ldots, b^*_{i-1}\) for \(1 \leq i \leq k\). It follows that \(b^*_1, b^*_2, \ldots, b^*_k\) is an orthogonal basis of \(\mathbb{R}^k\).
Definition 1. The basis \(b_1, b_2, \ldots, b_n\) for the lattice \(\mathcal{L}\) is called reduced if \[\begin{align} \| \mu_{i,j} \| &\leq \frac{1}{2}, \quad \text{for} \quad 1 \leq j < i \leq n,~~ \text{and}\\ \|b^*_{i}+\mu_{i,i-1} b^*_{i-1}\|^2 &\geq \frac{3}{4}\|b^*_{i-1}\|^2, \quad \text{for} \quad 1 < i \leq n, \end{align}\] where \(\| \cdot \|\) denotes the ordinary Euclidean length. The constant \({3}/{4}\) above is arbitrarily chosen, and may be replaced by any fixed real number \(y\) in the interval \({1}/{4} < y < 1\) (see [16], Section 1).
Let \(\mathcal{L}\subseteq\mathbb{R}^k\) be a \(k-\)dimensional lattice with reduced basis \(b_1,\ldots,b_k\) and denote by \(B\) the matrix with columns \(b_1,\ldots,b_k\). We define \[l\left( \mathcal{L},y\right):= \left\{ \begin{array}{c} \min_{x\in \mathcal{L}}||x-y|| \quad ;~~ y\not\in \mathcal{L}\\ \min_{0\ne x\in \mathcal{L}}||x|| \quad ;~~ y\in \mathcal{L} \end{array} \right.,\] where \(||\cdot||\) denotes the Euclidean norm on \(\mathbb{R}^k\). It is well known that, by applying the LLL–algorithm, it is possible to give in polynomial time a lower bound for \(l\left( \mathcal{L},y\right)\), namely a positive constant \(c_1\) such that \(l\left(\mathcal{L},y\right)\ge c_1\) holds (see [17], Section V.4).
Lemma 8. Let \(y\in\mathbb{R}^k\) and \(z=B^{-1}y\) with \(z=(z_1,\ldots,z_k)^T\). Furthermore,
if \(y\not \in \mathcal{L}\), let \(i_0\) be the largest index such that \(z_{i_0}\ne 0\) and put \(\sigma:=\{z_{i_0}\}\), where \(\{\cdot\}\) denotes the distance to the nearest integer.
if \(y\in \mathcal{L}\), put \(\sigma:=1\).
Finally, let \[c_2:=\max\limits_{1\le j\le k}\left\{\dfrac{||b_1||^2}{||b_j^*||^2}\right\}.\] Then, \[l\left( \mathcal{L},y\right)^2\ge c_2^{-1}\sigma^2||b_1||^2:=c_1^2.\]
In our application, we are given real numbers \(\eta_0,\eta_1,\ldots,\eta_k\) which are linearly independent over \(\mathbb{Q}\) and two positive constants \(c_3\) and \(c_4\) such that \[\begin{align} \label{2469} |\eta_0+a_1\eta_1+\cdots +a_k \eta_k|\le c_3 \exp(-c_4 H), \end{align}\tag{5}\] where the integers \(a_i\) are bounded as \(|a_i|\le A_i\) with \(A_i\) given upper bounds for \(1\le i\le k\). We write \(A_0:=\max\limits_{1\le i\le k}\{A_i\}\). The basic idea in such a situation, is to approximate the linear form 5 by an approximation lattice. So, we consider the lattice \(\mathcal{L}\) generated by the columns of the matrix \[\mathcal{A}=\begin{pmatrix} 1 & 0 &\ldots& 0 & 0 \\ 0 & 1 &\ldots& 0 & 0 \\ \vdots & \vdots &\vdots& \vdots & \vdots \\ 0 & 0 &\ldots& 1 & 0 \\ \lfloor C\eta_1\rfloor & \lfloor C\eta_2\rfloor&\ldots & \lfloor C\eta_{k-1}\rfloor& \lfloor C\eta_{k} \rfloor \end{pmatrix} ,\] where \(C\) is a large constant usually of the size of about \(A_0^k\) . Let us assume that we have an LLL–reduced basis \(b_1,\ldots, b_k\) of \(\mathcal{L}\) and that we have a lower bound \(l\left(\mathcal{L},y\right)\ge c_1\) with \(y:=(0,0,\ldots,-\lfloor C\eta_0\rfloor)\). Note that \(c_1\) can be computed by using the results of Lemma 8. Then, with these notations the following result is Lemma VI.1 in [17].
Lemma 9 (Lemma VI.1 in [17]). Let \(S:=\displaystyle\sum_{i=1}^{k-1}A_i^2\) and \(T:=\dfrac{1+\sum_{i=1}^{k}A_i}{2}\). If \(c_1^2\ge T^2+S\), then inequality 5 implies that we either have \(a_1=a_2=\cdots=a_{k-1}=0\) and \(a_k=-\dfrac{\lfloor C\eta_0 \rfloor}{\lfloor C\eta_k \rfloor}\), or \[H\le \dfrac{1}{c_4}\left(\log(Cc_3)-\log\left(\sqrt{c_1^2-S}-T\right)\right).\]
Suppose that \(n \le k+1\). By the properties of the \(k\)-Pell sequence, for \(2 \le n \le k+1\) we have \[P_n^{(k)} = F_{2n-1},\] where \(F_m\) denotes the \(m\)-th Fibonacci number (\(F_1=1, F_2=1, F_3=2, \dots\)). The same holds for \(m\) and \(p\) (since \(m,p < n\)).
Thus equation ?? becomes \[\label{eq6} F_{2n-1} = F_{2m-1} \cdot 10^{d} + F_{2p-1}.\tag{6}\]
This is a Diophantine equation involving only Fibonacci numbers. Banks and Luca [1] determined all Fibonacci numbers that are concatenations of two Fibonacci numbers. Their results show that the only solutions are \[F_8 = 21 = \overline{F_3F_2},\qquad F_{10} = 55 = \overline{F_5F_5}.\] However, these correspond to even indices (\(F_8\) and \(F_{10}\)), while 6 involves odd-index Fibonacci numbers \(F_{2n-1}, F_{2m-1}, F_{2p-1}\). A direct verification (or a short computer search for small indices) shows that the only possible solution would be \(2n-1=8\) (i.e. \(n=4.5\)) or \(2n-1=10\) (i.e. \(n=5.5\)), which are not integers. Hence no solution of 6 exists with \(n \ge 3\) (the case \(n=2\) gives trivial single-digit numbers).
Therefore, there are no solutions when \(n \le k+1\) and \(k \ge 3\). Consequently we may assume \(n \ge k+2\) in the rest of the proof.
Assume that ?? holds.
The number of digits of \(P_p^{(k)}\) is \[\label{eq7} d = \lfloor \log_{10} P_p^{(k)} \rfloor + 1.\tag{7}\] From the bounds \(\alpha^{p-2} \le P_p^{(k)} \le \alpha^{p-1}\) (Lemma 1) and the fact that \(\log_{10}\alpha \in (1/5, 1/3)\) for \(k\ge 3\), we obtain \[\label{eq8} \frac{p-2}{5} < d < \frac{p+2}{3}.\tag{8}\]
Using equation ?? together with the estimates \(\alpha^{n-2} \le P_n^{(k)} \le \alpha^{n-1}\) and the trivial bounds \(P_m^{(k)} \ge \alpha^{m-2}\), \(P_p^{(k)} \le \alpha^{p-1}\), after straightforward calculations we obtain \[\label{eq9} m + p - 3 < n < m + p + 6.\tag{9}\]
Now, using the Binet formula Lemma[1, (c)] with \(|e_n|<\frac{1}{2}\) and \(g_k(\alpha)>0.276\) in equation ?? we obtain
\[g_k(\alpha)\alpha^{n} + e_n = \bigl(g_k(\alpha)\alpha^{m} + e_m\bigr)10^{d} + g_k(\alpha)\alpha^{p} + e_p.\] Rearranging gives \[\label{eq:10} 1 - \frac{10^{d}}{\alpha^{n-m}} = \frac{e_m10^{d} + e_p - e_n}{g_k(\alpha)\alpha^{n}}.\tag{10}\]
Taking absolute values and using \(|e_n|,|e_m|,|e_p|<\frac{1}{2}\), we obtain \[\left|1 - \frac{10^{d}}{\alpha^{n-m}}\right| = \frac{|e_m10^{d} + e_p - e_n|}{g_k(\alpha)\alpha^{n}} \le \frac{0.5\cdot10^{d} + 1}{g_k(\alpha)\alpha^{n}}.\]
From \(d \le \log_{10} P_p + 1\) and \(P_p \le g_k(\alpha)\alpha^{p} + 0.5 < \alpha^{p}\) (for \(p\) large enough), we get \(10^{d} < 10\alpha^{p}\). Hence \[0.5\cdot10^{d} + 1 < 5\alpha^{p} + 1 < 6\alpha^{p} \quad\text{for } \alpha^{p}\ge 1.\] Thus \[\left|1 - \frac{10^{d}}{\alpha^{n-m}}\right| < \frac{6\alpha^{p}}{g_k(\alpha)\alpha^{n}} = \frac{6}{g_k(\alpha)\alpha^{n-p}}.\] Since \(g_k(\alpha) > 0.276\), we have \(\frac{6}{g_k(\alpha)} < 21.74\). So, we have that \[\label{eq:11} \left|1 - \frac{10^{d}}{\alpha^{n-m}}\right| < \frac{22}{\alpha^{n-p}}.\tag{11}\]
Set \(U_1 = \frac{10^{d}}{\alpha^{n-m}}-1\) and
\[\Lambda_1 = d\log10 - (n-m)\log\alpha\].
If \(\Lambda_1 = 0\), then \(10^{d} = \alpha^{n-m}\). The left‑hand side is an integer, while the right‑hand side is an algebraic number of degree \(k\ge 3\). The only rational power of \(\alpha\) is \(1\) (for exponent \(0\)), so we must have \(n=m\), contradicting \(n>m\). Hence \(\Lambda_1 \neq 0\).
For \(n-p \ge 10\) we have \(|U_1|<\frac{1}{2}\), hence \[\label{eq:11bis} |\Lambda_1| \le 2|U_1| < \frac{44}{\alpha^{n-p}}.\tag{12}\]
We apply Matveev’s theorem (Theorem 2) with \(t=2\), \(\gamma_1=\alpha\), \(\gamma_2=10\), \(b_1=-(n-m)\), \(b_2=d\). The number field is \(\mathbb{K}=\mathbb{Q}(\alpha)\) of degree \(D=k\). We take \[A_1 = \max\{D h(\alpha), |\log\alpha|, 0.16\} = \log\alpha,\] \[A_2 = \max\{D h(10), |\log10|, 0.16\} = k\log10.\] Set \(B = n-m\). Matveev gives \[\log|U_1| > -1.4\cdot30^{5}\cdot2^{4.5}\cdot k^2(1+\log k)(1+\log B)\cdot A_1 A_2.\]
Thus \[\log|U_1| > -7.70\times10^{8}\,k^2(1+\log k)(1+\log B)\cdot (\log\alpha)(k\log10).\] Since \(\log\alpha\log10 \approx 0.693\times2.3026 = 1.595\), we obtain \[\label{eq:12} \log|U_1| > -1.228\times10^{9}\,k^3(1+\log k)(1+\log B).\tag{13}\]
From 11 we have \(\log|U_1| < \log22 - (n-p)\log\alpha\). Comparing 13 with this upper bound gives \[(n-p)\log\alpha < 1.228\times10^{9}\,k^3(1+\log k)(1+\log B) + \log22.\] Neglecting \(\log22\) and using \(\log B \le \log n\), we obtain \[\label{eq:13} n-p < \frac{1.228\times10^{9}}{\log\alpha}\,k^3\log k\log n < 3.55\times10^{9}\,k^3\log k\log n.\tag{14}\]
Rewrite equation ?? as \[g_k(\alpha)\alpha^{n} - g_k(\alpha)\alpha^{p} - P_m^{(k)}10^{d} = e_p - e_n.\] Factoring \(g_k(\alpha)\alpha^{n}\) gives \[g_k(\alpha)\alpha^{n}\bigl(1-\alpha^{p-n}\bigr) - P_m^{(k)}10^{d} = e_p - e_n.\] Hence \[1 - \frac{P_m^{(k)}10^{d}}{g_k(\alpha)\alpha^{n}(1-\alpha^{p-n})} = \frac{e_p - e_n}{g_k(\alpha)\alpha^{n}(1-\alpha^{p-n})}.\] Taking absolute values and using \(|e_n|,|e_p|<\frac{1}{2}\), \(|1-\alpha^{p-n}| > \frac{1}{2}\) (since \(n-p\ge 1\)), and \(g_k(\alpha) > 0.276\), we obtain \[\label{mtigfzck} \left|1 - \frac{P_m^{(k)}10^{d}}{g_k(\alpha)\alpha^{n}(1-\alpha^{p-n})}\right| < \frac{6}{\alpha^{n}}.\tag{15}\]
Set \(U_2 = \frac{P_m^{(k)}10^{d}}{g_k(\alpha)\alpha^{n}(1-\alpha^{p-n})} - 1\) and \[\Lambda_2 = d\log10 - n\log\alpha + \log\frac{P_m^{(k)}}{g_k(\alpha)(1-\alpha^{p-n})}.\] If \(\Lambda_2 = 0\), then \[\frac{P_m^{(k)}10^{d}}{g_k(\alpha)(1-\alpha^{p-n})} = \alpha^{n}.\] Applying any Galois embedding \(\sigma_i\) (with \(\alpha\mapsto\alpha_i\), \(i\ge2\)) and taking absolute values gives \[\frac{P_m^{(k)}10^{d}}{|g_k(\alpha_i)|\cdot|1-\alpha_i^{p-n}|} = |\alpha_i|^{n} \le 1.\] Since \(|g_k(\alpha_i)|<1\) and \(|1-\alpha_i^{p-n}|<2\), the left‑hand side is at least \(\frac{P_m^{(k)}10^{d}}{2}\). For \(P_m^{(k)}\ge 2\) and \(d\ge1\), this is \(>1\), contradiction. Hence \(\Lambda_2 \neq 0\).
For \(n\ge 3\) we have \(|U_2|<\frac{1}{2}\), so \[\label{eq:13bis} \Lambda_2| \le 2|U_2| < \frac{12}{\alpha^{n}}.\tag{16}\]
We can now apply Matveev’s theorem again. Now \(t=3\). The additional algebraic number is \[\gamma_3 = \frac{P_m^{(k)}}{g_k(\alpha)(1-\alpha^{p-n})}.\] We need to bound the logarithmic height of \(\gamma_3\). Using \(h(g_k(\alpha))<4\log k\) and \(P_m^{(k)} < \alpha^{m-1}\), we have \[h(\gamma_3) \le h(P_m^{(k)}) + h(g_k(\alpha)) + h(1-\alpha^{p-n}) \le (m-1)\log\alpha + 4\log k + (n-p)\log\alpha + \log2.\] From 9 , \(m \le n-p+3\), so \[h(\gamma_3) \le (2n-2p+2)\log\alpha + 4\log k + \log2.\] Thus we take \[A_3 = 2k\bigl((2n-2p+2)\log\alpha + 4\log k + \log2\bigr).\] The other parameters are \(A_1 = k\log10\), \(A_2 = \log\alpha\), and \(B = n\).
Matveev gives \[\log|U_2| > -1.4\cdot30^{6}\cdot3^{4.5}\cdot k^2(1+\log k)(1+\log n)\cdot A_1 A_2 A_3.\]
Thus \[\log|U_2| > -1.432\times10^{11}\,k^2(1+\log k)(1+\log n)\cdot (k\log10)\cdot(\log\alpha)\cdot 2k\bigl((2n-2p+3)\log\alpha + 4\log k + \log2\bigr).\] The dominant term is \(2k\cdot (2n-2p)\log\alpha\). Therefore \[\log|U_2| > -1.432\times10^{11}\times2\log10\log\alpha \cdot k^4 (n-p)(1+\log k)(1+\log n).\] With \(\log10\log\alpha \approx 1.595\), we obtain \[\label{eq:14} \log|U_2| > -4.57\times10^{11}\,k^4(n-p)\log k\log n.\tag{17}\]
From 14 we have \(\log|U_2| < \log6 - n\log\alpha\). Hence \[n\log\alpha < 4.57\times10^{11}\,k^4(n-p)\log k\log n + \log6.\] Neglecting \(\log6\), we get \[\label{eq:15} n < 6.6\times10^{11}\,k^4(n-p)\log k\log n.\tag{18}\]
Inserting the upper bound on \(n-p\) obtained in 14 into 18 we have
\[n < 6.6\times10^{11}\,k^4\cdot 3.55\times10^{9}k^3\log k\log n\cdot \log k\log n < 2.34\times10^{21}\,k^7(\log k)^2(\log n)^2.\] Applying the Lemma 3 with \(e=2\) yields \[n < 4.6\times10^{23}\,k^7(\log k)^4.\] Thus we have a polynomial bound on \(n\) in terms of \(k\) for all solutions. We state it as a Lemma:
Lemma 10. If \((k,n,m,p)\) is a solution of equation ?? with \(k\geq 3\), \(n,m\geq 200\), then \[n < 4.6\times10^{23}\,k^7(\log k)^4.\]
We now assume \(2\le k \le 500\). Our goal is to reduce the large bound on \(n\) obtained in Lemma 10 to a small constant using continued fractions and the Baker–Davenport reduction.
From 11 we have \[\left|1 - \frac{10^{d}}{\alpha^{n-m}}\right| < \frac{22}{\alpha^{n-p}}.\] Let \(\tau = \frac{\log\alpha}{\log 10}\). Dividing by \(\log 10\) gives \[\label{eq:16} \left| \tau - \frac{d}{n-m} \right| < \frac{22}{(n-m)\,\alpha^{n-p}\,\log 10}.\tag{19}\]
Let \(q = n-m\). Let \([a_0;a_1,a_2,\dots]\) be the continued fraction expansion of \(\tau\). For the given \(q\), there exists an index \(i\) such that \(q_i \le q < q_{i+1}\). Define \(a_{\max} = \max\{a_0, a_1, \dots, a_i\}\). A classical result (see [16, Theorem 1.1.(iv)]) states that for any integer \(q\), \[\label{eq:17} \frac{1}{(a_{\max}+2)q^2} \le \left| \tau - \frac{d}{q} \right|.\tag{20}\]
Combining 19 and 20 yields \[\frac{1}{(a_{\max}+2)q^2} < \frac{22}{q\,\alpha^{n-p}\,\log 10}.\] Multiplying by \(q\) which is positive, gives \[\frac{1}{(a_{\max}+2)q} < \frac{22}{\alpha^{n-p}\,\log 10},\] hence \[\label{eq:18} \alpha^{n-p} < \frac{22 (a_{\max}+2) q}{\log 10}.\tag{21}\]
Taking natural logarithms, \[\label{eq:19} n-p < \frac{\log(22 (a_{\max}+2) q) - \log(\log 10)}{\log \alpha}.\tag{22}\]
For \(k \le 500\), numerical computation with Sagemath [18] shows that \(a_{\max}\) is bounded ( \(a_{\max} \le 1000\)). Using the initial bound \(n < 10^{45}\) (from Lemma 10), we have \(q \le n \le 10^{45}\). Then the right‑hand side of 22 is at most 200. Thus we obtain the uniform bound \[\label{eq:20} n-p \le 200 \qquad \text{for all } k \le 500.\tag{23}\]
We now turn to the second linear form. From 13 we have \[\left|1 - \frac{P_m^{(k)}10^{d}}{g_k(\alpha)\alpha^{n-1}(1-\alpha^{p-n})}\right| < \frac{6}{\alpha^{n-1}}.\] Set \[\Lambda_2 = d\log10 - (n-1)\log\alpha + \log\frac{P_m^{(k)}}{g_k(\alpha)(1-\alpha^{p-n})}.\] For \(n\ge 3\) we have \(|\Lambda_2| < \frac{12}{\alpha^{n-1}}\). Dividing by \(\log\alpha\) gives \[\label{uvmdhrki} 0 < \left| d\frac{\log 10}{\log\alpha} - (n-1) + \frac{\log\frac{P_m^{(k)}}{g_k(\alpha)(1-\alpha^{p-n})}}{\log\alpha} \right| < \frac{12}{\alpha^{n-1}\log\alpha}.\tag{24}\]
Define \[\tau_k = \frac{\log 10}{\log\alpha}, \qquad \mu_{k,m,p} = \frac{\log\frac{P_m^{(k)}}{g_k(\alpha)(1-\alpha^{p-n})}}{\log\alpha}.\] Then 23 becomes \[\label{eq:21} 0 < | d\,\tau_k - (n-1) + \mu_{k,m,p} | < \frac{16}{\alpha^{n-1}\log\alpha}.\tag{25}\]
We now apply the Baker–Davenport reduction as formulated by Dujella and Pethő (Lemma 5). For each fixed \(k\) and for each pair \((m, n-p)\) with \(1 \le m, n-p \le 200\) (since we already have \(n-p \le 200\) and \(m \le n \le 200\)), we compute the continued fraction expansion of \(\tau_k\). Let \(q_i\) be the denominators of its convergents. Choose a convergent such that \(q_i > 6M\), where \(M\) is an upper bound for \(d\) (e.g. \(M = n \le 200\)). Compute \[\epsilon = \| \mu_{k,m,p}\, q_i \| - M \| \tau_k q_i \|.\] If \(\epsilon \le 0\), we replace \(q_i\) by the next convergent \(q_{i+1}\) and repeat until \(\epsilon > 0\).
Once a convergent with \(\epsilon > 0\) is found, Lemma 5 gives an explicit bound: \[n-1 < \frac{\log( C_3 \, q_i / \epsilon )}{\log 2},\] where \(C_3 = \frac{16}{\log\alpha}\).
We carry out this procedure for each \(k = 2, 3, 4, \dots, 500\). The computation is performed with SageMath [18], using high precision (1000 digits). The results are as follows:
\[\begin{array}{c|c} k & \text{upper bound for } n \\ \hline 2 & 135 \\ 3 & 135 \\ 10 & 135 \\ 50 & 150 \\ 100 & 164 \\ 200 & 164 \\ 300 & 164 \\ 400 & 167 \\ 500 & 170 \end{array}\]
Thus for all \(2\le k\le 500\) we obtain \(n \le 170\).
With \(n \le 170\), we perform a direct computer search over \[2 \le k \le 500,\quad k+2 \le n \le 170,\quad 0 \le m,p < n,\] checking the original equation \(P_n^{(k)} = P_m^{(k)} 10^{d} + P_p^{(k)}\) (where \(d\) is the number of digits of \(P_p^{(k)}\)). The search reveals no solution for \(k \ge 3\). The only solutions come from the classical Pell case \(k=2\) where \(P_4 = 12 = 1\|2\).
We now assume \(k > 500\) and \(n \ge k+2\). Our goal is to show that the Diophantine equation ?? has no solutions in this range.
For \(k \ge 3\) and \(n < 2^{k/2}\), the Cooper–Howard expansion (Lemma 4) gives \[P_n^{(k)} = 2^{\,n-1} + \sum_{j=1}^{\ell-1} C_{n,j} \, 2^{\,n-(k+1)j-1},\] where the coefficients satisfy \(|C_{n,j}| \le \frac{2^{nj}}{j!}\). From this we obtain the estimate \[\label{klcmriov} P_n^{(k)} = 2^{\,n-1}\bigl(1 + \varepsilon_n\bigr),\qquad |\varepsilon_n| < \frac{2}{2^{k/2}}.\tag{26}\]
The same holds for \(P_m^{(k)}\) and \(P_p^{(k)}\).
Substituting equation 25 into the concatenation equation ?? yields \[\label{eq:22} 2^{\,n-1}(1+\varepsilon_n) = 2^{\,m-1}(1+\varepsilon_m)10^{d} + 2^{\,p-1}(1+\varepsilon_p).\tag{27}\]
We rewrite 27 as \[2^{\,n-1} - 10^{d}2^{\,m-1} = 10^{d}2^{\,m-1}\varepsilon_m + 2^{\,p-1}(1+\varepsilon_p) - 2^{\,n-1}\varepsilon_n.\] Dividing by \(2^{\,n-1}\) we obtain \[\label{eq:23} 1 - \frac{10^{d}}{2^{\,n-m}} = \frac{10^{d}2^{\,m-1}}{2^{\,n-1}}\varepsilon_m + \frac{2^{\,p-1}(1+\varepsilon_p)}{2^{\,n-1}} - \varepsilon_n.\tag{28}\]
Taking absolute values and using \(|\varepsilon_n|,|\varepsilon_m|,|\varepsilon_p| < 2/2^{k/2}\) gives \[\left|1 - \frac{10^{d}}{2^{\,n-m}}\right| \le \frac{10^{d}2^{\,m-1}}{2^{\,n-1}}\cdot\frac{2}{2^{k/2}} + \frac{2^{\,p-1}(1+2/2^{k/2})}{2^{\,n-1}} + \frac{2}{2^{k/2}}.\]
From \(10^{d} \le 10 P_p^{(k)} \le 10\cdot 2^{\,p-1}(1+\varepsilon_p) \le 20\cdot 2^{\,p-1}\) (since \(1+\varepsilon_p < 2\)), we have \[\frac{10^{d}2^{\,m-1}}{2^{\,n-1}} \le \frac{20\cdot 2^{\,p-1}\cdot 2^{\,m-1}}{2^{\,n-1}} = 20\cdot 2^{\,m+p-n-1}.\] Moreover, \(2^{\,p-1}/2^{\,n-1} = 2^{\,p-n}\). Hence \[\left|1 - \frac{10^{d}}{2^{\,n-m}}\right| \le 20\cdot 2^{\,m+p-n-1}\cdot\frac{2}{2^{k/2}} + 2^{\,p-n}\cdot 2 + \frac{2}{2^{k/2}}.\]
Using the relation \(n < m + p + 6\) from 9 , we get \(m+p-n-1 \ge -6\). Thus \(2^{\,m+p-n-1} \le 2^{\, -6} = 1/64\). Therefore \[20\cdot 2^{\,m+p-n-1} \le \frac{20}{64} = 0.3125.\]
Also \(2^{\,p-n} = 2^{-(n-p)}\). Hence \[\left|1 - \frac{10^{d}}{2^{\,n-m}}\right| \le 0.3125\cdot\frac{2}{2^{k/2}} + 2^{-(n-p)+1} + \frac{2}{2^{k/2}}.\]
Let \(\lambda = \min\{k/2-8,\; n-p-2\}\). Then we have \[\label{eq:24} \left|1 - \frac{10^{d}}{2^{\,n-m}}\right| < \frac{1}{2^{\lambda}}.\tag{29}\]
Indeed, if \(k/2-8 \le n-p-2\) then \(\lambda = k/2-8\) and the dominant term is \(0.3125\cdot 2/2^{k/2} + 2/2^{k/2} \le 2.3125/2^{k/2} < 1/2^{k/2-8}\) for \(k\) large. If \(n-p-2 \le k/2-8\) then \(\lambda = n-p-2\) and the dominant term is \(2^{-(n-p)+1} = 2^{-\lambda-1}\), which is less than \(1/2^{\lambda}\) for \(\lambda \ge 1\).
Thus 29 holds with \(\lambda = \min\{k/2-8,\; n-p-2\}\).
Set \[U_3 = \frac{10^{d}}{2^{\,n-m}} - 1\] and \(\Lambda_3 = d\log10 - (n-m)\log2\). For \(\lambda \ge 2\) we have \(|U_3| < 1/2\), hence \[|\Lambda_3| \le 2|U_3| < \frac{2}{2^{\lambda}}.\] One can easily show that \(\Lambda_3\neq 0\) using the same argument for \(\Lambda_1.\) We apply Matveev’s theorem with \(t=2\), \(\gamma_1=10\), \(\gamma_2=2\), \(b_1=d\), \(b_2=-(n-m)\). The field is \(\mathbb{Q}\) (\(D=1\)). We take \(A_1 = \log10\), \(A_2 = \log2\), \(B = n\). Matveev gives \[\log|U_3| > -1.4\cdot30^{5}\cdot2^{4.5}\cdot(1+\log n)\cdot\log10\cdot\log2.\] Thus \[\label{eq:25} \log|U_3| > -1.229\times10^{9}\,(1+\log n).\tag{30}\]
From 29 we have \(\log|U_3| < \log 2 - \lambda\log 2\). Comparing with 30 yields \[-\lambda\log 2 > -1.229\times10^{9}\,(1+\log n),\] so \[\label{eq:26} \lambda < \frac{1.229\times10^{9}}{\log 2}\,(1+\log n) < 1.773\times10^{9}\,(1+\log n).\tag{31}\]
Now use the polynomial bound \(n < 4.6\times 10^{23} k^7 (\log k)^4\) (Lemma 10) to replace \(\log n\): \[\log n < \log C_0 + 7\log k + 4\log(\log k) < \log C_0 + 11\log k \quad (k\ge 3).\] With \(C_0=4.6\times 10^{23}\) and \(\log C_0 \approx 61.5\), we obtain \[\lambda < 1.773\times10^{9}\,(62.5 + 11\log k) < 1.11\times10^{11} + 1.95\times10^{10}\log k.\] For simplicity we write \[\label{eq:27} \lambda < 2\times10^{10}\log k.\tag{32}\] We distinguish two different cases:
In this case, 32 becomes \[\frac{k}{2} - 8 < 2\times10^{10}\log k.\] So \[k<1.15\cdot{10}^{12}.\]
In this case we have \(\lambda = n-p-2\) and from 32 we know \[\lambda < 2\times10^{10}\log k,\] hence \[\label{eq:28} n-p < 2\times10^{10}\log k + 2.\tag{33}\]
We now use the second linear form which led to the inequality \[\label{eq:29} \left|1 - \frac{10^{d}}{2^{\,n-m}(1-2^{\,p-n})}\right| < \frac{10}{2^{k/2}}.\tag{34}\]
We set \(U_4 = \frac{10^{d}}{2^{\,n-m}(1-2^{\,p-n})} - 1\) and \[\Lambda_4 = d\log10 - (n-m)\log2 - \log(1-2^{p-n}).\] For \(k\) large enough we have \(|U_4| < 1/2\), thus \[\label{eq:30} |\Lambda_4| < \frac{20}{2^{k/2}}.\tag{35}\]
We have that \(\Lambda_4\neq 0\) using the same argument with \(\Lambda_2.\) We apply Matveev’s theorem with \(t=3\), \(\gamma_1=10\), \(\gamma_2=2\), \(\gamma_3=1-2^{p-n}\). The field is \(\mathbb{Q}\) (\(D=1\)). We take \[A_1 = \log10,\quad A_2 = \log2,\quad A_3 = h(1-2^{p-n}) = (n-p)\log2.\] We have for large \(k\): \[A_3 < 2\times10^{10}\log2 \cdot \log k.\]
Let \(B = n\). By applying matveev, we have that \[\log|U_4| > -4.295\times10^{12}\,(1+\log n)\,A_1 A_2 A_3.\]
Using \(\log n < 10\log k\) (for \(k\) large) and \(A_1A_2 \approx 1.106\), we obtain \[\label{eq:31} \log|U_4| > -1.90\times10^{24}\,(\log k)^2.\tag{36}\]
From 34 we have the upper bound \[\log|U_4| < \log10 - \frac{k}{2}\log2.\]
For large \(k\) the term \(-\frac{k}{2}\log2\) dominates, so \[-\frac{k}{2}\log2 > -1.90\times10^{24}\,(\log k)^2,\] hence \[\label{eq:32} k < \frac{2}{\log2}\times 1.90\times10^{24}\,(\log k)^2 \approx 5.48\times10^{24}\,(\log k)^2.\tag{37}\]
Solving this inequality numerically gives \[k < 10^{31}.\] Consequently, we obtain from Lemma 10 that \[n<10^{252}.\]
We now proceed to reduce the above bounds obtained on \(k\) and \(n\). We now apply the continued fraction method to the inequality \[\label{eq:33} \left| \frac{\log2}{\log10} - \frac{d}{n-m} \right| < \frac{2}{(n-m)\cdot 2^{\min\{k/2-8,\; n-p-2\}}\cdot\log10}.\tag{38}\]
Let \(\tau = \frac{\log2}{\log10}\) and let \(p_i/q_i\) be its convergents. The denominators \(q_i\) grow exponentially; it is known that \(q_i \approx \phi^{i}\) where \(\phi = (1+\sqrt5)/2\). From the bound \(n-m < 10^{252}\) (obtained from Lemma 10), we need \(q_i > 6(n-m) > 6\times10^{252}\). Solving \(\phi^{i} > 6\times10^{252}\) gives \(i > 1200\). Thus we consider the convergent with index \(i = 1200\); its denominator satisfies \(q_{1200} > 10^{252}\) and the maximum partial quotient \(a_{\max}\) up to this index is bounded by \(10^4\) (a known fact from the continued fraction expansion of \(\tau\)).
Using the standard lower bound for continued fractions, \[\frac{1}{(a_{\max}+2)q^2} \le \left| \tau - \frac{p}{q} \right|,\] and comparing with 38 we obtain \[\frac{1}{(a_{\max}+2)(n-m)^2} < \frac{2}{(n-m)\cdot 2^{\min\{k/2-8,\; n-p-2\}}\cdot\log10}.\] Multiplying by \((n-m)\) gives \[\frac{1}{(a_{\max}+2)(n-m)} < \frac{2}{2^{\min\{k/2-8,\; n-p-2\}}\cdot\log10},\] hence \[2^{\min\{k/2-8,\; n-p-2\}} < \frac{2(a_{\max}+2)(n-m)}{\log10}.\]
With \(n-m < 10^{252}\) and \(a_{\max} \le 10^4\), we get \[2^{\min\{k/2-8,\; n-p-2\}} < \frac{2\cdot(10^4+2)\cdot10^{252}}{\log10} < 3\times10^{256}.\] Taking \(\log_2\), \[\min\{k/2-8,\; n-p-2\} < \log_2(3\times10^{256}) = \log_2 3 + 256\log_2 10 \approx 1.58 + 256\cdot3.3219 \approx 852.\]
Thus \(\min\{k/2-8,\; n-p-2\} < 852\). We now distinguish two cases.
If \(\min\{k/2-8,\; n-p-2\} = k/2-8\), then \(k/2-8 < 852\), so \(k < 1720\).
If \(\min\{k/2-8,\; n-p-2\} = n-p-2\), then \(n-p-2 < 852\), so \(n-p < 854\).
These bounds are explicit and finite. They will be used in the next step to further reduce the variables.
Applying the bound obtained on \(k\) to the inequality in Lemma 10), we get \(n < 2\times10^{60}\). We revisit 34 and recall that \[V_4 := \frac{10^{d}}{2^{\,n-m}(1-2^{\,p-n})} - 1.\] Since we have already shown that \(V_4 \neq 0\), it follows that \(\Lambda_4 \neq 0\). Moreover, because \(k > 500\), we obtain the inequality \(|e^{|\Lambda_4|} - 1| = |V_4| < 0.5\). Hence, we arrive at \[\label{eq:34} | (n-p)\log2 - d\log10 + (n-m)\log2 | < \frac{200}{2^{k/2}}, \qquad\tag{39}\]
We now apply the LLL algorithm to the linear form 39 with coefficients bounded by \(n < 10^{252}\). We construct the lattice \[A_2 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ \lfloor C\log2\rfloor & \lfloor C\log(1/10)\rfloor & \lfloor C\log2\rfloor \end{pmatrix},\] with \(C = 10^{760}\). After LLL reduction we obtain the parameters \[\delta = 3.3\times10^{252},\quad S = 5.12\times10^{504},\quad T = 2.4\times10^{252}.\] Using Lemma 4 with \(c_3 = 200\) and \(c_4 = \log2\), we get \[\frac{k}{2} < \frac{\log\bigl(200 q / \epsilon\bigr)}{\log2} < 1538,\] hence \(k < 3076\). From the polynomial bound Lemma 10, this gives \(n < 2.0\times10^{56}\).
Now we apply the continued fraction method again with this new bound \(n < 2.0\times10^{56}\). We find that \[\min\{k/2-8,\; n-p-2\} < 194.\] If \(\min = k/2-8\), then \(k < 404\), contradicting \(k > 500\). If \(\min = n-p-2\), then \(n-p < 196\).
After applying the LLL algorithm, we obtain the bound \(n < 10^{60}\). We then take \(M = 2 \times 10^{60}\) as an upper bound for \(n\) (and hence for \(d\)). From 38 we have \[\label{eq:35} \left| d\frac{\log10}{\log2} - (n-m) - \frac{1-2^{p-n}}{\log2} \right| < \frac{200}{2^{k/2}}.\tag{40}\]
We apply Lemma 5 with \[\tau = \frac{\log10}{\log2}, \qquad \mu_{n-p} = \frac{1-2^{p-n}}{\log2}, \qquad A = \frac{200}{\log2}, \qquad B = 2.\] The continued fraction expansion of \(\tau\) is \[\tau = [3;3,9,2,2,4,6,2,2,\dots].\] The convergent \(p_{124}/q_{124}\) given by \[\frac{p_{124}}{q_{124}} = \frac{59709183646229903017509733728189314625139620328694917340547}{17974255294124444596871803224395333592038752850416569230287}\] satisfies \(q_{124} \approx 1.8\times10^{61} > 6M\). For all admissible values of \(n-p\), a computer calculation with SageMath [18] gives \[\epsilon = \| \mu_{n-p} q_{124} \| - M \| \tau q_{124} \| > 0.49693.\] Hence Lemma 5 yields \[\frac{k}{2} \le \frac{\log\!\left(\frac{200}{\log2} \cdot \frac{q_{124}}{\epsilon}\right)}{\log2} \approx 212.7,\] so \(k < 426\). This contradicts the assumption \(k > 500\). Therefore, there are no solutions for \(k > 500\).