# On Homogeneous Combinations of Linear Recurrence Sequences
## 3 The Proof of Theorem 1  
Suppose that the pair \((n,t_{n})\) is solution of Equation (4) for all \(n\) belonging to an infinite set \(\mathcal{N}\). Thus, we have \(P(G_{n+1},\ldots,G_{n+\ell})=G_{t_{n}}\), for all \(n\in\mathcal{N}\). By the formula in (5), we have that \(G_{m}=O(r_{1}^{m})\) (where as usual, \(O\) denotes the 'big-oh' Landau symbol). Thus the equality \(P(G_{n+1},\ldots,G_{n+\ell})=G_{t_{n}}\) yields \(O(r_{1}^{ns})=O(r_{1}^{n})\) and so \(t_{n}=O(n)\). Now, we rewrite \(P(G_{n+1},\ldots,G_{n+\ell})=G_{t_{n}}\) as  
\[\sum_{I\in\mathcal{I}}a_{I}(G_{n+1}^{i_{1}}\cdots(G_{n+\ell-1})^{i_{\ell-1}}+( G_{n+\ell})^{s}=G_{t_{n}},\]  
where the \(a_{I}\)'s are the coefficients of \(P\) and the sum runs over a subset \(\mathcal{I}\subseteq\mathcal{I}_{s}\subseteq\mathbb{Z}^{\ell-1}\). We divide the above equality by \(r_{1}^{sn}\) and by using that \(i_{1}+\cdots+i_{\ell-1}=s\), we arrive at  
\[\sum_{I\in\mathcal{I}}a_{I}\prod_{j=1}^{\ell-1}\frac{(G_{n+j})^{i_{j}}}{r_{1} ^{i_{j}}}+\frac{(G_{n+\ell})^{s}}{r_{1}^{sn}}=\frac{G_{t_{n}}}{r_{1}^{sn}}.\]  
Now, when \(n\) tends to infinity (in \(\mathcal{N}\)), we use Lemma 3 to get  
\[\lim_{ \begin{subarray}{c}n\to\infty\\ n\in\mathcal{N}\end{subarray} }\frac{G_{t_{n}}}{r_{1}^{ns}}=\sum_{I\in \mathcal{I}}g_{1}^{s}r_{1}^{i_{1}+2i_{2}+\cdots+(\ell-1)i_{\ell-1}}+(g_{1})^{ s}r_{1}^{ds}. \tag{7}\]  
On the other hand,  
\[\frac{G_{t_{n}}}{r_{1}^{ns}}=g_{1}r_{1}^{t_{n}-ns}+g_{2}(t_{n})\frac{r_{2}^{t_ {n}}}{r_{1}^{ns}}+\cdots+g_{t}(t_{n})\frac{r_{t}^{t_{n}}}{r_{1}^{ns}}.\]  
Now, we use that any exponential function of \(n\) dominates any polynomial function of \(n\), together with the fact that \(t_{n}=O(n)\) and \(|r_{1}|>|r_{j}|\), for \(j=2,3,\ldots,t\), to conclude that  
\[\lim_{ \begin{subarray}{c}n\to\infty\\ n\in\mathcal{N}\end{subarray} }\frac{g_{j}(t_{n})}{r_{1}^{ns/2}}\cdot\frac{r_{j }^{t_{n}}}{r_{1}^{ns/2}}=0,\,\text{for}\,\,j=2,3,\ldots,t.\]  
So  
\[\lim_{ \begin{subarray}{c}n\to\infty\\ n\in\mathcal{N}\end{subarray} }\frac{G_{t_{n}}}{r_{1}^{ns}}=g_{1}\lim_{ \begin{subarray}{c}n\in\mathcal{N}\end{subarray} }r_{1}^{t_{n}-ns}.\]  
Since \(t_{n}-ns\) is an integer and \(r_{1}>1\), then, by (7), \(t_{n}-ns\) must be constant with respect to \(n\), say \(\
u\), for all \(n\in\mathcal{N}\) sufficiently large. Hence, we have  
\[\sum_{I\in\mathcal{I}}a_{I}g_{1}^{s}r_{1}^{i_{1}+2i_{2}+\cdots+(\ell-1)i_{\ell- 1}}+(g_{1})^{s}  
\[|(g_{1})^{1-s}r_{1}^{\
u-\ell s}-1| \leq \frac{L(P)}{r_{1}^{\ell s-(b_{1}+2b_{2}+\cdots+(\ell-1)b_{\ell-1})}},\]  
where \(L(P)\) denotes the length of \(P\) (which is the sum of the absolute value of its coefficients) and \((b_{1},\ldots,b_{\ell-1})\in\mathcal{I}\) is such that \(b_{1}+2b_{2}+\cdots+(\ell-1)b_{\ell-1}\) is maximal. Now, we will get a lower bound for \(\ell s-(b_{1}+2b_{2}+\cdots+(\ell-1)b_{\ell-1})\). In fact, we see that \(b_{\ell-1}\leq s,b_{\ell-2}+b_{\ell-1}\leq s,\ldots,b_{1}+\cdots+b_{\ell-1}\leq s\) and by summing these inequalities, we arrive at \(b_{1}+2b_{2}+\cdots+(\ell-1)b_{\ell-1}\leq(\ell-1)s\) and so \(\ell s-(b_{1}+2b_{2}+\cdots+(\ell-1)b_{\ell-1})\geq s\). Therefore,  
\[|(g_{1})^{1-s}r_{1}^{\
u-\ell s}-1|\leq\frac{L(P)}{r_{1}^{\ell}}. \tag{9}\]  
In order to apply Lemma 1, we can suppose that \(t\
eq\ell s\) (otherwise, we would have a linear form in one logarithms and the proof would follows in the same way). Then, we choose  
\[\gamma_{1}=g_{1},\ \ \gamma_{2}=r_{1},\ \ b_{1}=1-s,\ \ b_{2}=\
u-\ell s.\]  
Before applying the lemma, let us see that we can suppose that \((g_{1})^{1-s}r_{1}^{\
u-\ell s}\
eq 1\). In fact, otherwise we would get a multiplicative relation between \(g_{1}\) and \(r_{1}\) (which are multiplicatively independent), this would imply in \(s=1\) (and the proof would be done). So, we can suppose that \(|(g_{1})^{1-s}r_{1}^{\
u-\ell s}-1|\
eq 0\). Note that \(D=[\mathbb{Q}(\gamma_{1},\gamma_{2}):\mathbb{Q}]\leq k!\).