# On Homogeneous Combinations of Linear Recurrence Sequences
## 2 Auxiliary Results  
Before proceeding further, we shall need some key results. The first tool is a lower bound for a linear form in logarithms _a la [PERSON]_. From the main result of [PERSON] [20], we deduce the following lemma.  
Lemma 1: _Let \(\gamma_{1},\ldots,\gamma_{t}\) be real algebraic numbers and let \(b_{1},\ldots,b_{t}\) be nonzero rational integer numbers. Let \(D\) be the degree of the number field \(\mathbb{Q}(\gamma_{1},\ldots,\gamma_{t})\) over \(\mathbb{Q}\) and let \(A_{j}\) be a real number satisfying_  
\[A_{j}\geq\max\{Dh(\gamma_{j}),|\log\gamma_{j}|,0.16\},\text{ for }j=1,\ldots,t.\]  
_Assume that_\[B\geq\max\{|b_{1}|,\ldots,|b_{t}|\}.\]  
_If \(\gamma_{1}^{b_{1}}\cdots\gamma_{t}^{b_{t}}\
eq 1\), then_  
\[|\gamma_{1}^{b_{1}}\cdots\gamma_{t}^{b_{t}}-1|\geq\exp(-1.4\cdot 30^{t+3}\cdot t^{ 4.5}\cdot D^{2}(1+\log D)(1+\log B)A_{1}\cdots A_{t}).\]  
The _logarithmic height_ of an \(s\)-degree algebraic number \(\gamma\) is defined as  
\[h(\gamma)=\frac{1}{s}(\log|a|+\sum_{j=1}^{s}\log\max\{1,|\gamma^{(j)}|\}),\]  
where \(a\) is the leading coefficient of the minimal polynomial of \(\gamma\) (over \(\mathbb{Z}\)) and \((\gamma^{(j)})_{1\leq j\leq s}\) are the conjugates of \(\gamma\) (over \(\mathbb{Q}\)).  
The next lemmas play an important role in the proof of Theorem 1.  
Lemma 2: _Let \((G_{n})_{n}\) be a linear recurrence whose characteristic polynomial has a simple positive dominant root. Then the dominant polynomial of \((G_{n})_{n}\) is a positive constant._  
Proof: We know that (2) holds, where each \(r_{j}\) has multiplicity \(m_{j}\), and each \(g_{j}(n)\) is a non-zero polynomial with degree \(\leq m_{j}-1\). Suppose that \(r_{1}\) is the dominant root, since it is simple, we have immediately \(m_{1}=1\) and then the degree of \(g_{1}(n)\) is at most \(m_{1}-1=0\), so it is a constant, say \(g_{1}\). Now, dividing \(G_{n}\) by \(r_{1}^{n}\), we get  
\[0<\frac{G_{n}}{r_{1}^{n}}=g_{1}+\sum_{j=2}^{\ell}\frac{g_{j}(n)}{\kappa_{j}^{n }}, \tag{5}\]  
where \(\kappa_{j}=r_{1}/r_{j}\). Since \(|\kappa_{j}|>1\), we have  
\[\lim_{n\to\infty}\frac{g_{j}(n)}{\kappa_{j}^{n}}=0,\,\mbox{for all}\,\,2\leq j \leq\ell,\]  
and so  
\[0\leq\limsup_{n\to\infty}\frac{G_{n}}{r_{1}^{n}}=g_{1}.\]  
Therefore, \(g_{1}>0\) as \(g_{1}\
eq 0\).  
Lemma 3: _Let \((G_{n})_{n}\) be a linear recurrence with a dominant root. For any nonnegative integers \(i\) and \(r\), we have_  
\[\lim_{n\to\infty}\frac{G_{i}^{r}+i}{r_{1}^{n}}=g_{1}^{r}r_{1}^{ir}. \tag{6}\]  
Proof: For any \(i\geq 0\), we use (2) together the multinomial theorem, to obtain  
\[G_{n+i}^{r}=\sum_{\pi\in\mathcal{I}_{r}}\frac{r!}{\alpha_{1}!\cdots\alpha_{t}!}\prod_{j=1}^{t}g_{j}(n+t)^{\alpha_{j}}r_{1}^{\alpha_{1}(n+i)}\cdots r_{t}^{ \alpha_{t}(n+i)},\]  
where \(\mathcal{I}_{r}=\{\pi=(\alpha_{1},\ldots,\alpha_{t})\in\mathbb{Z}^{t}:\alpha_ {i}\geq 0\text{ and }\sum_{i=1}^{t}\alpha_{i}=r\}.\) A straightforward computation shows that  
\[\frac{G_{n+i}^{r}}{r_{1}^{n}}=(g_{1}(n+i))^{s}r_{1}^{ir}+\sum_{\pi\in \mathcal{I}_{r}\setminus\{r_{1}\}}\frac{r!}{\alpha_{1}!\cdots\alpha_{t}!} \prod_{j=1}^{t}g_{j}(n+t)^{\alpha_{j}}\frac{r_{1}^{\alpha_{1}^{1}i}r_{2}^{ \alpha_{2}(n+i)}\cdots r_{t}^{\alpha_{t}(n+i)}}{r_{1}^{(r-\alpha_{1})n}},\]where \(\epsilon_{1}=(1,0,\ldots,0)\). Since \(r-\alpha_{1}=\alpha_{2}+\cdots+\alpha_{\ell}>0\) and \(g_{1}(n+i)=g_{1}\) (by Lemma 2), we can write  
\[\frac{G_{n+i}^{r}}{r_{1}^{n}}=g_{1}^{r}r_{1}^{ir}+\sum_{\pi\in\mathcal{I}_{r} \setminus\{r_{1}\}}\frac{r!}{\alpha_{1}!\cdots\alpha_{\ell}!}\prod_{j=2}^{t} \left(\frac{g_{j}(n+i)}{\kappa_{j}^{n}}\right)^{\alpha_{j}}g_{1}^{\alpha_{1}}r _{1}^{\alpha_{1}i}r_{2}^{\alpha_{2}i}\cdots r_{t}^{\alpha_{\ell}i},\]  
where \(\kappa_{j}=r_{1}/r_{j}\). Since \(|\kappa_{j}|>1\), then \(\lim_{n\to\infty}g_{j}(n+i)/\kappa_{j}^{n}=0\) and so each term in the previous summation tends to zero as \(n\to\infty\), because \(\alpha_{j}>0\), for some \(2\leq j\leq t\). So,  
\[\lim_{n\to\infty}\frac{G_{n+i}^{r}}{r_{1}^{n}}=g_{1}^{r}r_{1}^{ir}.\]  
Now, we are ready to deal with the proof of Theorem 1.