Let $E$ be the set of all even numbers in the sequence and $m$ is the biggest element of $E$. Clearly: $$(*)\,\,\,\,\,\,\,\,\,\, a_{n+1}=a_{n-k}+e \text{ for some } e\in E \text{ and } k\in \{0,1,\dots,m\}. $$We denote $\Delta_n := a_{n+1}-a_n$. By $(*)$ it follows $\Delta_n$ is bounded. Hence, for any fixed $s$, the set of all possible $s$-tupples $D_n:=(\Delta_{n},\Delta_{n+1},\dots,\Delta_{n+s-1}), n\in\mathbb{N},n>m$ is finite. Let $s:= m+1$ and so, there exist two equal $s$-tupples: $D_k=D_{\ell}$. We claim $\Delta_n$ is periodic with period $\ell-k$, starting from $n=k$. The fact $D_k=D_{\ell}$ means $a_{\ell}, a_{\ell+1},\dots, a_{\ell+s}$ is obtained by $a_{k}, a_{k+1},\dots, a_{k+s}$ by shifting it $a_{\ell}-a_k$ units to the right. Suppose $a_{k+s+1}=a_{k+i}+e, i\in\{1,2,\dots,s\}, e\in E$. Let us see that $a_{\ell+s+1}=a_{\ell+i}+e$. Indeed, if we assume $x=a_{\ell+i}+e$ can be represented as $x=a_{\ell+i_1}+e_1$ then $i_1\in \{1,2,\dots,s\}$ and $a_{k+i_1}+e_1=a_{k+i}+e=a_{k+s+1}$, which contradicts with the choice of $a_{k+s+1}$. Similarly, assuming $a_{\ell+s}<a_{\ell+i_1}+e_1<x$ for some $i_1,e_1\in E$ will bring us to a contradiction with the minimal property of $a_{k+s+1}$. Thus, we obtained $a_{\ell+s+1}-a_{\ell+s}= a_{k+s+1}-a_{k+s}$ i.e. $\Delta_{\ell+s}=\Delta_{k+s}$. In the same way we consecutively get $\Delta_{\ell+r}=\Delta_{k+r}$ for any $r>s$.

2012 China TST