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Rafinha wrote: Let $a_1,a_2, \cdots$ be a sequence of integers that satisfies: $a_1=1$ and $a_{n+1}=a_n+a_{\lfloor \sqrt{n} \rfloor} , \forall n\geq 1 $. Prove that for all positive $k$, there is $m \geq 1$ such that $k \mid a_m$. Suppose on the contrary that $a_n\not\equiv 0\pmod k$ $\forall n$ Let $m_k(x)=x-k\left\lfloor\frac{x}k\right\rfloor$ the function which associates to a positive integer $x$ the remainder of its division by $k$. Let $a=\min_{n\ge k}m_k(a_n)$ with $a\in\{1,2,...,k-1\}$ Let one integer $p\ge k$ such that $m_k(a_p)=a$ Let $b=m_k(a_{p^2})$ with $b\in\{a,a+1,a+2,...,k-1\}$ Note that $a_{p^2+i}=a_{p^2}+ia_p$ $\forall i\in\{0,1,...2p\}$ And so $m_k(a_{p^2+i})=m_k(b+ia)$ $\forall i\in\{0,1,...2p\}$ Let $j=\left\lfloor\frac{k-1-b}a\right\rfloor$ so that $j\le k-2$ and $b+(j+1)a>k-1\ge b+ja$ And so $a-1\ge b+(j+1)a-k\ge 0$ And so $m_k(b+(j+1)a)<a$ And $j+1\le k-1<k<2k\le 2p$ And so $m_k(a_{p^2+j+1})<a$ and contradiction. Hence the claim.