Let $$a_{1}=1,a_{n+1}-a_{n}=a_{\lfloor\frac{n+1}{2}\rfloor}$$ if $a_{n}=\left(\overline{b_{1}b_{2}\cdots b_{n}}\right)_{10}$ show that the sequence $\{b_{n}\}$ can't periodic sequence
A sketch of my thoughts:
$a_{1},$so $ b_{1}=1$
$a_{2}-a_{1}=a_{1},a_{2}=2,$ so $b_{2}=2$
$a_{3}=a_{2}+a_{1}=3$ so $b_{3}=3$
$a_{4}=a_{3}+a_{2}=3+2=5$,so $b_{4}=5$
$a_{5}=a_{4}+a_{2}=7,b_{5}=7$
$a_{6}=a_{5}+a_{3}=7+3=10$ so $b_{6}=0$
$a_{7}=a_{6}+a_{3}=10+3=13$ so $b_{7}=3$
$a_{8}=a_{7}+a_{4}=13+5=18$,so $b_{8}=8$
$a_{9}=a_{8}+a_{4}=18+5=23$,so $b_{9}=3$
$a_{10}=a_{9}+a_{5}=23+7=30$,so $b_{10}=0$
$a_{11}=a_{10}+a_{5}=30+7=37$,so $b_{11}=7$ $$\cdots\cdots$$
Assume $a_n\bmod 10$ is (eventually) periodic with period $p$. Then so is $a_{n+1}-a_n\bmod 10$, i.e., $a_{\lfloor(n+1)/2\rfloor}\bmod 10$. If $p$ is even, this implies that $a_n\bmod 10$ is eventually periodic with period $\frac p2$. Therefore, if the sequence is eventually periodic then its minimal periof $p$ is odd. This odd $p$ is also a period of $a_{n+2}-2a_{n+1}+a_{n}\bmod 10 = a_{\lfloor (n+2)/2\rfloor}-a_{\lfloor (n+1)/2\rfloor}\bmod 10$. As this latter sequence is $0$ for all odd $n$, we conclude that it is zero for almost all $n$. This means that $a_{n+1}-a_n\bmod 10$ is eventually constant, and hence $a_{\lfloor(n+1)/2\rfloor}\bmod 10$ is eventually zero. But in order to have $a_n\equiv 0\pmod{10}$ for all $n\ge N$, we need $a_n\equiv 0\pmod{10}$ for all $n\ge \lfloor \frac{N+1}2\rfloor$. Hence the minimal such $N$ has the property $N\le \lfloor \frac{N+1}2\rfloor$, or $N\le 1$. We verify that $a_1\not\equiv 0\pmod{10}$, hence the original assumption of eventual periodicity is false.