I first thought that this is impossible, but the hardest part should be to see that the factor of the geometric series must not be integral Idea: take $b_0=c^n , b_1 = c^{n-1}(c+1) , b_3 = c^{n-2}(c+1)^2 , ... , b_n=(c+1)^n$. Then the sequence generated by $a_0=b_1-1 , a_1=b_2-1$, so with difference $b_2-1-b_1+1=c^{n-2}(c+1)$, will work: Since the differences $b_{i+1}-b_i$ increase with $i$, this works already if $a_{n} > b_{n}$, so $(c+1)c^{n-1}+nc^{n-2}(c+1) > (c+1)^n \iff c^{n-2}(c+n) \geq (c+1)^{n-1}$. By using the binomial theorem to expand the RHS we see that the LHS grows faster with $c$ than the RHS (the expansion is $c^{n-1}+nc^{n-2} > c^{n-1}+(n-1)c^{n-2}+...+1$), so we can take a big $c$ such that this is fulfilled and we get our sequences (with length $n+1$). When using $c=4=n$ we get already our sequence with $5$ terms: $256 < 319 < 320 < 399 < 400 < 479 < 500 < 559 < 625 < 639$