# A question about generalized Fibonacci sequences

Can the following lemma be proved?

Lemma (Rokach-Goldstein)

Let $$x_i$$ be a finite generalized Fibonacci sequence of positive integer numbers, $$x_0,x_1,x_2,\ldots,x_m$$ such that for every $$2\le i\le m$$, we have $$x_i=x_{i-1}+x_{i-2}$$, and where $$x_0$$ and $$x_1$$ are coprime to each other.

In this case, there exists some $$n$$, $$0\le n\le m$$, such that $$x_n$$ is coprime to the sum of all numbers in the sequence.

• What exactly do you have in mind when you say "the sum of all [the] numbers in the sequence"? Commented Nov 14, 2019 at 17:02
• Further, supposing you have a sensible way to define "the sum of all [the] numbers in the sequence" you can take x=1, can't you? Commented Nov 14, 2019 at 17:38
• I fixed the formatting for you, but you should learn some very basic TeX so that you can make your posts more readable. As Jose indicated, it's not clear what you're asking. Maybe you want $x_n$ to be coprime to the sum of the previous terms $x_0+x_1+\cdots+x_{n-1}$. Finally, when you say "can it be proved", you shouldn't call it a lemma with your name attached. Do you know how to prove it? If not, ask it as a question. If so, say so, and then what are you asking? Commented Nov 14, 2019 at 17:39
• We have $x_n=x_1F_n+x_0F_{n-1}$. From there you can derive an explicit formula for $x_0+\dots+x_{n-1}$ and compare it to $x_n$. Commented Nov 14, 2019 at 17:55
• Thanks a lot for the editing. The question referrs to a finite sequence. It seems that there must be at least one number in it thst is co prime to the sum of all numbers in the finite sequence. I dont know how to prove it nut it seems true. Commented Nov 14, 2019 at 19:26

$$x_0=1874587$$, $$x_1=89$$. $$\gcd(x_0,x_1)=1$$. Let $$m=10$$. $$\sum_0^{10}x_i=166850970=2\times3\times5\times11\times13\times19\times23\times89$$.

$$x_0=1874587\equiv0\bmod11$$.

$$x_1=89\equiv0\bmod89$$.

$$x_2=x_1+x_0=1874676\equiv0\bmod2$$.

$$x_3=x_2+x_1=1874765\equiv0\bmod5$$.

$$x_4=x_3+x_2=3749441\equiv0\bmod19$$.

$$x_5=x_4+x_3=5624206\equiv0\bmod2$$.

$$x_6=x_5+x_4=9373647\equiv0\bmod3$$.

$$x_7=x_6+x_5=14997853\equiv0\bmod13$$.

$$x_8=x_7+x_6=24371500\equiv0\bmod2$$.

$$x_9=x_8+x_7=39369353\equiv0\bmod23$$.

$$x_{10}=x_9+x_8=63740853\equiv0\bmod11$$.

It may be worth checking my arithmetic.