I define a recurrence relation as follow:
$$ω_{n+1}=f(ω_{n})$$ for some function $f$.
Assume that $f:ℝ→ℝ$ is a continuous, strictly increasing (so it is bijective) and a contraction function. Then $(ω_{n})$ converges to the fixed point of $f$, i.e., $ω_{n}$→$x$ with $f(x)=x$ and this is the only type of solutions for the above recurrence equation. Now, assume we know that $ω_{n}=a_{n}$ (with $a_{n+1}=f(a_{n}$)) for some values of $n$ (the two sequences are equal without any law) where $a_{n}$ is a known sequence.
My question is: Can we deduce that $ω_{n}=a_{n}$ for all $n$.
As said in the comments it is not enough, for instance if you take $a_n$ defined as follow : $a_{2n}$ = $\omega_{2n}$ and $ a_{2n+1} = 0$ then your premises will be verified but (unless $(\omega)$ is null after a certain $n$) the two sequences are not equal.
Edit : You said in the comments that $(a_n)$ also verifies the property $a_{n+1} = f(a_n)$, now you can simply start at the smallest $n_0$ for which they are equal and prove by induction that : $\forall n \geq n_0 , a_n = \omega_n$. To prove it for $n < n_0$ you will at least need to assume $f$ is an injection.