Does $P/O(1)$ equal to $P$ if solver needs to consider smaller inputs?

Suppose that $F$ is a problem such that for every $n$, there is a program of length $O(1)$, running in polynomial time to $n$, that solves $F$ correctly on all instances of size less than $n$. Can $F$ be solved in polynomial time? That is, does there exist a single polynomial time algorithm that solves $F$ on all instances?

Suppose that $F$ is a problem such that for every $n$, there is a program of length at most $C$, running in polynomial time, that solves $F$ correctly on all instances of size less than $n$.

Let $P_n$ be a program of length at most $C$, running in polynomial time, that solves $F$ correctly on all instances of size less than $n$. Since there are only finitely many programs of length at most $C$, there must be a single program $P$ that occurs infinitely often as $P_n$. This program solves $F$ in polynomial time for all instances.

If we change $O(1)$ to any bound $f(n) = \omega(1)$ computable in $\mathit{poly}(n)$ time, then we can easily construct undecidable problems which are computable in this model. We can assume that $f$ is monotone and has no jumps (otherwise take $g(n) = \min(\max(f(0),\ldots,f(n)),g(n-1)+1)$). For an undecidable $S \subseteq \mathbb{N}$, consider the following language:
$$
L = \{ x \in \{0,1\}^* : f(|x|) \in S \}.
$$
Since $f$ is monotone, if $|x| \leq n$ then $f(|x|) \leq f(n)$. Hence there is a program of length $f(n) + O(1)$ that solves $L$ correctly on inputs of length at most $n$. Conversely, since $f$ has no jumps, for each $m$ we can find a value $n$ such that $f(n) = m$, and this gives a computable reduction from $S$ to $L$, showing that $L$ is uncomputable.

StackExchange Computer Science Q#118590, answer score: 4