I'm looking for a slow growing integer sequence that is periodic modulo primes.
Essentially, the absolute value of each digit should be small.
EFFECTIVENESS
I've devised a measure of how effective the sequence is. We take the $n$th value, and find for $f(n)$, the $n$th number:
$$f(n) = 2^{mn}$$
We measure the period of the sequence modulo $p$. For example, the period modulo $p$ could be $p^2$. We'll call this value, whatever it is, $f(p)$, which in the example is $f(p)=p^2$.
The measure is then:
$$f(p)^m$$
Ideally, this should be less than $p$.
AN EXAMPLE
Consider the Fibonacci numbers. They are given by the exact formula, as given in Wikipedia:
$$\text{Fibonacci}(n) = \frac{\varphi^n-\psi^n}{\varphi-\psi}$$
where $\varphi \approx 1.61$ and $\psi \approx -0.62$. Thus this function is bounded above by $O(1.61^n) = O(2^{\log_2{(1.61)}}) = O(2^{0.68n})$
We also know that the Fibonacci numbers are periodic modulo each prime $p$, and depend on the previous 2 values. Thus there are at most $p^2$ combinations of previous values modulo $p$, and so the period is at most $p^2$ modulo $p$.
Thus, the effectiveness measure gives $m=0.68$ since $f(n) = 2^{0.68n}$. It gives $f(p) = p^2$ as given above. Combined, this gives $f(p)^m = (p^2)^{0.68} \approx p$.
WHAT I'M LOOKING FOR
Something that gives an effectiveness measure < $p$. For example, a measure of $p/\log{(p)}$ would tie the best known result in the problem I'm working on.
The function $f(n)=n$ has period $p$ modulo every prime $p$, and grows so slowly that it should give you a suitably small effectiveness measure.