TITLE: Prime one heap Nim
QUESTION [12 upvotes]: I have been working on an interesting problem my lecturer mentioned recently. Prime Nim is a variant of the Nim game where you have a single pile with an arbitrary number $n\in \Bbb N+\{0\}$ of elements and players can take away a prime count of elements every round. Now I want to find a way to decide whether we can ensure victory in a given position (and the winning strategy, of course).
What I did so far:
$0$ and $1$ are clearly lost positions. On the contrary, any prime $n$ and $n+1$ are winning positions. For all other $n$ we can say that if there is no prime $p<n$ such that $n-p \notin \{n'|n'<n \land n' \text{ is lost}\}$, then $n$ is a losing position. The set of losing positions can be more formally expressed recursively based on previous such sets for smaller $n$. (In other words, this is an application of the very basic idea that a position form which we can make no move to a losing position is a losing position). All other positions are winning positions. Very simple and general.
Losing positions can be described as a sequence - http://oeis.org/A025043 (quite interestingly a subsequence of http://oeis.org/A093513). This uses the recursive nature of the problem.
An algorithm starting from the recursion edge case ($0$) and building up the sequence progressively can give us both the answer and the prospective winning strategy generated as a side-product in polynomial time.
My lecturer said the strategy-finding algorithm is probably optimal, but he also suggested there might be a simple formula to decide whether a given position can be won using an optimal strategy.
And I am stuck on it now. I guess the formula can't possibly be that simple, since it has to, in my opinion, include at least primality testing. I tried to determine a simple way to decide whether a number is in the sequence of losing positions, but I had no success so far. Basically, I always encounter the impenetrable problem of generating $n$th prime.
Any ideas on approaching this differently?

REPLY [4 votes]: The lack of additional info with the OEIS sequence may indicate that the strategy is not "simple". However, here is some heuristics:
From the OEIS data, 10000 numbers up to 67673 are losers, that is more that $\frac17$, and practically  all of them are odd (I see only the exceptions $0, 10, 34, 100, 310$). Heuristically, this odd overweight will continue: 
Assume that there are $a$ odd and $b$ even loser numbers $x_i<N$ with $a\gg b$. Then if $N$ is even, it is a loser if none of the $a$ odd numbers $N-x_i$ is prime. This probability might be expected to be something like $$p_{\text{even}}=\left(1-\frac2{\ln N}\right)^a\approx e^{-\frac{2a}{\ln N}}.$$
On the other hand, if $N$ is odd, it is a loser only if $N-2$ is not a loser (which we will ignore) and none of the $b$ odd numbers $N-x_i$ is prime, which should be something like $$p_{\text{odd}}=\left(1-\frac2{\ln N}\right)^b\approx e^{-\frac{2b}{\ln N}}.$$
From the empirical data, it seems that $b$ is of size comparable to $\ln N$, hence $p_{\text{odd}}$ is of "reasonable" size (not really close to $0$), hence $a\approx p_{\text{odd}} N$ of size comparable to $N\gg\ln N$, which makes $p_{\text{even}}$ almost negligibly small.
I guess a similar argument might indicate that "most" loser numbers are not divisible by 3 or in fact any prime $p\ll N$ (maybe rather $p\ll \ln N$?).
But all this is far from giving a simple definite answer to whether a particulyr $N$ is a loser or a winner ...