Solving daxim's problem

solution.txt

G(0)                               = 1
G(1) = mex(G(0))                   = 0
G(2) = mex(G(0), G(1))             = 2
G(3) = mex(G(0), G(1), G(2))       = 3
G(4) = mex(G(0), G(1), G(2), G(3)) = 4
G(5) = mex(G(1), G(2), G(3), G(4)) = 1
G(6) = mex(G(2), G(3), G(4), G(5)) = 0
G(7) = mex(G(3), G(4), G(5), G(6)) = 2
                                   = 3
                                   = 4

So, G( $n ) := $m < 2 ?? 1 - $m !! $m
               where $m = $n % 5

G(21) = 0 by this rule. Don't move first in that game.

The 5 there is simply $MAX + 1, where $MAX was 4 in this case.
Let's try and generalize for $MIN as well. Here's one case where
$MIN == 2 and $MAX == 5:

G(0)                                = 1
G(1)                                = 1
G(2)                                = 0
G(3) = mex(G(1))                    = 0
G(4) = mex(G(1), G(2))              = 2
G(5) = mex(G(1), G(2), G(3))        = 2
G(6) = mex(G(1), G(2), G(3), G(4))  = 3
G(7) = mex(G(2), G(3), G(4), G(5))  = 1
G(8) = mex(G(3), G(4), G(5), G(6))  = 1
                                    = 0
                                    = 0
                                    = 2
                                    = 2

So, the period is 7, which is $MIN + $MAX. The G($n) numbers occur in pairs,
except the 3, which is cut off by the modulus. The parising is probably due to
$MIN being 2. So i postulate the following formula, which covers all the data so far:

G( $n ) := $d < 2 ?? 1 - $d !! $d
           where $d = floor($m / $MIN)
             and $m = $n % ($MIN + $MAX)

In other words, try to be the next on turn for all games except those where
floor(($n % ($MIN + $MAX)) / $MIN) == 1.