Correct solutions were submitted by
Dan Dima, Dennis Muniz, Juan Carlos Marivela and Ron Welch.
Set f(x) := (x+1) p(x) - x.
Then f(x) = 0 for x = 0, 1, 2, ..., n, so there is a number c such that
f(x) = c x (x-1) (x-2) ... (x-n).
By evaluating f(-1) two ways, we find c = (-1)n+1 / (n+1)!.
Notice that f(n+1) = (-1)n+1.
Since p(x) = (f(x) + x) / (x + 1),
we get p(n+1) = ((-1)n+1 + n+1) / (n+2).
Put another way, p(n+1) is 1 for odd n and n / (n+2) for even n.
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