- e(0) := 2
- e(n+1) := (e(n) - 1) ⋅ e(n) + 1
Indeed,
- e(n+1) := 1 + ∏k=1...n e(k)
Let (p(k) : k=0 1 ...) be the increasing sequence of all prime numbers. We see that
- p(n) ≤ e(n) for every n=0 1...
- p(0) = e(0) = 2
- p(1) = e(1) = 3
- p(2) = 5 < 7 = e(2)
- p(n) < e(n) for every n=2 3...
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