Sylvester’s sequence

Construct an Egyptian fraction equal to 1.

$$\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{43}+\frac{1}{1807}+\mathrm{\cdots }$$

The denominators form the sequence 2, 3, 7, 43, 1807, … This is Sylvester’s sequence (listed in A58 of Sloane’s On-Line Encyclopedia of Integer Sequences), after the mathematician James Joseph Sylvester. The sequence can be calculated from the recurrence relation $a_n=1+{(a_{n-1})}^2-a_{n-1}$, with $a_0=2$. Knowing the terms up to $n-1$ one can calculate $a_n$ with the formula

$$a_n=1+\prod _{i=0}^{n-1}{a}_i$$

If the sequence was meant to construct an Egyptian fraction equal to 2, then it would be 1, 2, 3, 7, 43, 1807, … and could still be calculated by multiplying the previous terms and adding 1, but the recurrence relation given above would have to be reformulated.

Whatever the definition, the sequence consists of coprime terms, and thus can be used in Euclid’s proof of the infinity of primes. For this reason, these numbers are sometimes called Euclid numbers.

This sequence is useful in finding solutions to Znám’s problem.