# Sylvester's sequence

Sylvester's sequence $a_1,a_2,a_3,\ldots$ is defined recursively by setting $a_1=2$ and $a_k = a_1 \ldots a_{k-1}+1$ for all subsequent k, thus the sequence begins
There is a connection to the finding primes project: It is a result of Odoni that the number of primes less than n that can divide any one of the $a_k$ is $O(n / \log n \log\log\log n)$ rather than $O(n / \log n)$ (the prime number theorem bound). If we then factor the first k elements of this sequence, we must get a prime of size at least $k\log k \log \log \log k$ or so.
It is also conjectured that this sequence is square-free; if so, $a_k, a_k-1$ form a pair of square-free integers, settling a toy problem in the finding primes project.
Proposed proof of the above conjecture: Let $n=a_1\ldots a_{k-2} \lt\math\gt. Then \ltmath\gt a_k=n^2+n+1 \lt\math\gt. Since \ltmath\gt n^2 \lt n^2+n+1 \lt (n+1)^2 \lt\math\gt, \ltmath\gt a_k \ltmath\gt cannot be a square, and the sequence is squarefree. * [[wikipedia:Sylvester's_sequence|The Wikipedia entry for this sequence]]$