Sylvester's sequence
Sylvester's sequence [math]\displaystyle{ a_1,a_2,a_3,\ldots }[/math] is defined recursively by setting [math]\displaystyle{ a_1=2 }[/math] and [math]\displaystyle{ a_k = a_1 \ldots a_{k-1}+1 }[/math] for all subsequent k, thus the sequence begins
- 2, 3, 7, 43, 1807, 3263443, 10650056950807, 113423713055421844361000443 (sequence A000058 in OEIS).
The elements of this sequence are mutually coprime, so after factoring k of them, one is guaranteed to have at least k prime factors.
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 [math]\displaystyle{ a_k }[/math] is [math]\displaystyle{ O(n / \log n \log\log\log n) }[/math] rather than [math]\displaystyle{ O(n / \log n) }[/math] (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 [math]\displaystyle{ k\log k \log \log \log k }[/math] or so.
It is also conjectured that this sequence is square-free; if so, [math]\displaystyle{ a_k, a_k-1 }[/math] form a pair of square-free integers, settling a toy problem in the finding primes project.
Proposed proof of the above conjecture: Let <math>n=a_1\ldots a_{k-2} <\math>. Then <math> a_k=n^2+n+1 <\math>. Since <math> n^2 < n^2+n+1 < (n+1)^2 <\math>, <math> a_k <math> cannot be a square, and the sequence is squarefree.
