# Difference between revisions of "Sylvester's sequence"

(Proposed simple proof the sequence is square-free, which would tie up a loose end in the primes project. Hope this helps someone. This is my first edit.) |
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Proposed proof of the above conjecture: | Proposed proof of the above conjecture: | ||

− | Let <math>n=a_1\ldots a_{k-2} < | + | 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. |

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* [[wikipedia:Sylvester's_sequence|The Wikipedia entry for this sequence]] | * [[wikipedia:Sylvester's_sequence|The Wikipedia entry for this sequence]] |

## Revision as of 20:20, 23 August 2010

**Sylvester's sequence** [math]a_1,a_2,a_3,\ldots[/math] is defined recursively by setting [math]a_1=2[/math] and [math]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]a_k[/math] is [math]O(n / \log n \log\log\log n)[/math] rather than [math]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]k\log k \log \log \log k[/math] or so.

It is also conjectured that this sequence is square-free; if so, [math]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 \lt n^2+n+1 \lt (n+1)^2 [/math], [math] a_k [/math] cannot be a square, and the sequence is squarefree.