Polymath Wiki - User contributions [en]

Sylvester's sequence

2010-08-24T00:56:50Z

Zomega: I...made a gigantic slip, forgot what squarefree meant. Sorry.

'''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 [http://www.research.att.com/~njas/sequences/A000058 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.

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

Sylvester's sequence

2010-08-23T20:20:57Z

Zomega:

'''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 [http://www.research.att.com/~njas/sequences/A000058 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 < n^2+n+1 < (n+1)^2 </math>, <math> a_k </math> cannot be a square, and the sequence is squarefree.

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

Sylvester's sequence

2010-08-23T20:18:45Z

Zomega: 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.

'''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 [http://www.research.att.com/~njas/sequences/A000058 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 < n^2+n+1 < (n+1)^2 <\math>, <math> a_k <math> cannot be a square, and the sequence is squarefree.

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