Sylvester's sequence - Revision history

Charles R Greathouse IV: fix OEIS link

2015-09-30T04:30:11Z

fix OEIS link

← Older revision		Revision as of 21:30, 29 September 2015
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	'''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		'''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).		: 2, 3, 7, 43, 1807, 3263443, 10650056950807, 113423713055421844361000443 (sequence [http://oeis.org/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.		The elements of this sequence are mutually coprime, so after factoring k of them, one is guaranteed to have at least k prime factors.

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

2010-08-24T00:56:50Z

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

← Older revision		Revision as of 17:56, 23 August 2010
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	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.		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]]		* [[wikipedia:Sylvester's_sequence\|The Wikipedia entry for this sequence]]

Zomega at 20:20, 23 August 2010

2010-08-23T20:20:57Z

← Older revision		Revision as of 13:20, 23 August 2010
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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} <\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.		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]]		* [[wikipedia:Sylvester's_sequence\|The Wikipedia entry for this sequence]]

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.

2010-08-23T20:18:45Z

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.

← Older revision		Revision as of 13:18, 23 August 2010
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	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.		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]]		* [[wikipedia:Sylvester's_sequence\|The Wikipedia entry for this sequence]]

Asdf at 23:58, 19 August 2009

2009-08-19T23:58:43Z

← Older revision		Revision as of 16:58, 19 August 2009
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	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		'''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).		: 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.		The elements of this sequence are mutually coprime, so after factoring k of them, one is guaranteed to have at least k prime factors.
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	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.		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.

	# [~~http~~:~~//en.wikipedia.org/wiki/~~Sylvester~~%27s_sequence~~ The Wikipedia entry for this sequence]		* [[wikipedia:Sylvester's_sequence\|The Wikipedia entry for this sequence]]

Teorth: New page: 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 sequenc...

2009-08-09T14:21:04Z

New page: 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 sequenc...

New page

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.

# [http://en.wikipedia.org/wiki/Sylvester%27s_sequence The Wikipedia entry for this sequence]