https://asone.ai/polymath/api.php?action=feedcontributions&user=Zomega&feedformat=atom Polymath Wiki - User contributions [en] 2022-07-06T07:24:21Z User contributions MediaWiki 1.23.5 https://asone.ai/polymath/index.php?title=Sylvester%27s_sequence Sylvester's sequence 2010-08-24T00:56:50Z <p>Zomega: I...made a gigantic slip, forgot what squarefree meant. Sorry.</p> <hr /> <div>'''Sylvester's sequence''' &lt;math&gt;a_1,a_2,a_3,\ldots&lt;/math&gt; is defined recursively by setting &lt;math&gt;a_1=2&lt;/math&gt; and &lt;math&gt;a_k = a_1 \ldots a_{k-1}+1&lt;/math&gt; for all subsequent k, thus the sequence begins<br /> <br /> : 2, 3, 7, 43, 1807, 3263443, 10650056950807, 113423713055421844361000443 (sequence [http://www.research.att.com/~njas/sequences/A000058 A000058] in OEIS).<br /> <br /> The elements of this sequence are mutually coprime, so after factoring k of them, one is guaranteed to have at least k prime factors.<br /> <br /> 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 &lt;math&gt;a_k&lt;/math&gt; is &lt;math&gt;O(n / \log n \log\log\log n)&lt;/math&gt; rather than &lt;math&gt;O(n / \log n)&lt;/math&gt; (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 &lt;math&gt;k\log k \log \log \log k&lt;/math&gt; or so. <br /> <br /> It is also conjectured that this sequence is square-free; if so, &lt;math&gt;a_k, a_k-1&lt;/math&gt; form a pair of square-free integers, settling a toy problem in the finding primes project.<br /> <br /> * [[wikipedia:Sylvester's_sequence|The Wikipedia entry for this sequence]]</div> Zomega https://asone.ai/polymath/index.php?title=Sylvester%27s_sequence Sylvester's sequence 2010-08-23T20:20:57Z <p>Zomega: </p> <hr /> <div>'''Sylvester's sequence''' &lt;math&gt;a_1,a_2,a_3,\ldots&lt;/math&gt; is defined recursively by setting &lt;math&gt;a_1=2&lt;/math&gt; and &lt;math&gt;a_k = a_1 \ldots a_{k-1}+1&lt;/math&gt; for all subsequent k, thus the sequence begins<br /> <br /> : 2, 3, 7, 43, 1807, 3263443, 10650056950807, 113423713055421844361000443 (sequence [http://www.research.att.com/~njas/sequences/A000058 A000058] in OEIS).<br /> <br /> The elements of this sequence are mutually coprime, so after factoring k of them, one is guaranteed to have at least k prime factors.<br /> <br /> 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 &lt;math&gt;a_k&lt;/math&gt; is &lt;math&gt;O(n / \log n \log\log\log n)&lt;/math&gt; rather than &lt;math&gt;O(n / \log n)&lt;/math&gt; (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 &lt;math&gt;k\log k \log \log \log k&lt;/math&gt; or so. <br /> <br /> It is also conjectured that this sequence is square-free; if so, &lt;math&gt;a_k, a_k-1&lt;/math&gt; form a pair of square-free integers, settling a toy problem in the finding primes project.<br /> <br /> Proposed proof of the above conjecture:<br /> Let &lt;math&gt; n = a_1\ldots a_{k-2} &lt;/math&gt;. Then &lt;math&gt; a_k=n^2+n+1 &lt;/math&gt;. Since &lt;math&gt; n^2 &lt; n^2+n+1 &lt; (n+1)^2 &lt;/math&gt;, &lt;math&gt; a_k &lt;/math&gt; cannot be a square, and the sequence is squarefree.<br /> <br /> <br /> * [[wikipedia:Sylvester's_sequence|The Wikipedia entry for this sequence]]</div> Zomega https://asone.ai/polymath/index.php?title=Sylvester%27s_sequence Sylvester's sequence 2010-08-23T20:18:45Z <p>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.</p> <hr /> <div>'''Sylvester's sequence''' &lt;math&gt;a_1,a_2,a_3,\ldots&lt;/math&gt; is defined recursively by setting &lt;math&gt;a_1=2&lt;/math&gt; and &lt;math&gt;a_k = a_1 \ldots a_{k-1}+1&lt;/math&gt; for all subsequent k, thus the sequence begins<br /> <br /> : 2, 3, 7, 43, 1807, 3263443, 10650056950807, 113423713055421844361000443 (sequence [http://www.research.att.com/~njas/sequences/A000058 A000058] in OEIS).<br /> <br /> The elements of this sequence are mutually coprime, so after factoring k of them, one is guaranteed to have at least k prime factors.<br /> <br /> 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 &lt;math&gt;a_k&lt;/math&gt; is &lt;math&gt;O(n / \log n \log\log\log n)&lt;/math&gt; rather than &lt;math&gt;O(n / \log n)&lt;/math&gt; (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 &lt;math&gt;k\log k \log \log \log k&lt;/math&gt; or so. <br /> <br /> It is also conjectured that this sequence is square-free; if so, &lt;math&gt;a_k, a_k-1&lt;/math&gt; form a pair of square-free integers, settling a toy problem in the finding primes project.<br /> <br /> Proposed proof of the above conjecture:<br /> Let &lt;math&gt;n=a_1\ldots a_{k-2} &lt;\math&gt;. Then &lt;math&gt; a_k=n^2+n+1 &lt;\math&gt;. Since &lt;math&gt; n^2 &lt; n^2+n+1 &lt; (n+1)^2 &lt;\math&gt;, &lt;math&gt; a_k &lt;math&gt; cannot be a square, and the sequence is squarefree.<br /> <br /> * [[wikipedia:Sylvester's_sequence|The Wikipedia entry for this sequence]]</div> Zomega