How long can a primal egyptian fraction be, that optimally approaches unity? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T05:38:28Z http://mathoverflow.net/feeds/question/111173 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/111173/how-long-can-a-primal-egyptian-fraction-be-that-optimally-approaches-unity How long can a primal egyptian fraction be, that optimally approaches unity? John Bentin 2012-11-01T16:40:41Z 2013-01-20T23:36:51Z <p>Thus, do there exist $n$ distinct primes whose summed reciprocals fall short of $1$ by the reciprocal of their product, for some $n\geqslant6$? I can get as far as $n=5$: $$\dfrac{1}{2}=1-\dfrac{1}{2},$$ $$\dfrac{1}{2}+\dfrac{1}{3}=1-\dfrac{1}{2\cdot3},$$ $$\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{7}=1-\dfrac{1}{2\cdot3\cdot7},$$ $$\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{1}{43}=1-\dfrac{1}{2\cdot3\cdot7\cdot43},$$ $$\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{11}+\dfrac{1}{23}+\dfrac{1}{31}=1-\dfrac{1}{2\cdot3\cdot11\cdot23\cdot31}.$$ But I can't see the way beyond that. If there is a fraction for some $n\geqslant6$, then one may naturally ask: is there a bound on such $n$?</p> http://mathoverflow.net/questions/111173/how-long-can-a-primal-egyptian-fraction-be-that-optimally-approaches-unity/111177#111177 Answer by Anonymous for How long can a primal egyptian fraction be, that optimally approaches unity? Anonymous 2012-11-01T17:00:46Z 2012-11-01T17:28:36Z <p>What about tacking on 1/47059 to your last example? This extension is suggested by the definition of Sylvester's sequence. Of course we got lucky that 2*3*11*23*31 + 1 = 47059 was prime.</p> <p><a href="http://en.wikipedia.org/wiki/Sylvester" rel="nofollow">http://en.wikipedia.org/wiki/Sylvester</a>'s_sequence. </p> http://mathoverflow.net/questions/111173/how-long-can-a-primal-egyptian-fraction-be-that-optimally-approaches-unity/111184#111184 Answer by Stefano Pascolutti for How long can a primal egyptian fraction be, that optimally approaches unity? Stefano Pascolutti 2012-11-01T17:37:02Z 2012-11-01T17:37:02Z <p>$$\frac{1}{2} + \frac{1}{3} + \frac{1}{11} + \frac{1}{23} + \frac{1}{31} + \frac{1}{2\cdot 3\cdot 11\cdot 23\cdot 31+1} = 1 - \frac{1}{2214502422}$$ and $2\cdot 3\cdot 11\cdot 23\cdot 31+1$ is prime (and 2214502422 is the product of the denominators).</p> http://mathoverflow.net/questions/111173/how-long-can-a-primal-egyptian-fraction-be-that-optimally-approaches-unity/111220#111220 Answer by Gerry Myerson for How long can a primal egyptian fraction be, that optimally approaches unity? Gerry Myerson 2012-11-01T21:59:12Z 2012-11-01T21:59:12Z <p>This may be related to <a href="http://oeis.org/A007850" rel="nofollow">Giuga numbers</a>: composite numbers $n$ such that $p$ divides $(n/p)-1$ for every prime divisor $p$ of $n$. A 10-factor Giuga number is given in the comments on that page: \eqalign{&amp;420001794970774706203871150967065663240419575375163\cr&amp;060922876441614 2557211582098432545190323474818\cr&amp;= 2 \cdot 3 \cdot 11 \cdot 23 \cdot 31 \cdot 47059 \cdot 2217342227 \cdot 1729101023519 \cdot \cr&amp;8491659218261819498490029296021 \cdot 58254480569119734123541298976556403\cr}</p> <p>Also possibly related to <a href="http://oeis.org/A054377" rel="nofollow">primary pseudoperfect numbers</a>. </p>