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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 $n$?

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Are you also interested in approximation from above? Would you accept 1/2 + 1/3 + 1/5 = 1 + 1/30? I don't have an answer for that, I am just curious. –  Stefano Pascolutti Nov 1 '12 at 21:17
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3 Answers

up vote 2 down vote accepted

This may be related to Giuga numbers: 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: $$\eqalign{&420001794970774706203871150967065663240419575375163\cr&060922876441614 2557211582098432545190323474818\cr&= 2 \cdot 3 \cdot 11 \cdot 23 \cdot 31 \cdot 47059 \cdot 2217342227 \cdot 1729101023519 \cdot \cr&8491659218261819498490029296021 \cdot 58254480569119734123541298976556403\cr}$$

Also possibly related to primary pseudoperfect numbers.

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I think this is the "solution" from above (see my comment). By the way, is you take the first eight factors of that number, you get a solution for the original problem for $n = 8$, which a step further. –  Stefano Pascolutti Nov 1 '12 at 22:12
    
I missed the part about primary pseudoperfect numbers. That will do the trick. –  Stefano Pascolutti Nov 1 '12 at 22:14
    
Thank you, Gerry. This problem is indeed just that of finding the primary pseudoperfect numbers. From Wikipedia, the (unique) answer is known for each $n=1,\dots,8$. It is unknown whether there are any, finitely many, or infinitely many other solutions. –  John Bentin Nov 2 '12 at 8:50
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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.

http://en.wikipedia.org/wiki/Sylvester's_sequence.

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...and unlucky that $2 \cdot 3 \cdot 11 \cdot 23 \cdot 31 \cdot 47059 + 1 = 7^2 45193927$ is not. –  Noam D. Elkies Jan 21 '13 at 0:24
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$$\frac{1}{2} + \frac{1}{3} + \frac{1}{11} + \frac{1}{23} + \frac{1}{31} + \frac{1}{2\cdot 3\cdot 11\cdot 23\cdot 31+1} = 1 - \frac{1}{2214502422}$$ and $2\cdot 3\cdot 11\cdot 23\cdot 31+1$ is prime (and 2214502422 is the product of the denominators).

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