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 (x!+1)/(x+1) = prime number (Posted on 2010-07-30)
For a positive integer x drawn at random between 1 and 20000 inclusively, determine the probability that (x!+1)/(x+1) is a prime number.

 No Solution Yet Submitted by K Sengupta Rating: 2.3333 (3 votes)

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here are two pages showing that the only ones below 30941 are
4,6,10,28,772,1320, and 2620
http://www.research.att.com/njas/sequences/index.html?q=5,7,11,29,773&language=english&go=Search

also, at the bottom of the second page there is a nice argument suggesting that no higher values exist due to probability of such primes existing above a number X tends to zero as X tends to infinity.

also, thank you Steve for pointing out the flaw in my proof, below is a repaired version of said proof:

f(x) = ( x! + 1 ) / (x+1)

now for f(x) to be integer we need
x! = -1 mod (x+1)
but if x+1 is composite then its positive divisors are less than x and thus a factor of x! and thus
GCD(x!, x+1) > 1 and thus we can not have
x! = -1 mod (x+1)
thus f(x) is not an integer when x is composite

 Posted by Daniel on 2010-07-31 20:29:48

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