A repunit is a number consisting solely of ones (such as 11
or 11111).
Let us call p(n) a 10base integer represented by a
string of n ones, e.g. p(1)=1, p(5)=11111 etc.
Most of the repunit numbers are composite.
2, 19,23,317 are
the first four indices of prime repunits.
Prove: For a prime repunit p(n) to be prime, n has to
be prime.
(In reply to
re: Another proof  not a proof by Ady TZIDON)
I don't understand the problem. In this case, "q=p(y) with x1 zeroes placed between each pair of adjacent ones" results in q depending on x and y.
If you set x=3, y=2, then p(x)=111 as you mention. However q=1001 in this case, as p(y)=11 and placing x1=2 zeroes between each pair of ones results in 1001.

Posted by Gamer
on 20110111 14:16:32 