Any divisor of an X2+-X+1 number must have -3 as a quadratic residue. (Because 4N = 4X2+-4X+4 = (2X+-1)^2 + 3.) This excludes half the primes (the 6K-1 set). --Rich --------- Quoting Scott Huddleston <c.scott.huddleston@gmail.com>:
Nice proof! Is it obvious (or known) if all primes will or won't eventually appear as a factor of some member? (In general, and specifically for Sylvester's sequence.)
On Thu, Feb 26, 2015 at 2:45 PM, Bill Gosper <billgosper@gmail.com> wrote:
Hi William, you could simplify the p16 proof of the infinitude of the primes. Put any positive integer (e.g.1) in a bag. Repeat forever: Include in the bag the number formed by adding 1 to the product of all the numbers already in the bag.
Each new inclusion is relatively prime to all the rest, so no prime appears more than once among the factorizations. This requires an infinitude of primes.
If we start with 1, it's sort of neat that the successive inclusions are 2, 3, 7, 43, ..., A000058, Sylvester's sequence, whose reciprocals sum to 1 by being the denominators of the greedy Egyptian expansion of 1-1/?. --Bill Gosper _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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