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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
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
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
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
What are all the zeroes of the [analytic continuation of] f(s) := Prod 1/(1 - 1/q^s) q where q ranges over all the Sylvester primes ? --Dan
On 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
More seriously: What accounts for the self-similarish oscillations on p. 29 (not counting the upper left graph), of ----- the number of multiplicatively odd numbers between 2 and X minus the number of multiplicatively even numbers between 2 and X? ----- --Dan P.S. In the paragraph containing the above passage, the symbol "X" is used in two different ways.
On Feb 26, 2015, at 3:35 PM, Dan Asimov <asimov@msri.org> wrote:
What are all the zeroes of the [analytic continuation of]
f(s) := Prod 1/(1 - 1/q^s) q
where q ranges over all the Sylvester primes ?
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.
Tweak: "... no prime appears in more than one of the factorizations." Primes might repeat within one of the factorizations. E.g. if we started with 4, vs 1. Man, I wonder what the first squareful Sylvester number is. --rwg
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
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.
Tweak: "... no prime appears in more than one of the factorizations." Primes might repeat within one of the factorizations. E.g. if we started with 4, vs 1. Man, I wonder what the first squareful Sylvester number is. --rwg
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
participants (4)
-
Bill Gosper -
Dan Asimov -
rcs＠xmission.com -
Scott Huddleston