Primes mod 4
 

Are they as likely to ==1 as ==3?

PS thought I'd ask here rather than the proper maths thread
to avoid embarrassment:smile:

Yes and no. ;)

What do you mean by "as likely"?

[quote=Zeta-Flux;124801]Yes and no. ;)

What do you mean by "as likely"?[/quote]
Are there the ~same number of each below 100M?

[url]http://www.dms.umontreal.ca/~andrew/PDF/PrimeRace.pdf[/url]

is likely to tell you all you want to know about this, and then a little more besides.

It's a well-known nice problem.

[url]http://www.math.umn.edu/~focm/c_/Martin.pdf[/url] is a slightly more sophisticated article for people with a small amount of analytic number theory background (that is, who know a Dirichlet L-function from a hole in the ground), amongst whom I once counted myself but now don't.

[quote=fivemack;124804][URL]http://www.dms.umontreal.ca/~andrew/PDF/PrimeRace.pdf[/URL]

is likely to tell you all you want to know about this, and then a little more besides.

It's a well-known nice problem.

[URL]http://www.math.umn.edu/~focm/c_/Martin.pdf[/URL] is a slightly more sophisticated article for people with a small amount of analytic number theory background (that is, who know a Dirichlet L-function from a hole in the ground), amongst whom I once counted myself but now don't.[/quote]

THX

I take it the answer wasn't "yes" then:smile:

OTOH If I interpret "nice" as meaning "subtle", then I think
that "yes" is good enough for my present purposes.

[quote=fivemack;124804](that is, who know a Dirichlet L-function from a hole in the ground), amongst whom I once counted myself but now don't.[/quote]

I'm old enough to remember Bernard Cribbins' "Hole in the ground"
but I'm sure Ernst would find something wittier to say.

[QUOTE=davieddy;124806]THX

I take it the answer wasn't "yes" then:smile:[/QUOTE]Yes, unless you are using a logarithmic measure of how often one count is ahead of another (as defined on page 18 of the first link). In that sense 3 mod 4 beats 1 mod 4 soundly.

[quote=davieddy;124806]
"yes" is good enough for my present purposes.[/quote]

My "purpose" was to investigate the Wagstaff conjecture.
The probability of 2^p-1 being prime involves ln(ap) where
a=2 if p==3 mod 4 and a=6 if p==1 mod 4.

Commonly we take ln(ap)~ln(p) , true for huge p.
But for p in the GIMPS range I calculate ln(ap)~1.07 ln(p),
an appreciable discrepancy.

[URL]http://primes.utm.edu/mersenne/heuristic.html[/URL]

[quote=davieddy;124895]
a=2 if p==3 mod 4 and a=6 if p==1 mod 4.

But for p in the GIMPS range I calculate ln(ap)~1.07 ln(p),
[/quote]

ln(65M)~18
ln(2)~0.7
ln(6)~1.8

2.5 is 7% of 36.

[quote=Zeta-Flux;124825]Yes, unless you are using a logarithmic measure of how often one count is ahead of another (as defined on page 18 of the first link). In that sense 3 mod 4 beats 1 mod 4 soundly.[/quote]
I think page 2 "The prime number theorem for arithmetic progressions"
told me what I needed to know.

David

[url]http://mathworld.wolfram.com/ChebyshevBias.html[/url]