Re: Pattern to the primes



quasi wrote:

On Sat, 16 Feb 2008 10:51:56 -0800 (PST), Rob Barry
<rob.barry@xxxxxxxxx> wrote:

Hi -- can anybody explain to me why the following
equations do not
yield a pattern for the distribution of prime
numbers?

For all x E integers:

(1) cos(pi * x! / x^2) + cos(pi / x) = 0

ive seen this before somewhere ...



It's nice.

Let f(x) = cos(pi*(gamma(x+1)/x^2)) + cos(pi/x)

for real part > 1 writing z might be intresting too.

as the OP mentioned the complex.



Here's how it looks numerically for x from 2 to 31,
for integer values
of x.

g(2) = 0.
g(3) = 0.
g(4) = 0.7071067810
g(5) = 0.
g(6) = 1.866025404
g(7) = 0.
g(8) = 1.923879532
g(9) = 1.939692621
g(10) = 1.951056516
g(11) = 0.
g(12) = 1.965925826
g(13) = 0.
g(14) = 1.974927912
g(15) = 1.978147601
g(16) = 1.980785280
g(17) = 0.
g(18) = 1.984807753
g(19) = 0.
g(20) = 1.987688341
g(21) = 1.988830826
g(22) = 1.989821442
g(23) = 0.
g(24) = 1.991444861
g(25) = 1.992114701
g(26) = 1.992708874
g(27) = 1.993238358
g(28) = 1.993712210
g(29) = 0.
g(30) = 1.994521895
g(31) = 0.

notice the rising of the nonzero terms.


As to why it works, I'm not sure.

It's easy to show that if x is a composite integer
and x > 4, then

cos(pi * x! / x^2) = 1

and

cos(Pi/x) > -1

so f(x) is nonzero for all composite integers x.

One can verify directly that f(4) is nonzero, hence
f(x) is nonzero
for all composite integers x.

It seems evident from the data that f(x) = 0 for all
primes x. I'm
sure there is an easy proof, but I don't see it off
hand.

Anyway, it's a cool pattern.

a plot would be nice.


Regarding f as a continuous function on the domain
[1,infinity), it's
clear that f has other zeros besides the set of
primes. Of course,
these other zeros are non-integers.

dont forget complex z.


As to what their
pattern is
somewhat mysterious. Considering the set S of
non-integer zeros of f,
we can ask ...

(1) What is the density of S intersect [1,x] as a
function of x?

well , cos(pi / x) tends to 1 for large x.

so when the other terms hits about -1 we get g(x) = 0.

this should help you.



(2) Do the elements of S have any "prime-like"
properties? For
example, is it true that the ratio of two distinct
elements of S is
never an integer?

very likely.



quasi

intresting function.

regards
tommy1729
.



Relevant Pages

  • Re: A strange ring, and some strange ideals...
    ... quasi wrote: ... R is clearly a commutative ring. ... If detis nonzero, x^is in R, hence x is a unit in R and thus ... It follows that R/I is a field, ...
    (sci.math)
  • Re: Integral extension of Integral Domains
    ... Here's a counterexample ... ... It's not hard to verify that A is integrally closed in B. ... For every b in B (nonzero) there is an a in A such that a/b is in B ...
    (sci.math)
  • Re: polynomials over Z_p
    ... can assume, without loss of generality, that w=0. ... Simply complete the square. ... Then, assuming a is nonzero, we can ...
    (sci.math)
  • Re: quadratic quadratic non-residue
    ... quasi wrote: ... I believe that others are asserting ... in Z_p, and a is nonzero. ... Regard f as a function from Z_p to Z_p. ...
    (sci.math)
  • Comment on math/0507611
    ... the proof is rather of functional analytical type: ... where gis nonzero everywhere. ... Then fcan have zeros only at Re= 1/2. ... Prev by Date: ...
    (sci.math)