RE : [math-fun] Re: Prime's position in S

[Eric Angelini]

Hello Dean,
a warm thanks for the computations and the Mathematica code.
I will complete OEIS's ecord A121053 at the end of the month.
Best and thank you again !
É.
 

________________________________

De: math-fun-bounces+eric.angelini=kntv.be at mailman.xmission.com de la part de Dean Hickerson
Date: sam. 12/08/2006 0:01
À: seqfan at ext.jussieu.fr; math-fun at mailman.xmission.com
Objet : [math-fun] Re: Prime's position in S



Mostly to Eric Angelini:

> Hello SeqFan and math-fun,
> a somehow self-describing sequence here, with a question:
> http://www.cetteadressecomportecinquantesignes.com/PrimePos.htm
> Best,
> Éric A.

In the limit, exactly half of the terms are prime.

Here's a formula, found empirically, for a(n) for n>=5:

Let pi(n) be the number of primes <= n and p(n) be the n'th prime.  Then:

    if n is prime or (n is composite and n+pi(n) is even) then

                                            a(n) = p(floor((n+pi(n))/2));

    if n is composite and n+pi(n) is odd and n+1 is composite then a(n) = n+1;

    if n is composite and n+pi(n) is odd and n+1 is prime then a(n) = n+2.

Also, for n>=5, n is in the sequence iff either n is prime or n+pi(n) is even.

I'm sure this can all be proved by induction on n; I'll leave it to you to
check all of the cases if you want to.

It follows from this that, for n>=4, the number of primes among a(1), ...,
a(n) is exactly  floor((n+pi(n))/2.  Since  pi(n)/n -> 0  as  n -> infinity,
this is asymptotic to n/2.

Here's some Mathematica code that can be used to check this.  a[n] is computed
from your definition; b[n] is computed from the formula above:

    $RecursionLimit=Infinity;

    a[1]=2;
    inseq[n_]:=False;  (* inseq[n] will be True if n is in the sequence *)
    inseq[2]=True;

    a[n_]:=a[n]=
    Module[{m},
      a[n-1];
      For[m=1, inseq[m] || ((inseq[n] || m==n) && !PrimeQ[m]) ||
               (m<n && !PrimeQ[a[m]]),
          m++, Null];
      inseq[m] = True;
      Return[m] ];

    b[n_]:=
    If[n<=4, {2,3,5,1}[[n]],
    If[PrimeQ[n] || EvenQ[n+PrimePi[n]], Prime[Floor[(n+PrimePi[n])/2]],
    If[PrimeQ[n+1], n+2, n+1] ] ];

    a/@Range[1000] == b/@Range[1000]  (* Should return True *)


    countprimes[n_]:=Length[Select[a/@Range[n],PrimeQ[#]&]];

    cp[n_]:=Floor[(n+PrimePi[n])/2];

    countprimes/@Range[4,1000] == cp/@Range[4,1000]  (* Should return True *)

Eric, feel free to use any or all of this in your sequence entry.

Dean Hickerson
dean at math.ucdavis.edu

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