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Re: "Substract one and add one" algorithm
*To*: mathgroup at smc.vnet.net
*Subject*: [mg59076] Re: "Substract one and add one" algorithm
*From*: Peter Pein <petsie at dordos.net>
*Date*: Thu, 28 Jul 2005 02:26:46 -0400 (EDT)
*References*: <dc776m$k2h$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
Gilmar schrieb:
> Dear Mathematica Users Forum Friends:
>
> I want to build a function h[n] that does the following:
>
> For n even greater and equal than 4:
>
> Case 1: If m=n/2 is prime then h[n]={n/2,n/2}. Done.
>
> Case 2: If m =n/2 is not prime; let p[1]=n/2 -1 and q[1]=n/2+1.
>
> If both p[1], and q[1] are prime then,
> h[n]={p[1],q[1]}. Done.
>
> If either one or both p[1] and q[1] are not prime;
> let p[2] =p[1]-1, and q[2]=q[1]+1.
>
> If both p[2], and q[2] are prime then
> h[n]={p[2],q[2]}. Done.
>
> If either one or both p[2] and q[3] are not prime;
> let p[3] =p[2]-1, and q[3]=q[2]+1.
>
> etc.
>
> I want to test empirically that a value h[n] = {p[k],q[k]}
> (for an appropriate integer k) exists.
>
> A few examples:
>
>
> n=4
> n/2=2 is prime; so h[4]={2,2}.
>
> n=6
> n/2=3 is prime; so h[6]={3,3}.
>
> n=8 is not prime; so p[1]=n/2 -1 =3 is prime and q[1]=n/2+1=5 is prime;
> so h[8]={3,5}.
>
> n=10
> n/2=5 is prime; so h[10]={5,5}.
>
> n=12
> n/2=6 is not prime; so p[1]=n/2-1=5 is prime and q[1]=n/2+1=7 is prime;
> so h[12]={5,7}.
>
>
> n=14
> n/2=7 is prime; so h[14]={7,7}.
>
> n=16
> n/2=8 is not prime;
> so p[1]=n/2-1=7 is prime but, q[1]=n/2+1=9 is not prime,
> so p[2]=7-1=6 is not prime, and q[2]=9+1=10 is not prime,
> so p[3]=6-1=5 is prime, and q[3]=10+1=11 is prime,
> so h[16]={5,11}.
>
> Thank you for your help!
>
Hi Gilmar,
In[1]:=
h[(n_Integer)?EvenQ] :=
Module[{p = n/2, k},
For[k = 0, k < p - 1 &&
!And @@ PrimeQ /@ {p + k, p - k},
k++];
If[k == p - 1, Fail, p + {-k, k}]]
In[2]:=
h /@ Range[2, 16, 2]
Out[2]=
{{2, 2}, {3, 3}, {3, 5}, {5, 5}, {5, 7}, {7, 7}, {5, 11}}
--
Peter Pein
Berlin
http://people.freenet.de/Peter_Berlin/
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