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/