number of irreductible fractions

[spx2]

so,i got this informatics problem actually.
i want to reformulate in math therms

it's like,i have this and
i want to find the number of all fractions with
so that these fractions are irreductible,this meaning that there isn't any
so that and (ofcourse )

so,for we have the following fractions that satisfy our
conditions:


i probably will have to go about an approach with but remember that it will have to use little operations to be a fast method, a fast algorithm actually.
i thank anyone who can help me in advance

[Myth]

I have some doubts there is a fast algorithm to find .

[spx2]

i really remember that i saw in a book a formula
for

in the mean-time look at what i've done to compute it into a c++ program


#include <stdio.h>
#include <math.h>
long N,i,j,k,nr;
char E[1000000]; //here i mark the numbers
//this is eratosthenes
//sieve
long phi[1000000]; //here phi[i] is generated
int main(){
FILE *f=fopen("fractii.in","r");fscanf(f,"%d",&N);//takes N from file
long nr=0;
phi[1]=0;
phi[2]=1;
phi[3]=2;
phi[4]=2;
phi[5]=4;
phi[6]=1;
phi[7]=6;
phi[8]=4;
phi[9]=6;
phi[10]=4;//these are precalculated values
for(i=2;i<sqrt(N)+1;i++)
{
long t=i;
while(t<=N)
E[t]=1,t+=i;
}//here i've marked the multiples
//of i and the j's with E[j]=0
//that remain are prime
//according to eratosthenes
for(i=11;i<sqrt(N)+1;i++)
{
if(E[i]==0)
{
long t=i;
if(E[i]==0)phi[i]=i-1;//if i is prime i know
else //that E[i]=0
while(t<=N) //i also know that
phi[t]*=(i-1)/i,t+=i;//for a prime i
} //,phi[i]=i-1;
}// here i calculate phi[t] step by step
// as i see every prime number by the formula
// phi(n)=n*(1-1/p1)*(1-1/p2)*...*(1-1/pn)
// primes generated through the sieve
for(i=1;i<=N;i++)
nr+=2*phi[i];
//now i multiply phi[i] by two
//because for example if i have
//ireductible fraction 2/3
//i also have ireductible fraction
//3/2 so i have to multiply by two this number
FILE *fo=fopen("fractii.out","w+");
fprintf(fo,"%d",nr);
return 0;
}

[Myth]

Are you participating in a programming contest?

BTW, there is no any appropriate formulae for .

[spx2]

tell me what do you think about the algorithm.
try it(compile+run).

[Myth]

I am sure you are able to compile it yourself

Some aspects of your algorithm are not clear for me.

[spx2]

after i've done erathosthenes for the interval [1,N]
i have


E[i]=
/ 0,if i is prime
|
\1,if its composite

now, i know the prime numbers between 1 and N.

so now i need to calculate phi[i] for all
for the sum i am searching for is really what you
stated before:



then i calculate phi(k) for each k and add up to s

[jmerry]

We're probably more interested in your algorithm than in the exact details of how it is written.

Your algorithm: start by finding all primes less than using the sieve of Eratosthenes. This takes computations, if done right.

During this pass through the array, we build an array of values of . I'm not quite sure if what you're doing makes sense here. I would do it as:
(initialize phi(k)=k, notprime(k)=0,sum=1)
(let i be an index variable, starting at 2 and increasing to N)
.....(if notprime(i)=0, execute following; else do nothing)
..........(let j=i)
..........(multiply phi(j) by (i-1)/i)
...............(increase j by i, exit if j>N)
...............(set notprime(j)=1)
...............(multiply phi(j) by (i-1)/i)
.....(increase sum by 2*phi(i))

This will produce correct values of and , in time. The only drawback is the large amount of storage space required. As far as I know, there's no faster way, and is very fast. In particular, the sieve is the fastest way to find all of the primes between 1 and .

The nice formula is that . That doesn't help here.
If you want an algorithm that doesn't use the storage space, you'll have to settle for much slower. Using the Euclidean algorithm to determine whether a pair is relatively prime gets you ; using factoring algorithms to find without the sieve gets you something like .

By the way, . It's a special case anyway; we add 1 instead of 2 in our sum. Also, . I don't see why you need precalculated values; your algorithm should be able to calculate them, except for the special case .
How close is the sum you obtain to ?


To anyone: Is there any way to format spacing in "normal" mode? That pseudocode needs indents, and I can't do them with spaces.

[spx2]

i've done it afterall.
it takes 40% of tests but i'm staisfied
it's now mathematically correct.
thanks all

#include <stdio.h>
//#include <math.h>
//#include <time.h>
#define mare unsigned long long
mare N,i,j,k,nr;
int E[1000000]; //here i mark the numbers
//this is eratosthenes
//sieve
mare phi[1000000]; //here phi[i] is generated
int main(){
//-----
//clock_t start=clock();
//-----
FILE *f=fopen("fractii.in","r");fscanf(f,"%d",&N);//takes N from file
mare nr=0;

for(i=2;i<=N/2;i++)
{
if(E[i]==0)
for(j=2*i;j<=N;j+=i)
E[j]=1;//ciur
};
for(i=1;i<=N;i++)if(E[i]==1)phi[i]=i;

for(i=2;i<=N;i++)
{
if(E[i]==0)
{
phi[i]=i-1;
for(j=2*i;j<=N;j+=i)
phi[j]/=i,phi[j]*=(i-1);
};
};
//phi[1..N];
//phi[i]*2=nr fractii ireductibile a/b | 1<=a,b<=i
for(i=2;i<=N;i++)
nr+=2*phi[i];
nr++;
FILE *fo=fopen("fractii.out","w+");
fprintf(fo,"%u",nr);
//-----------------
//clock_t end=clock();
//fprintf(fo,"\n%f",(double)(end-start)/CLOCKS_PER_SEC);
//-----------------
return 0;
}

[jmerry]

It's not mathematically correct- you're missing .

Also, you could combine the loops as I wrote to slightly increase efficiency.

[spx2]

i am counting because of that last nr++ if you look better.
it's at the end of the code

jmerry you are smart.
its verry interesting what you say there in the last part
of your post,it particularly screamed at me.
that that sum i'm searching for is almost

i remember i saw that somwhere....where have i saw
it ...oh oh yes ....it had sumthin to do with partial
sums of phi function.
i just skimmed it a little bit and that approx.
remained in my head...
but still...the problem asks for concrete value,
wich brings me to the question is there any posibility
of going around that value until we get to the right one.
i mean can we make a set of steps to get from the
approx. to the concrete value?

i'll send attached a pdf on my hdd where i saw them
before
man...how would i want to have the time to read that stuff
i'm still in highschool,i would have to attend college
but at the rythme i'm studying at my maternal language
and physics it seems to me ill never get to make somethin
out of my life.
anyway i would want to read this stuff but i'm in way
over my head with school...i barely have time to post
but now i'm in some hollydays...yeaaaaaaaaaa!!!
i am indeed verry stupid

[spx2]

this is where i saw the approx

[jmerry]

[fedja]

[jmerry]

Any further improvement on fedja's algorithm is minor- you actually only need the values for square-free, but this is asymptotically a constant fraction.

In the following, {a/b} denotes integer division.
Here's how I would do it. Start by choosing (slightly) greater than .
(Create arrays notprime[n], phi[n], sumphi[n] of size A^2)
(Create array mu[n] of size A-1)
(Initialize notprime[n]=0, phi[n]=n)
(Initialize mu[n]=0)
(Let S=0)
(set sumphi[1]=1)
(let i be an index variable, starting at 2 and increasing to A^2)
.....(if notprime[i]=0, execute following; else do nothing)
..........(let j=i)
..........(multiply phi[j] by (i-1)/i)
...............(increase j by i, exit if j>N)
...............(set notprime[j]=1)
...............(multiply phi[j] by (i-1)/i)
.....(set sumphi[i]=sumphi[i-1]+2*phi[i])
(set mu[1]=1)
(let i be an index variable, starting at 1 and increasing to A-1)
.....(if mu[i]=0 do nothing; else execute following)
..........(Increase S by {N/i}^2*mu[i])
..........(Decrease S by mu[i]*sumphi[1]*({N/i}-{N/2i}))
..........(let j be an index variable, starting at 2 and increasing until j^2*i>N)
...............(if i*j < A decrease mu[i*j] by mu[i])
...............(else decrease S by mu[i]*sumphi[{N/ij}])
...............(Decrease S by mu[i]*sumphi[j]*({N/ij}-{N/i(j+1)}
(Output S)

This is very hard to understand, but it uses far less space and time. The first part takes time, and the second takes time, for overall time. Alsom the storage is only .

We could calculate somewhat more efficiently, but it doesn't really matter here.

[spx2]

thank you all for remarks and help