variation of Euler Totient

to calculate the number of integers co-prime to N and less than N we can simply calculate its ETF( Euler's totient function - Wikipedia, the free encyclopedia . )However to calculate the number of integers co-prime to N but less then X where X < N , how can we modify / calculate it ? I know how to calcuate the ETF but can't proceed how to modify it to get the required result.

Thanks

Re: variation of Euler Totient

Hey pranay.

Being co-prime to an integer means the gcd(a,b) = 1.

But in terms of the totient function, this means that totient(a,b) = totient(a)*totient(b) (which is on the first page of the wiki).

So you know how many numbers up to n-1 are co-prime by using totient and you also have a condition for when two numbers are co-prime. Can you modify this to get an expression to "remove" the specific numbers of integers that are co-prime but equal or larger than x?

Re: variation of Euler Totient

pranay,

I'm afraid that I don't see how to use chiro's hint. However, your function is very close to what's known as the *Legendre Totient Function*. It's denoted and is defined to be the number of positive integers less than or equal to X that are coprime to N. Your function can be written as since you define your function to be the number of integers (positive, I assume) that are less than X and coprime to N. The following formula expresses the Legendre Totient function in terms of the Moebius function and the floor function :

where the summation runs over the positive divisors of N. I can provide a proof and/or some references. Just ask or let me know if you have any questions..

HTH

Petek

Re: variation of Euler Totient

Thanks a lot Petek, i too got that formula, but could you please explain as to why this works ?

Re: variation of Euler Totient

The formula in my previous post follows from the inclusion-exclusion principle and the observation that the number of positive integers less than or equal to X and divisible by d equals .

For example, let's compute :

The positive divisors of 20 are 1, 2, 4, 5, 10 and 20.

By direct inspection, the integers 1, 3, 5, 7 and 9 are less than or equal to 11 and coprime to 20, so .

The total number of positive integers less than or equal to 11 is .

From this we have to subtract the number of positive integers less than or equal to 11 that are divisible by 2, which is , and the number of positive integers less than or equal to 11 that are divisible by 5, which is .

However, we subtracted the number of positive integers less than or equal 11 that are divisible by 10 twice, so we have to add back .

Therefore,

,

where we used and .

(Note that , so the term with divisor = 4 doesn't actually appear in the formula. The same is true of the divisor 20, since .)

Hope you find this explanation useful.