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Wikipedia:Reference desk/Archives/Mathematics/2013 May 30

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May 30

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Generalized totient function

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Is there such a thing as a generalized totient function? Specifically, a function φi(n) that is the count of positive integers less than n with exactly i prime divisors. This means that Euler's totient function φ(n) is φ1(n) in the more general form (i.e., it is the number of positive integers less than n with only one prime divisor). Likewise, φ2(n) is the number of positive composite integers less than n with only two prime divisors, e.g., numbers from the set {4,6,9,10,14,15,21,22,25,...,n}, which includes all the squares of primes. φ3(n) includes {8,12,18,27,28,30,...,n}, and so forth. I dimly recall seeing something about Ramanujan studying something similar to this(?). Perhaps such a thing might also be related to the Riemann hypothesis? — Loadmaster (talk) 17:24, 30 May 2013 (UTC)[reply]

Do you mean prime counting function rather than totient function? Otherwise I'm very confused. Sławomir Biały (talk) 17:29, 30 May 2013 (UTC)[reply]
Yes, that's what I meant. So (replacing φ above with π): Is there such a thing as a generalized prime counting function? Specifically, a function πi(n) that is the count of positive integers less than n with exactly i prime divisors. — Loadmaster (talk) 18:48, 30 May 2013 (UTC)[reply]
Yes, but these can be expressed in terms of the functions pi[n^(1/i)] using Mobius inversion. Count Iblis (talk) 19:54, 30 May 2013 (UTC)[reply]

With your help, I would like to find out a mathematical relationship between complete elliptic integrals of the first kind

and gaussian integrals

all of which are known to possess the following property

where

is half of the harmonic mean between m and n, and the entire above expression is equal to the product between 1 + + and the beta function of arguments 1 + and 1 + .

It also goes on without saying that the factorial of every positive number is the gaussian integral of its reciprocal or multiplicative inverse

79.118.171.165 (talk) 18:33, 30 May 2013 (UTC)[reply]

Resolved
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