# Identity for frequency of integers with smallest prime(n) divisor

An identity for A038110 and A038111: $$\frac{\phi(e^{\psi(p_{n}-1)})}{e^{\psi(p_{n})}}=\frac{\prod _k^{n-1} \left(1-\frac{1}{p_k}\right)}{p_n},$$ where $\psi(\cdot)$ is the second Chebyshev function and $\phi(\cdot)$ is Euler's totient function. RHS is Euler's Prime Product from OEIS.

Is this a known identity? (Still an open question, though I think Kronecker knew about it.)

Edit How I found this identity: I created a palindromic divisor sequence to see if I could find a solution to Oppermann's conjecture by evaluating the divisors of the composite numbers in the attempt to isolate the primes. An example: $$\left( \begin{array}{cccc} 1 & 2 & 3 & 4 \\ 1 & 3 & 1 & 4 \\ 3 & 2 & 1 & 4 \\ \end{array} \right),\left( \begin{array}{ccccc} 1 & 2 & 3 & 4 & 5 \\ 3 & 1 & 4 & 3 & 5 \\ 1 & 4 & 1 & 2 & 5 \\ 4 & 1 & 3 & 1 & 5 \\ 3 & 2 & 1 & 4 & 5 \\ 2 & 3 & 4 & 1 & 5 \\ 1 & 4 & 3 & 2 & 5 \\ 4 & 1 & 2 & 3 & 5 \\ 1 & 3 & 1 & 4 & 5 \\ 2 & 1 & 4 & 1 & 5 \\ 3 & 4 & 1 & 3 & 5 \\ 4 & 3 & 2 & 1 & 5 \\ \end{array} \right),$$ where we have two palindromic sequences when we drop the last elements. Notice that the count of ones corresponds to Euler's Totient function. The left hand matrix corresponds to the numerator and the right hand side to the denominator. Edit2 The palindrome for the LHS is: $\{1,2,3,4,1,3,1,4,3,2,1,4\}$, which is the underlying value of the right-most column of the RHS before the $5$'s are overlaid. So, to determine the significance of this observation, I crafted the fraction at the top of the post, created a short sequence of denominators, searched OEIS, and found this identity. (Thanks to Gerry Myerson who crafted the proof below.)

The divisor sequence appears to be related to Mertens's theorem and the Riemann's Hypotheses. This is unexpected, so discovering the identity was serendipitous. I have no interest in proving this regarding Riemann. My only interest is to show that the symmetic divisor sequence is predictable.

Mathematica code:

n = 2; Product[1 - 1/Prime[k], {k, n - 1}]/Prime[n] ==
EulerPhi[Exp[Sum[MangoldtLambda[m], {m, 1, Prime[n] - 1}]]]/
Exp[Sum[MangoldtLambda[m], {m, 1, Prime[n]}]]

• What do you mean that this expression is an "identity for" these two sequences? Also, in whatever way you answer the previous question, is this a conjecture of your own, or something that you have seen elsewhere? If the former, what sort of evidence do you have for it, and how did you come by it? If the latter, where have you seen it? (And please edit your question as opposed to adding such information in the comments.) – user642796 Jul 14 '14 at 19:19
• @Arthur asked you, what identity? You came back to edit, but didn't answer. Voting to close as unclear. – Gerry Myerson Jul 15 '14 at 9:29
• You are completely missing the point. You asked, "Is this a known identity?" What I'm asking, and I think what Arthur is asking, is "Is WHAT a known identity?" An answer to that question, preferably in the body of the question rather than in a comment, might just add something to the "discussion". – Gerry Myerson Jul 15 '14 at 10:44

The definition is $$\psi(x)=\sum_{p^k\le x}\log p$$ where $p$ runs through primes. That is, it's the sum of $\log p$ over all prime powers $p^k$ not exceeding $x$. So $$e^{\psi(p_n)}=\prod_{p^k\le p_n}p=p_n\prod_{p^k\le p_n-1}p=p_ne^{\psi(p_n-1)}$$ where $p_n$ is the $n$th prime. So, the left side of the identity is $${1\over p_n}{\phi(u)\over u}$$ where $u=e^{\psi(p_n-1)}$. Since $\phi(m)=m\prod_{p\mid m}(1-p^{-1})$, the identity is immediate.