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 Apr 24 comment origin of analogy “primes as the atoms of number theory/ arithmetic” In 1968, P.M. Cohn introduced the term atom for an irreducible element of an integral domain (i.e., a nonunit that cannot be written as a product of nonunits). Thus, the atoms of $\mathbb{Z}$ are the (positive and negative) primes, as in your question. See the Wikipedia article en.wikipedia.org/wiki/Atomic_domain. This terminology seems popular among commutative algebraists who study factorization theory. A more recent, related invention is the playful term "antimatter domain" for an integral domain with no irreducibles (such as the ring of all algebraic integers). Apr 23 comment How to show this bound? The relation you write down is true for reasons having nothing to do with modular forms. First, $(1-1/p)^{-1} = 1+1/p + 1/p^2+\dots \ge (1+1/p)$, and so the left hand-side is at least as large as the right. On the other hand, $(1-1/p)^{-1}/(1+1/p) = (1-1/p^2)^{-1}$, and the product of $(1-1/p^2)^{-1}$ over all primes $p$ is $\zeta(2)=\pi^2/6$. So the left-hand side is at most $\pi^2/6$ times the right-hand side. Feb 10 revised If the natural density (relative to the primes) exists, then the Dirichlet density also exists, and the two are equal bei, not by ! Feb 9 comment If the natural density (relative to the primes) exists, then the Dirichlet density also exists, and the two are equal @SalvoTringali: Since nothing about their particular set is used, it's obvious that Erdos and Davenport knew the general case. (See also the exposition in Hall's monograph Sets of Multiples.) But I agree that they did not spell out the general case. Feb 9 comment If the natural density (relative to the primes) exists, then the Dirichlet density also exists, and the two are equal The observation that analytic density ==> logarithmic density appears (but is not emphasized) in the proof of Theorem 1 of this 1936 paper of Erdos and Davenport: renyi.hu/~p_erdos/1936-04.pdf. They cite a 1914 theorem of Hardy and Littlewood. Feb 8 comment If the natural density (relative to the primes) exists, then the Dirichlet density also exists, and the two are equal OK, I went ahead and updated the answer. Feb 8 revised If the natural density (relative to the primes) exists, then the Dirichlet density also exists, and the two are equal added reference to Landau Feb 7 comment If the natural density (relative to the primes) exists, then the Dirichlet density also exists, and the two are equal You want p. 225 in the 1st edition, and p. 236 in the 2nd. Feb 7 answered If the natural density (relative to the primes) exists, then the Dirichlet density also exists, and the two are equal Feb 5 answered Asymptotics of product of Euler's totient function (A001088)? Feb 4 awarded Enlightened Feb 4 awarded Nice Answer Feb 4 answered Does the sum $\sum_{n=1}^{\infty}\frac{1}{p_n(p_{n+1}-p_n)}$ converge? Feb 4 comment Elementary prime-generating sequences Have you seen this paper? Shapiro, Harold N.; Sparer, Gerson H. Composite values of exponential and related sequences. Comm. Pure Appl. Math. 25 (1972), 569–615. Feb 4 comment Does the sum $\sum_{n=1}^{\infty}\frac{1}{p_n(p_{n+1}-p_n)}$ converge? The convergence of your series follows from upper bound sieve estimates. In fact, a somewhat stronger result is shown in this paper: Erdös, Paul(H-AOS); Nathanson, Melvyn B.(1-CUNY7) On the sum of the reciprocals of the differences between consecutive primes. Number theory (New York, 1991–1995), 97–101, Springer, New York, 1996. Dec 19 comment If $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form, is $n$ squarefree? Yes, that's clearer. I've made the edit.Thanks! Dec 19 revised If $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form, is $n$ squarefree? Edited as per Lucia's suggestion. Dec 18 answered If $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form, is $n$ squarefree? Dec 9 comment What is wrong with this counterexample to the Weak Bunyakovsky's conjecture and reformulation of Bunyakovsky's conjecture? I do not understand the comment: "searching the web, couldn't find out how to contact the author of the paper". If you delete the last part of your URL, you are led directly to the author's website, which lists his email address. Dec 1 awarded Nice Answer