Ramanujan Primes and Bertrand's Postulate

Abstract

In a two-page paper [8; 9, pp. 208-209] published one year before his death in 1920 at the age of 32, the Indian mathematical genius Srinivasa Ramanujan wrote: Landau in his Handbuch [5], pp. 89-92, gives a proof of a theorem the truth of which was conjectured by Bertrand: namely that there is at least one prime p such that x < p ≤ 2x , if x ≥ 1. Landau's proof is substantially the same as that given by Tschebyschef. The following is a much simpler one.Ramanujan then ``uses simple properties of the Γ-function'' (P. Ribenboim [10, p. 188]) to prove the theorem, which is known as Bertrand's postulate or Tschebyschef's theorem.(For an exposition of Ramanujan's proof, see Shapiro [14, Section 9.3B].) 2. RAMANUJAN PRIMES. In [6, p. 178] W. J. LeVeque explains that the theorem is called Bertrand's ``postulate'' rather than ``conjecture'' because he took it as a working tool in his study of a problem in group theory. This must have seemed entirely safe, considering the actual density of primes in the tables. There is not merely one prime between 500,000 and l,000,000, say, there are 36,960 of them!

Primality, Fractality, and Image Analysis

“…The cardinality of a set A will be denoted by # A. For the convenience of the reader and for brevity the preliminaries are reported below without proofs [8,17,18], thus making our exposition self-contained.…”

On generalized Ramanujan primes

“…Since r(k) is decreasing, from r(745.8) ≤ 2.999966 we get that r(k) ≤ 3 and therefore X 4 = 3k for every k ≥ 745.8. Since π(3k) + 1 ≤ k for every k ≥ 745.8, we obtain N (k) ≤ π(3k) + 1 by (18). Finally, we apply (17).…”

New upper bounds for Ramanujan primes

“…Let R n = p s , where p i denotes the i th prime. Sondow [7] showed that p 2n < R n < p 4n for all n, and conjectured that R n < p 3n for all n. This conjecture was proved by Laishram [4], and the upper bound p 3n improved by various authors ( [1], [8]). Subsequently, Srinivasan [9] and Axler [1] improved these bounds by showing that for every ǫ > 0, there exists an integer N such that R n < p [2n(1+ǫ)] for all n > N. * Using the method in [9] (outlined below), a further improvement was presented by Srinivasan and Nicholson, who proved that s < 2n 1 + 3 log n + log(log n) − 4 for all n > 241.…”