OFFSET
0,2
COMMENTS
In A192824 Noe defines 0-Ramanujan primes to be simply primes, and 1-Ramanujan primes to be Ramanujan primes. Define the k-th 2-Ramanujan prime to be the smallest number R'_k (the notation in Paksoy 2012) with the property that the interval (x/2,x] contains at least k 1-Ramanujan primes, for any x >= R'_k. Continuing inductively, define n-Ramanujan primes in terms of (n-1)-Ramanujan primes.
Only the first three terms 0, 2, 11 are proved (by Chebyshev, Ramanujan, and Paksoy, respectively). The rest are conjectural--see the 2nd comment in A192821.
See A104272 for additional comments, references, links, and cross-refs.
Is it true that for every n there exists K = K(n) such that for all k > K, the k-th n-Ramanujan prime is greater than half of the (k+1)-th n-Ramanujan prime? (Equivalently, is there a largest n-Ramanujan prime that is less than half of the next n-Ramanujan prime?) It is true for n = 0 by Bertrand's Postulate (see A062234), and for n = 1 by a theorem of Paksoy. Is it even true that if n is fixed, then (k-th n-Ramanujan prime) ~ ((k+1)-th n-Ramanujan prime) as k -> infinity? - Jonathan Sondow, Dec 16 2013
LINKS
Murat Baris Paksoy, Derived Ramanujan primes: R'_n, arXiv:1210.6991 [math.NT], 2012.
EXAMPLE
By Bertrand's Postulate (proved by Chebyshev), prime(k+1) < 2*prime(k) for all k, so a(0) = 0.
Ramanujan proved that the Ramanujan primes begin 2, 11, ..., so a(1) = 2.
Paksoy proved that the 2-Ramanujan primes begin 11, 41,..., so a(2) = 11.
It appears that the 3-Ramanujan primes begin 41, 149, ...; if true, then a(3) = 41.
It appears that the 4-Ramanujan primes begin 569, 571, 587, 1367 ...; if true, then a(4) = 587.
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Jonathan Sondow, Jun 08 2013
STATUS
approved