OFFSET
1,1
COMMENTS
For k>1 (not necessarily integer), we call a Ramanujan k-prime R_n^(k) the prime a_k(n) which is the smallest number such that if x >= a_k(n), then pi(x)- pi(x/k) >= n. Note that, the sequence of all primes corresponds to the case of "k=oo". These numbers possess the following properties: R_n^(k)~p_((k/(k-1))n) as n tends to the infinity; if A_k(x) is the counting function of the Ramanujan k-primes not exceeding x, then A_k(x)~(1-1/k)\pi(x) as x tends to the infinity; let p be a Ramanujan k-prime, such that p/k is in the interval (p_m, p_(m+1)), where p_m>=3 and p_n is the n-th prime. Then the interval (p, k*p_(m+1)) contains a prime. Conjecture. For every k>=2 there exist non-Ramanujan k-primes, which possess the latter property. For example, for k=2, the smallest such prime is 109. Problem. For every k>2 to estimate the smallest non-Ramanujan k-prime which possesses the latter property. [From Vladimir Shevelev, Sep 01 2009]
All Ramanujan 3-primes are in the sequence.
LINKS
Vladimir Shevelev, Charles R. Greathouse IV, Peter J. C. Moses, On intervals (kn, (k+1)n) containing a prime for all n>1, Journal of Integer Sequences, Vol. 16 (2013), Article 13.7.3. arXiv:1212.2785
EXAMPLE
If p=61, the p/3 is in the interval (19, 23); we see that the interval (61,69) contains a prime (67).
MATHEMATICA
nn=1000; t=Table[0, {nn}]; s=0; Do[If[PrimeQ[k], s++]; If[PrimeQ[k/3], s--]; If[s<nn, t[[s+1]]=k], {k, Prime[3*nn]}]; t=t+1
CROSSREFS
KEYWORD
nonn
AUTHOR
Vladimir Shevelev, Sep 01 2009
EXTENSIONS
Extended and edited by T. D. Noe, Nov 22 2010
Comments edited by Jonathan Sondow, Aug 27 2011
STATUS
approved