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 A164952 Primes p with the property: if p/3 is in the interval (p_m, p_(m+1)), where p_m>=3 and p_k is the k-th prime, then the interval (p, 3p_(m+1)) contains a prime. 8
 2, 3, 11, 17, 23, 29, 41, 43, 59, 61, 71, 73, 79, 97, 101, 103, 107, 131, 137, 149, 151, 163, 167, 179, 191, 193, 223, 227, 229, 239, 251, 257, 269, 271, 277, 281, 311, 331, 347, 349, 353, 359, 367, 373, 383, 397, 419, 421, 431, 433, 439, 457, 461, 463, 479, 491 (list; graph; refs; listen; history; text; internal format)
 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

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Last modified October 23 07:11 EDT 2019. Contains 328336 sequences. (Running on oeis4.)