

A168421


Small Associated Ramanujan Prime, p_(in)


9



2, 7, 11, 17, 23, 29, 31, 37, 37, 53, 53, 59, 67, 79, 79, 89, 97, 97, 127, 127, 127, 127, 127, 137, 137, 149, 157, 157, 179, 179, 191, 191, 211, 211, 211, 223, 223, 223, 233, 251, 251, 257, 293, 293, 307, 307, 307, 307, 307, 331, 331, 331
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OFFSET

1,1


COMMENTS

a(n) is the smallest prime p_(k+1n) on the left side of the Ramanujan Prime Corollary, 2*p_(in) > p_i for i > k, where the nth Ramanujan Prime R_n is the kth prime p_k. [Comment clarified and shortened by Jonathan Sondow, Dec 20 2013]
Smallest prime number, a(n), such that if x >= a(n), then there are at least n primes between x and 2x exclusively.
This is very useful in showing the number of primes in the range [p_k, 2*p_(in)] is greater than or equal to 1. By taking into account the size of the gaps between primes in [p_(in),p_k], one can see that the average prime gap is about ln(p_k) using the following R_n / (2*n) ~ ln(R_n).
Proof of Corollary: See Wikipedia link
The number of primes until the next Ramanujan prime, R_(n+1), can be found in A190874.
Not the same as A124136.
A084140(n) is the smallest integer where ceiling ((A104272(n)+1)/2), a(n) is the next prime after A084140(n).  John W. Nicholson, Oct 09 2013
If a(n) is in A005382(k) then A005383(k) is a twin prime with the Ramanujan prime, A104272(n) = A005383(k)  2, and A005383(k) = A168425(n). If this sequence has an infinite number of terms in A005382, then the twin prime conjecture can be proved.  John W. Nicholson, Dec 05 2013
Except for A000101(1)=3 and A000101(2)=5, A000101(k) = a(n). Because of the large size of a gap, there are many repeats of the prime number in this sequence.  John W. Nicholson, Dec 10 2013
For some n and k, we see that a(n) = A104272(k) as to form a chain of primes similar to a Cunningham chain. For example (and the first example), a(2) = 7, links A104272(2) = 11 = a(3), links A104272(3) = 17 = a(4), links A104272(4) = 29 = a(6), links A104272(6) = 47. Note that the links do not have to be of a form like q = 2*p+1 or q = 2*p1.  John W. Nicholson, Dec 14 2013
Srinivasan's Lemma (2014): p_(kn) < (p_k)/2 if R_n = p_k and n > 1. Proof: By the minimality of R_n, the interval ((p_k)/2,p_k] contains exactly n primes, so p_(kn) < (p_k)/2.  Jonathan Sondow, May 10 2014
In spite of the name Small Associated Ramanujan Prime, a(n) is not a Ramanujan prime for many values of n.  Jonathan Sondow, May 10 2014


LINKS

T. D. Noe, Table of n, a(n) for n = 1..10000
J. Sondow, Ramanujan primes and Bertrand's postulate, Amer. Math. Monthly 116 (2009) 630635.
J. Sondow, J. W. Nicholson, and T. D. Noe, Ramanujan Primes: Bounds, Runs, Twins, and Gaps, J. Integer Seq. 14 (2011) Article 11.6.2
J. Sondow, J. W. Nicholson, and T. D. Noe, Ramanujan Primes: Bounds, Runs, Twins, and Gaps, J. Integer Seq. 14 (2011) Article 11.6.2
Anitha Srinivasan, An upper bound for Ramanujan primes, Integers, 19 (2014), #A19
Wikipedia, Ramanujan Prime


FORMULA

a(n) = prime(primepi(A104272(n)) + 1  n).
a(n) = nextprime(A084139(n+1)), where nextprime(x) is the next prime > x. Note: some A084139(n) may be prime, therefore nextprime(x) not equal to x.  John W. Nicholson, Oct 11 2013
a(n) = nextprime(A084140(n)).  John W. Nicholson, Oct 11 2013


EXAMPLE

For n=10, the n'th Ramanujan prime is A104272(n)= 97, the value of k = 25, so i is >= 26, in >= 16, the in prime is 53, and 2*53 = 106. This leaves the range [97, 106] for the 26th prime which is 101. In this example, 53 is the small associated Ramanujan prime.


CROSSREFS

Cf. A104272, A168425, A179196, A190874.
Cf. A165959 (range size), A230147 (records).
Sequence in context: A063205 A090613 A063097 * A038942 A175283 A124136
Adjacent sequences: A168418 A168419 A168420 * A168422 A168423 A168424


KEYWORD

nonn


AUTHOR

John W. Nicholson, Nov 25 2009


EXTENSIONS

Extended by T. D. Noe, Nov 22 2010


STATUS

approved



