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The sequence is infinite. Conjecture. For n>=2, all terms are in A001359. This conjecture (weaker than the conjecture in comment to A195325) also implies the twin prime conjecture.

The sequence is unbounded.

Conjecture. In the supposition that there are infinitely many twin primes, every term beginning with the 15th is 2 or in A001359 (lesser of twin primes).

A generalization. Consider sequence A_r: "A_r(n) is the smallest prime(m) such that the interval (prime(m)*n, prime(m+1)*n) contains exactly r>=0 primes". Then the sequence is unbounded and, in the supposition that there are infinitely many twin primes, we conjecture that there exists number N=N(r)>=2, such that for n>=N every term A_r(n) is 2 or in A001359.

Proof of unboundedness. If the sequence is bounded,then for some k and for arbitrary n, a(n) is in the set {p_1,p_2,...,p_k}, where p_i=prime(i). This means that for all n there exists p_i<=p_k such that interval (n*p_i, n*p_(i+1)) contains exactly r primes. However, from the PNT it evidently follows that pi(n*p_(i+1))-pi(n*p_i) tends to infinity as n goes to infinity, i.e., for sufficiently large n we obtain a contradiction for every i<=k.

Note that from these arguments it follows more: for every fixed r>=0, A_r(n) tends to infinity as n goes to infinity. Thus a fixed prime which is in the sequence can repeat only a finite number of times.

In addition, note that the condition "in the supposition that there are infinitely many twin primes" means that, if after a large number N_(tw) there are no twin primes, then this sequence, existing after N_(tw), of course, cannot have any term in A001359.

From Peter Munn, Oct 19 2017: (Start)

This is also a list of the leaf node labels in the tree of primes described in A290183.

For k > 0, the earliest run of k adjacent primes in this sequence starts with the least prime greater than A215238(k+1)/2. Thus we see that A215238(3) = 1637 corresponds to 821 followed by 823 being the first run of 2 adjacent primes in this sequence.

(End)

From Peter Munn, Nov 02 2017: (Start)

If p is in A005384 (a Sophie Germain prime), 2p+1 is therefore a prime, so p cannot be in this sequence. Similarly, any prime p in A023204 has a corresponding prime 2p+3, which (if p>2) likewise implies its absence (and if p=2 it is in A005384).

If p is the lesser of twin primes it is in this sequence if it is neither Sophie Germain nor in A023204.

Conjecture: a(n)/A000040(n) is asymptotic to 3. Reason: I expect the distribution of terms in A102820 to converge to a geometric distribution with mean value 2.

(End)

Conjecture. For n>=3, every a(n) is the lesser of a pair of cousin primes p and p+4, cf. A023200.

The limit of a(n) as n goes to infinity is infinity.

The limit of a(n) as n goes to infinity is infinity.

Conjectures: (1) If q is the nearest prime>a(n), then q-a(n)=4 or 6 and both of these cases occur infinitely many times. (2) If q-a(n)=4 then q is the lesser of twin primes.

Thus, if the conjectures are true, then there exist infinitely many triples of primes of the form {p,p+4,p+6}.

The limit of a(n) as n goes to infinity is infinity.

Conjectures: (1) If q is the nearest prime > a(n), then q-a(n) = 2 or 6 and both of these cases occur infinitely many times. (2) If q-a(n) = 2, then also q is lesser of a pair of cousin primes q and q+4, see A023200.

Thus, if the conjectures are true, then there exist infinitely many triples of primes of the form {p,p+2,p+6}.

The limit of a(n) as n goes to infinity is infinity.

Conjecture: for n >= 12, every a(n) is the lesser of a pair of cousin primes p and p+4, (see A023200).

For n<=20000, the largest a(n) is a(49)=23. a(n)=0 for n = 1, 2, 3, 4, 6, 43, 37890, 606457, ... corresponding to the twin primes (p, p+2) with p=3, 5, 11, 17, 41, 1277, 5995727, 143556431, ....