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A198469 a(n) is the smallest prime(m) such that the interval (prime(m)*n, prime(m+1)*n) contains exactly seven primes.

%I A198469 #36 Feb 13 2013 23:58:29
%S A198469 1129,113,139,139,23,47,7,7,37,67,67,37,127,3,3,5,41,11,17,5,5,5,29,
%T A198469 71,11,101,2,2,2,101,107,2,2,71,71,191,191,227,239,281,2,197,227,107,
%U A198469 29,569,281,821,599,1031,521,641,659,1229,569,1061,1481,2657,641
%N A198469 a(n) is the smallest prime(m) such that the interval (prime(m)*n, prime(m+1)*n) contains exactly seven primes.
%C A198469 The sequence is unbounded.
%C A198469 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).
%C A198469 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.
%C A198469 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.
%C A198469 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.
%C A198469 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.
%H A198469 Alois P. Heinz, <a href="/A198469/b198469.txt">Table of n, a(n) for n = 2..200</a>
%e A198469 Let n=20, and consider intervals of the form (20*prime(m), 20*prime(m+1)).
%e A198469 For 2, 3, 5, ..., the intervals (40,60), (60,100), (100,140), (140,220), (220,260), (260,340), (340,380), ... contain 5, 8, 9, 13, 8, 13, 7, ... primes. Hence the smallest such prime is 17.
%t A198469 a[n_] := Catch[ For[m = 1, True, m++, p = Prime[m]; If[PrimePi[n*Prime[m + 1]] - PrimePi[n*p] == 7, Throw[p]]]]; Table[a[n], {n, 2, 60}] (* _Jean-François Alcover_, Jan 18 2013 *)
%Y A198469 Cf. A195871, A187809, A187810, A187812, A198195, A198319, A195325.
%K A198469 nonn
%O A198469 2,1
%A A198469 _Vladimir Shevelev_ and _Peter J. C. Moses_, Jan 07 2013