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A210467 Let p_(3,2)(m) be the m-th prime == 2(mod 3). Then a(n) is the smallest p_(3,2)(m) such that the interval(p_(3,2)(m)*n, p_(3,2)(m+1)*n) contains exactly one prime == 2 (mod 3). 3

%I #20 Feb 13 2013 23:58:30

%S 2,2,101,263,1097,251,311,461,641,941,1601,2351,2543,5003,2837,4787,

%T 5711,4283,7901,10331,8831,2687,7877,54287,5711,5501,5303,56087,69827,

%U 15641,63611,138581,106427,91571,69827,266177,142421,177533,179687,309311,55691,119291,509543,593987,1393913

%N Let p_(3,2)(m) be the m-th prime == 2(mod 3). Then a(n) is the smallest p_(3,2)(m) such that the interval(p_(3,2)(m)*n, p_(3,2)(m+1)*n) contains exactly one prime == 2 (mod 3).

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

%C 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.

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

%t bPrime=Select[Table[Prime[n], {n, 1000000}], Mod[#, 3]==2&];

%t binarySearch[lst_, find_]:=Module[{lo=2, up=Length[lst], v}, (While[lo<=up, v=Floor[(lo+up)/2]; If[lst[[v]]-find==0, Return[v]]; If[lst[[v]]<find, lo=v+1, up=v-1]]; 0)];

%t bPrimeQ[n_]:=binarySearch[bPrime, n];

%t nextBPrime[n_, offset_Integer:1]:=bPrime[[bPrimeQ[NextPrime[n, NestWhile[#1+1&, 1, !bPrimeQ[NextPrime[n, #1]]>0&]]]+offset-1]];

%t z=1; (*example for "contains exactly ONE b-

%t primes"*)Table[bPrime[[NestWhile[#1+1&, 1, !((nextBPrime[n bPrime[[#1]], z]<n bPrime[[#1+1]]&&nextBPrime[n bPrime[[#1]], z+1]>n bPrime[[#1+1]]))&]]], {n, 2, 20}]

%Y Cf. A195325, A210465, A207820, A210475, A210476.

%K nonn

%O 2,1

%A _Vladimir Shevelev_ and _Peter J. C. Moses_, Jan 22 2013

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