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A210476 Let p_(4,3)(m) be the m-th prime == 3 (mod 4). Then a(n) is the smallest p_(4,3)(m) such that the interval(p_(4,3)(m)*n, p_(4,3)(m+1)*n) contains exactly one prime == 3(mod 4). 2

%I #18 Feb 13 2013 23:58:30

%S 7,67,43,67,67,191,883,43,643,379,739,103,463,643,487,883,1303,3847,

%T 1447,13963,1087,8863,1999,8167,7687,8443,2707,2203,11083,3463,7687,

%U 31387,8419,15919,12979,10099,26683,22027,46687,79687,15439,65839,46723,44683,14887,58963,13879,26947,77587

%N Let p_(4,3)(m) be the m-th prime == 3 (mod 4). Then a(n) is the smallest p_(4,3)(m) such that the interval(p_(4,3)(m)*n, p_(4,3)(m+1)*n) contains exactly one prime == 3(mod 4).

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

%C Conjecture: every a(n), except for a(7) = 191, is the lesser of a pair of cousin primes p and p+4, (see A023200).

%t myPrime=Select[Table[Prime[n],{n,3000000}],Mod[#,4]==3&];

%t binarySearch[lst_,find_]:=Module[{lo=1,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 myPrimeQ[n_]:=binarySearch[myPrime,n];

%t nextMyPrime[n_,offset_Integer:1]:=myPrime[[myPrimeQ[NextPrime[n,NestWhile[#1+1&,1,!myPrimeQ[NextPrime[n,#1]]>0&]]]+offset-1]];

%t z=1;(*contains exactly ONE myPrime in the interval*)

%t Table[myPrime[[NestWhile[#1+1&,1,!((nextMyPrime[n myPrime[[#1]],z+1]>n myPrime[[#1+1]]))&]]],{n,2,30}]

%Y Cf. A195325, A207820, A210465, A210467, A210475.

%K nonn

%O 2,1

%A _Vladimir Shevelev_ and _Peter J. C. Moses_, Jan 23 2013

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