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Let p_(4,1)(m) be the m-th prime == 1 (mod 4). Then a(n) is the smallest p_(4,1)(m) such that the interval(p_(4,1)(m)*n, p_(4,1)(m+1)*n) contains exactly one prime == 1 (mod 4).
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%I #16 Feb 13 2013 23:58:30

%S 13,13,29,13,193,97,97,277,457,1193,109,229,937,397,349,1597,2137,937,

%T 5569,5737,2833,1549,6733,7477,5077,3457,877,4153,12277,11113,8689,

%U 14029,11113,5233,24109,14737,26713,1297,77797,12097,51577,57973,33409,30493,49429,112237,10333,143137

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

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

%C Conjecture: for n >= 12, every a(n) 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]==1&];

%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.

%K nonn

%O 2,1

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