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).

7, 67, 43, 67, 67, 191, 883, 43, 643, 379, 739, 103, 463, 643, 487, 883, 1303, 3847, 1447, 13963, 1087, 8863, 1999, 8167, 7687, 8443, 2707, 2203, 11083, 3463, 7687, 31387, 8419, 15919, 12979, 10099, 26683, 22027, 46687, 79687, 15439, 65839, 46723, 44683, 14887, 58963, 13879, 26947, 77587

Offset

2,1

Comments

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

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

Links

Table of n, a(n) for n=2..50.

Mathematica

myPrime=Select[Table[Prime[n], {n, 3000000}], Mod[#, 4]==3&];
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)];
myPrimeQ[n_]:=binarySearch[myPrime, n];
nextMyPrime[n_, offset_Integer:1]:=myPrime[[myPrimeQ[NextPrime[n, NestWhile[#1+1&, 1, !myPrimeQ[NextPrime[n, #1]]>0&]]]+offset-1]];
z=1; (*contains exactly ONE myPrime in the interval*)
Table[myPrime[[NestWhile[#1+1&, 1, !((nextMyPrime[n myPrime[[#1]], z+1]>n myPrime[[#1+1]]))&]]], {n, 2, 30}]

Crossrefs

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

Sequence in context: A223889 A197744 A052351 * A217095 A106111 A261184

Adjacent sequences: A210473 A210474 A210475 * A210477 A210478 A210479

Keyword

nonn

Author

Vladimir Shevelev and Peter J. C. Moses, Jan 23 2013

Status

approved