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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
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).
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
KEYWORD
nonn
AUTHOR
STATUS
approved