OFFSET
1,1
COMMENTS
Minimal distance between prime(i) and prime(i+3) is 12 if all three consecutive prime gaps are different.
There are 6 possible consecutive prime gap configurations:
{2,4,6}, {2,6,4}, {4,2,6}, {4,6,2}, {6,2,4}, and {6,4,2}.
Least prime quartets with such gap configurations are:
{17,19,23,29}->{2,4,6}
{29,31,37,41}->{2,6,4}
{67,71,73,79}->{4,2,6}
{19,23,29,31}->{4,6,2}
{1601,1607,1609,1613}->{6,2,4}
{31,37,41,43}->{6,4,2}.
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
MATHEMATICA
p = Prime[Range[1000]]; First /@ Select[Partition[p, 4, 1], Last[#] - First[#] == 12 &] (* T. D. Noe, May 23 2011 *)
PROG
(Magma) [NthPrime(i): i in [2..60000] | NthPrime(i+3)-NthPrime(i) eq 12]; // _Bruno Berselli-, May 20 2011
(PARI) is(n)=if(!isprime(n), return(0)); my(p=nextprime(n+1), q); if(p-n>6, return(0)); q=nextprime(p+1); q-n<11 && nextprime(q+1)-n==12 \\ Charles R Greathouse IV, Sep 14 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Zak Seidov, May 20 2011
STATUS
approved