OFFSET
1,1
COMMENTS
Theorem: If in the triple (3n-2,n,3n+2) all numbers are primes, then n=5 or the decimal representation of n ends in 3 or 7. Proof: Similar to A136191. Alternative Mathematica proof: Table[nn = 10k + r; Intersection (AT)(AT) (Divisors[CoefficientList[(3nn - 2) nn(3nn + 2), k]]), {r, 1, 9, 2}]; This gives {{1, 5}, {1}, {1, 5}, {1}, {1, 5}}. Therefore only r=3 and r=7 allow nontrivial divisors (excluding nn=5 itself).
LINKS
Robert Israel, Table of n, a(n) for n = 1..10000
MAPLE
filter:= proc(n) isprime(n) and isprime(3*n-2) and isprime(3*n+2) end proc:
select(filter, [seq(i, i=7..10^4, 10)]); # Robert Israel, Nov 20 2023
MATHEMATICA
TPrimeQ = (PrimeQ[ # - 2] && PrimeQ[ #/3] && PrimeQ[ # + 2]) &; Select[Select[Range[100000], TPrimeQ]/3, Mod[ #, 10] == 7 &]
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Carlos Alves, Dec 21 2007
STATUS
approved