A133313 Primes p such that 3p-2 and 3p+2 are primes (see A125272) and its decimal representation finishes with 3.

3, 13, 23, 43, 103, 163, 293, 313, 433, 523, 953, 1013, 1063, 1153, 1283, 1303, 1483, 1693, 1723, 1783, 1913, 2003, 2333, 3533, 3823, 3943, 4003, 4013, 4093, 4943, 5483, 6043, 6133, 6173, 6473, 6803, 7523, 7573, 7603, 7673, 7853, 7993, 8513, 9283, 9343

Offset: 1

Author

Carlos Alves, Dec 21 2007

Keywords

• nonn
• base

Comments

Theorem: If in the triple (3n-2,n,3n+2) all numbers are primes, then n=5 or the decimal representation of n finishes with 3 or 7. Proof: Similar to A136191. Alternative Mathematica proof: Table[nn = 10k + r; Intersection @@ (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

Crossrefs

Cf. A136204 (finishing with 7), A136191, A136192, A125272.

Programs

Maple

filter:= proc(n) isprime(n) and isprime(3*n-2) and isprime(3*n+2) end proc:
select(filter, [seq(i,i=3..10^4,10)]); # Robert Israel, Nov 20 2023

Mathematica

TPrimeQ = (PrimeQ[ # - 2] && PrimeQ[ #/3] && PrimeQ[ # + 2]) &; Select[Select[Range[100000], TPrimeQ]/3, Mod[ #, 10] == 3 &]
Select[Prime[Range[1200]],Mod[#,10]==3&&AllTrue[3#+{2,-2},PrimeQ]&] (* The program uses the AllTrue function from Mathematica version 10 *) (* Harvey P. Dale, Oct 18 2019 *)