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

1,1

Comments

Theorem: If in the triple (3n-2,n,3n+2) all numbers are primes, then n=5 or the decimal representation of n 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

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 *)

Crossrefs

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

Sequence in context: A030431 A090146 A348559 * A230026 A260798 A102010

Adjacent sequences: A133310 A133311 A133312 * A133314 A133315 A133316

Keyword

nonn,base

Author

Carlos Alves, Dec 21 2007

Status

approved