A210512 Primes formed by concatenating k, k and 3 for k >= 1.

113, 223, 443, 773, 883, 10103, 11113, 14143, 25253, 26263, 28283, 32323, 35353, 41413, 50503, 61613, 68683, 71713, 77773, 80803, 83833, 85853, 88883, 97973, 1001003, 1011013, 1101103, 1131133, 1161163, 1181183, 1221223, 1241243, 1281283, 1331333, 1361363, 1391393

Offset: 1

Author

Abhiram R Devesh, Jan 26 2013

Keywords

• base
• nonn
• easy

Comments

This sequence is similar to A030458, A052089 and A210511.

k must not be a multiple of 3, otherwise the concatenation of k, k and 3 will also be a multiple of 3 and therefore not prime. This is a necessary but not sufficient condition.

Some of the terms can be found with this simple process: 5 - 3 = 2 = 1 + 1 giving 113; 7 - 3 = 4 = 2 + 2 giving 223; 11 - 3 = 8 = 4 + 4 giving 443; 17 - 3 = 14 = 7 + 7 giving 773; 19 - 3 = 16 = 8 + 8 giving 883. - J. M. Bergot, Jul 25 2022

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A030458, A052089, A210511.

Programs

Magma

[nn3: n in [1..140] | IsPrime(nn3) where nn3 is Seqint([3] cat Intseq(n) cat Intseq(n))]; // Bruno Berselli, Jan 30 2013

Mathematica

Select[Table[FromDigits[Flatten[{IntegerDigits[n], IntegerDigits[n], {3}}]], {n, 100}], PrimeQ] (* Alonso del Arte, Jan 27 2013 *)

Python

import numpy as np
from functools import reduce
def factors(n):
    return reduce(list._add_, ([i, n//i] for i in range(1, int(n**0.5) +1) if n % i == 0))
for i in range(1, 1000):
    p1=int(str(i)+str(i)+"3")
    if len(factors(p1))<3:
        print(p1, end=',')

Python

from sympy import isprime
def xf(n): return int(str(n)*2+'3')
def ok(n): return isprime(xf(n))
print(list(map(xf, filter(ok, range(1, 140))))) # Michael S. Branicky, May 21 2021