A385481 Primes whose decimal expansion consists of the concatenation of ij, iijj, iiijjj,..., and m i’s followed by m j’s, i != j, where 1<= i, j <= 9 and m > 0.

13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 31331133311133331111, 37337733377733337777, 43443344433344443333, 49449944499944449999

Offset: 1

Author

Gonzalo Martínez, Jun 30 2025

Keywords

• nonn
• base

Comments

a(25) has 812 digits and starts with 292299222999...999, where the last concatenated strings have 28 2's followed by 28 9's.

Since each term is prime, then j = 1, 3, 7, and 9 and i + j == 1, 2 (mod 3).

m == 1 (mod 3). Proof. If k is a term whose last concatenated string has m i's followed by m j's, then it has 2(1 + 2 + ... + m) = m(m + 1) digits, whose sum is (i + j)*m(m + 1)/2, so that if m == 0, 2 (mod 3), then the sum of digits of k is a multiple of 3 and so k is not prime.

Are there infinite primes of this form?

From Michael S. Branicky, Jul 01 2025: (Start)

A probabilistic argument suggests the sequence is finite.

a(26), if it exists, has m > 321 and > 103362 digits. (End)

Examples

			for i = 3, j = 1 and m = 4, by concatenating 31, 3311, 333111, 33331111 the prime 31331133311133331111 is obtained.
		

Links

• Gonzalo Martínez, Table of n, a(n) for n = 1..25

Crossrefs

Cf. A059170, A034845.

Programs

Python

from gmpy2 import is_prime, mpz
from itertools import count, islice, product
def agen(): yield from (p for m in count(1) for i in "123456789" for j in "1379" if i != j and is_prime(p:=int(mpz("".join(i*k+j*k for k in range(1, m+1))))))
print(list(islice(agen(), 24))) # Michael S. Branicky, Jun 30 2025