A348559 Primes where every other digit is 3 starting with the rightmost digit, and no other digit is 3.

3, 13, 23, 43, 53, 73, 83, 313, 353, 373, 383, 1303, 1373, 2383, 2393, 4363, 4373, 5303, 5323, 5393, 6323, 6343, 6353, 6373, 7393, 8353, 8363, 9323, 9343, 30313, 30323, 31393, 32303, 32323, 32353, 32363, 34303, 34313, 35323, 35353, 35363, 35393, 36313, 36343

Offset: 1

Author

Lars Blomberg, Oct 22 2021

Keywords

• nonn
• base

Links

• Michael S. Branicky, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000040, A348558, A348560, A348561.

Programs

Magma

f3:=func;  fc:=func; [p:p in PrimesUpTo(40000)|f3(p) and fc(p)]; // Marius A. Burtea, Oct 22 2021

Mathematica

Select[Prime@Range@10000,(n=#;s={EvenQ,OddQ};t=Take[IntegerDigits@n,{#}]&/@Select[Range@i,#]&/@If[EvenQ[i=IntegerLength@n],s,Reverse@s];Union@Flatten@First@t=={3}&&FreeQ[Flatten@Last@t,3])&] (* Giorgos Kalogeropoulos, Oct 22 2021 *)

Python

from sympy import primerange as primes
def ok(p):
    s = str(p)
    if not all(s[i] == '3' for i in range(-1, -len(s)-1, -2)): return False
    return all(s[i] != '3' for i in range(-2, -len(s)-1, -2))
print(list(filter(ok, primes(1, 36344)))) # Michael S. Branicky, Oct 22 2021

Python

# faster version for generating large initial segments of sequence
from sympy import isprime
from itertools import product
def eo3(maxdigits): # generator for every other digit is 3, no other 3's
    yield 3
    for d in range(2, maxdigits+1):
        if d%2 == 0:
            for f in "12456789":
                f3 = f + "3"
                for p in product("012456789", repeat=(d-1)//2):
                    yield int(f3 + "".join(p[i]+"3" for i in range(len(p))))
        else:
            for p in product("012456789", repeat=(d-1)//2):
                yield int("3" + "".join(p[i]+"3" for i in range(len(p))))
print(list(filter(isprime, eo3(5)))) # Michael S. Branicky, Oct 22 2021