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A348559
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Primes where every other digit is 3 starting with the rightmost digit, and no other digit is 3.
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4
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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
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OFFSET
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1,1
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LINKS
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MATHEMATICA
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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 *)
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PROG
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(Magma) f3:=func<n|forall{i:i in [1..#Intseq(n) by 2]| Intseq(n)[i] eq 3}>; fc:=func<n| forall{i:i in [2..#Intseq(n) by 2]| Intseq(n)[i] ne 3}>; [p:p in PrimesUpTo(40000)|f3(p) and fc(p)]; // Marius A. Burtea, 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))
(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))))
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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