A046392 Palindromes with exactly 2 distinct prime factors.

6, 22, 33, 55, 77, 111, 141, 161, 202, 262, 303, 323, 393, 454, 505, 515, 535, 545, 565, 626, 707, 717, 737, 767, 818, 838, 878, 898, 939, 949, 959, 979, 989, 1111, 1441, 1661, 1991, 3113, 3223, 3443, 3883, 7117, 7447, 7997, 9119, 9229, 9449, 10001

Offset: 1

Author

Patrick De Geest, Jun 15 1998

Keywords

• nonn
• base

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A046328, A046408.

Intersection of A002113 and A006881.

Programs

Maple

revdigs:= proc(n) local L,i;
  L:= convert(n,base,10);
  add(L[-i]*10^(i-1),i=1..nops(L))
end proc:
f:= proc(n) local F;
  F:= ifactors(n)[2];
  if nops(F) = 2 and F[1,2]=1 and F[2,2]=1 then n fi
end proc:
N:=5: # for terms of up to N digits.
Res:= 6:
for d from 2 to N do
  if d::even then
    m:= d/2;
    Res:= Res, seq(f(n*10^m + revdigs(n)), n=10^(m-1)..10^m-1);
  else
    m:= (d-1)/2;
    Res:= Res, seq(seq(f(n*10^(m+1)+y*10^m+revdigs(n)), y=0..9), n=10^(m-1)..10^m-1);
  fi
od:
Res; # Robert Israel, Mar 24 2020

Mathematica

pdpfQ[n_]:=Module[{idn=IntegerDigits[n]},idn==Reverse[idn] && PrimeNu[n] == PrimeOmega[n] == 2]; Select[Range[11000],pdpfQ] (* Harvey P. Dale, Dec 16 2012 *)

Python

from sympy import factorint
from itertools import product
def pals(d, base=10): # all d-digit palindromes
    digits = "".join(str(i) for i in range(base))
    for p in product(digits, repeat=d//2):
        if d > 1 and p[0] == "0": continue
        left = "".join(p); right = left[::-1]
        for mid in [[""], digits][d%2]: yield int(left + mid + right)
def ok(pal): f = factorint(pal); return len(f) == 2 and sum(f.values()) == 2
print(list(filter(ok, (p for d in range(1, 5) for p in pals(d) if ok(p))))) # Michael S. Branicky, Jun 22 2021