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

1,1

Links

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

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 *)

Prog

(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

Crossrefs

Cf. A046328, A046408.

Intersection of A002113 and A006881.

Sequence in context: A020195 A035134 A075799 * A046408 A123017 A264043

Adjacent sequences: A046389 A046390 A046391 * A046393 A046394 A046395

Keyword

nonn,base

Author

Patrick De Geest, Jun 15 1998

Status

approved