A046392 - OEIS

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

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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);

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