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A046328
Palindromes with exactly 2 prime factors (counted with multiplicity).
15
4, 6, 9, 22, 33, 55, 77, 111, 121, 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
Giovanni Resta, Table of n, a(n) for n = 1..10000 (first 2000 terms from Zak Seidov, terms a(2001)-a(2816) from Michael De Vlieger)
EXAMPLE
111 is a palindrome and 111 = 3*37. 3 and 37 are primes.
MATHEMATICA
fQ[n_] := Block[{id = IntegerDigits[n]}, Plus @@ Last /@ FactorInteger[n] == 2 && id == Reverse[id]]; Select[ Range[ 10000], fQ[ # ] &] (* Robert G. Wilson v, Jun 06 2005 *)
Select[Range[10002], Reverse[x = IntegerDigits[#]] == x && PrimeOmega[#] == 2 &] (* Jayanta Basu, Jun 23 2013 *)
Select[Range[11000], PalindromeQ[#]&&PrimeOmega[#]==2&] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Apr 30 2018 *)
PROG
(PARI) ispal(n) = my(d=digits(n)); d == Vecrev(d) \\ A002113
for(k=1, 1e4, if(ispal(k)&&bigomega(k)==2, print1(k, ", "))) \\ Alexandru Petrescu, Jul 07 2022
(Python)
from sympy import factorint
from itertools import product
def ispal(n): s = str(n); return s == s[::-1]
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): return sum(factorint(pal).values()) == 2
print(list(filter(ok, (p for d in range(1, 6) for p in pals(d) if ok(p))))) # Michael S. Branicky, Aug 14 2022
CROSSREFS
Subsequence of A001358 and A046338.
Sequence in context: A046338 A118690 A084994 * A196104 A046376 A229129
KEYWORD
nonn,base
AUTHOR
Patrick De Geest, Jun 15 1998
STATUS
approved