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A046328
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Palindromes with exactly 2 prime factors (counted with multiplicity).
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15
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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
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OFFSET
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1,1
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LINKS
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EXAMPLE
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111 is a palindrome and 111 = 3*37. 3 and 37 are primes.
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MATHEMATICA
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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 *)
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PROG
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(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
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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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