A229971 Palindromes n whose product of proper divisors is a palindrome > 1 and not equal to n.
4, 9, 121, 212, 1001, 10201, 110011, 1100011, 10100101, 11000011, 101000101, 110000011, 1010000101, 1100000011, 10000000001, 10100000101, 1000000000001, 10000000000001, 10011100000111001, 10022212521222001, 10100101110100101, 10101100100110101
Offset: 1
Examples
The product of the proper divisors of 4 is 2 (also a palindrome, different from 4). So, 4 is a member of this sequence. The proper divisors of 1001 are 1, 7, 11, 13, 77, 91, and 143. 1*7*11*13*77*91*143 = 1001^3 = 1003003001 (also a palindrome, different from 1001). So, 1001 is a member of this sequence.
Programs
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Mathematica
palQ[n_] := Block[{d = IntegerDigits@ n}, d == Reverse@ d]; fQ[n_] := Block[{s = Times @@ Most@ Divisors@ n}, And[palQ@ s, s > 1, s != n]]; Select[Select[Range@ 1000000, palQ], fQ] (* Michael De Vlieger, Apr 06 2015 *) ppdpQ[n_]:=Module[{pp=Times@@Most[Divisors[n]]},AllTrue[{n,pp},PalindromeQ]&&pp>1&&pp!=n]; Select[Range[115*10^4],ppdpQ] (* The program generates the first 8 terms of the sequence. *) (* Harvey P. Dale, Sep 18 2022 *) -
PARI
pal(n)=d=digits(n);Vecrev(d)==d for(n=1,10^6,D=divisors(n);p=prod(i=1,#D-1,D[i]);if(pal(n)&&pal(p)&&p-1&&p-n,print1(n,", "))) \\ Derek Orr, Apr 05 2015
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Python
from sympy import divisors def PD(n): p = 1 for i in divisors(n): if i != n: p *= i return p def pal(n): r = '' for i in str(n): r = i + r return r == str(n) {print(n,end=', ') for n in range(1,10**6) if pal(n) and pal(PD(n)) and (PD(n)-1) and PD(n)-n} # Simplified by Derek Orr, Apr 05 2015 # Syntax error fixed by Robert C. Lyons, Mar 17 2023
Extensions
a(7)-a(22) from Giovanni Resta, Oct 06 2013
Name edited by Derek Orr, Apr 05 2015
Comments