20061011, 15:53  #1 
Oct 2006
2^{2}×5×13 Posts 
What's the next in the sequence?
1331; 12221; 112211; 145541; 167761; 201091
What's the next in the sequence, and how do you know? PS  how do you black out text?? 
20061011, 16:25  #2 
"Mike"
Aug 2002
1110101010011_{2} Posts 
Use the [ spoiler ] tag.

20061011, 16:26  #3 
∂^{2}ω=0
Sep 2002
República de California
2×3^{2}×619 Posts 
My first guess was "ordered sequence of integers of the form a*b*c, where a,b,c are primes whose decimal expansion is of the form 1...1," but in that case your sequence is missing some terms.

20061011, 16:40  #4 
Bamboozled!
May 2003
Down not across
10011100001000_{2} Posts 

20061011, 17:32  #5 
Jan 2005
Transdniestr
503 Posts 
This is pretty ambiguous but I have a few guesses
1) 347743 = 11*101*313 The previous three terms are multiples of 11*101 have 131,151 and 181 as high factors. 313 is the next palindromic prime after 181. 2) 1144411 = 11*101*10301 Same explanation as #1 above, except that only palindromic primes beginning and ending with 1 are allowed. 3) 188771 = 11*131*131 For the first 2 terms, the 2nd factor is 11. The next four, the 2nd factor is 101. So for the next 8 (or 6?) terms, the 2nd factor would be the next palindromic prime 101. The third factor would start with 131 and then skip to the next palindrome. 4) 1167216611 = 11 * 10301 *10301 Same explanation as #3, except that except that only palindromic primes beginning and ending with 1 are allowed. 
20061011, 22:32  #6 
Oct 2006
2^{2}·5·13 Posts 
To Grandpascorpion:
Your guess that the next in sequence would be the next palindromatic 'high' prime is correct. Honestly though, I didn't think someone would figure it out so quickly... PS, thanks xyzzy 
20061012, 01:55  #7 
Jan 2005
Transdniestr
503 Posts 
Roger, just curious, how did you come up with the first term? It doesn't really fit the rest of the sequence.

20061012, 22:53  #8 
Oct 2006
2^{2}·5·13 Posts 
Grandpascorpion,
The numbers were all the product of three palindromatic primes. Since 1digit numbers are only technically palindromatic, I started with 2digit primes. 1331=11^3. Next is 11^2*101 (next palindromatic prime), then 11*101^2, 11*101*131, and so on up in size. 
20061013, 00:25  #9 
Jan 2005
Transdniestr
503 Posts 
Sure, but couldn't the third term have been 11^2 * 103 instead of 11*101^2.
It seems a little arbitrary. 
20061013, 08:07  #10 
Bamboozled!
May 2003
Down not across
2^{3}×1,249 Posts 

20061013, 09:07  #11 
Jun 2003
4,591 Posts 
Last fiddled with by axn on 20061013 at 09:08 
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