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 2006-10-11, 15:53 #1 roger     Oct 2006 22×5×13 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??
 2006-10-11, 16:25 #2 Xyzzy     Aug 2002 20D816 Posts Use the [ spoiler ] tag.
 2006-10-11, 16:26 #3 ewmayer ∂2ω=0     Sep 2002 Repรบblica de California 266558 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.
2006-10-11, 16:40   #4
xilman
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3·5·743 Posts

Quote:
 Originally Posted by roger 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??
42 because it is the answer to the ultimate question of life, the universe and everything.

Paul

 2006-10-11, 17:32 #5 grandpascorpion     Jan 2005 Transdniestr 7678 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.
 2006-10-11, 22:32 #6 roger     Oct 2006 4048 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
 2006-10-12, 01:55 #7 grandpascorpion     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.
 2006-10-12, 22:53 #8 roger     Oct 2006 26010 Posts Grandpascorpion, The numbers were all the product of three palindromatic primes. Since 1-digit numbers are only technically palindromatic, I started with 2-digit 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.
 2006-10-13, 00:25 #9 grandpascorpion     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.
2006-10-13, 08:07   #10
xilman
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Quote:
 Originally Posted by grandpascorpion Sure, but couldn't the third term have been 11^2 * 103 instead of 11*101^2. It seems a little arbitrary.
I'm missing something. 103 is palindromic?

Paul

2006-10-13, 09:07   #11
axn

Jun 2003

527510 Posts

Quote:
 Originally Posted by xilman I'm missing something. 103 is palindromic?
Maybe he meant 131 ?

Last fiddled with by axn on 2006-10-13 at 09:08

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