What's the next in the sequence?

roger · 2006-10-11, 15:53

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??

Xyzzy · 2006-10-11, 16:25

Use the [ spoiler ] tag.

ewmayer · 2006-10-11, 16:26

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.

xilman · 2006-10-11, 16:40

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

grandpascorpion · 2006-10-11, 17:32

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.

roger · 2006-10-11, 22:32

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

grandpascorpion · 2006-10-12, 01:55

Roger, just curious, how did you come up with the first term? It doesn't really fit the rest of the sequence.

roger · 2006-10-12, 22:53

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.

grandpascorpion · 2006-10-13, 00:25

Sure, but couldn't the third term have been 11^2 * 103 instead of 11*101^2.

It seems a little arbitrary.

xilman · 2006-10-13, 08:07

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

axn · 2006-10-13, 09:07

Quote:

Originally Posted by xilman

I'm missing something. 103 is palindromic?

Maybe he meant 131 ?

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2006-10-11, 15:53	#1
roger Oct 2006 100000100₂ 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 2·7·13·47 Posts	Use the [ spoiler ] tag.

2006-10-11, 16:26	#3
ewmayer ∂²ω=0 Sep 2002 República de California 5·2,351 Posts	My first guess was "ordered sequence of integers of the form abc, 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, 17:32	#5
grandpascorpion Jan 2005 Transdniestr 503 Posts	This is pretty ambiguous but I have a few guesses 1) 347743 = 11101313 The previous three terms are multiples of 11101 have 131,151 and 181 as high factors. 313 is the next palindromic prime after 181. 2) 1144411 = 1110110301 Same explanation as #1 above, except that only palindromic primes beginning and ending with 1 are allowed. 3) 188771 = 11131131 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 2²×5×13 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

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 2²·5·13 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 11^2101 (next palindromatic prime), then 11101^2, 11101131, and so on up in size.

2006-10-13, 00:25	#9
grandpascorpion Jan 2005 Transdniestr 1F7₁₆ Posts	Sure, but couldn't the third term have been 11^2 * 103 instead of 11*101^2. It seems a little arbitrary.