First of the Math Ones
Monday, February 23, 2009

The following is a list of the first 106 (hint: note, not the first 100) terms of a sequence, {a(n)}.

1, 2, 2, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 16, 16, 17, 17, 17, 17, 17, 17, 17, 18, 18, 18, 18, 18, 18, 18, 19, 19, 19, 19, 19, 19, 19, 20, 20, 20, 20, 20, 20, 20, 20, 21, 21, 21, 21, 21, 21, 21, 21, . . .

The pattern this sequence employs, if it is to contain each of the positive integers, must take on the form above, with no variation. However, if you were to deviate slightly, and neglect that rule, the sequence would be constructed purely on how you define a(1).

Three possibly interesting, yet useless, characteristics:

  • The sequence is non-decreasing, that is, a(k)≤ a(k+1).
  • The limit of the sequence at infinity is infinity, that is, limn→∞a(n)=∞.
  • The sequences flatten as it approaches infinity, that is, in calculus speak, if a magical, continuous curve of best-fit could be drawn for the sequence, its derivative would be zero at infinity, limx→∞f'(x)=0, where f(x) is the function of the magical curve of best fit.

Describe the pattern of the sequence.

If you use a search engine, please don't submit it in the comments. Tomorrow, if no one's solved it, I'll post in the comments section a sequence that follows the same pattern, but not following the strict rule of containing all of the positive integers. I think you'll find it helpful. Wednesday night, I'll post in the comments section the solution.

If you ask a polar question (i.e., a yes/no question), I will probably answer it if it relates to this sequence.

I love math.

Posted byPeter Eddy
at9:44 PM

7 comments
 

7 comments:

Anonymous said...

Are these the next terms?

22, 22, 22, 22, 22, 22, 22, 22, 23, 23, 23, 23, 23, 23, 23, 23, 24, 24, 24, 24, 24, 24, 24, 24, 24, 25...

(Oh Peter... why are you doing this to me??)

February 23, 2009 at 11:11 PM  
Anonymous said...

I have the same question: Here's what I got:
1 – 1
2 – 2’s, 3’s - 2
3 – 4’s, 5’s - 2
4 – 6’s, 7’s, 8’s - 3
5 – 9’s, 10’s, 11’s, - 3
6-12’s, 13’s, 14’s, 15’s - 4
7 – 16’s, 17’s, 18’s, 19’s, - 4
8 – 20’s, 21’s …

February 23, 2009 at 11:16 PM  
Anonymous said...

and now that youve had 8 20's 21's, 22's, 23's, and 9 24's, youd also have 9 25's, 26's, 27's, then move to 10 28's so you have 3 of the -4's ...this way there will be infinity infinities and the derivative max at zero.

February 23, 2009 at 11:51 PM  
Peter Eddy said...
This comment has been removed by the author.
February 24, 2009 at 9:56 AM  
Peter Eddy said...

KWTsui,

I'm assuming you're hiding your identity. If you're not, you can use "Name/URL" next time to accomplish nearly the same thing as the "OpenID" but it gives you a little more freedom.

Anyway, when I first saw the sequence, I was only given the first 28 terms (i.e., up to the end of 9-terms). If I'd given fewer terms, you think it would have been tougher?

February 24, 2009 at 10:35 AM  
Peter Eddy said...

KWTsui: Yes. I'm doing this because I love you "(let the reader understand)" (Mk. 13:14).

Dan: I think you get it, but your notation is not obvious to me. And there will be nine occurrences of 28, not ten. I still think you get it, since you understand why it flattens at infinity.

February 24, 2009 at 10:39 AM  
Anonymous said...

I think 28 terms would have just enough to confirm it (we would have seen the 2 "pairs" (2s and 3s), 2 "triplets" (4s and 5s), and 3 "quadruplets" (6s, 7s and 8s) and the set of five 9s would have confirmed the pattern because we would have known that we have seen 3 complete sets of "n-tuplets", i.e. "pairs", "triplets" and "quadrulets"), so anything less than 28 terms would have made it tougher. I hope that made sense.

And thanks for the suggestion about Name/URL.

February 24, 2009 at 4:17 PM  

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