Monday, March 23, 2009

The following are the first ten terms of a sequence:

1, 1, 2, 1, 2, 2, 3, 1, 2, 2, . . .

It may be of interest, though probably unhelpful in determining the pattern, that the limit at infinity of the sequence does not exist, though maxnє{1,2,3,4, . . .}{a(n)}=. Therefore, it follows that the derivative of the mystical curve of best fit is also non-existent at infinity.

Same idea as the other sequences. Describe the pattern of the sequence. If you use a search engine, please don't submit it in the comments.

This one's easy, but, if it hasn't been solved by Thursday, I'll give a hint then. Also, if you ask a polar question (i.e., a yes/no question), I will probably answer it if it relates to this sequence.

I'll post the solution on next week's sequence.

Last week's sequence (1, 2, 3, 2, 5, 3, 7, 2, 3, 5, . . .) was the greatest prime that divided the placeholder, with the first term defined as 1, that is, a(1)=1. E.g., a(30)=5, since the primes that divide 30 are 2, 3 and 5, and 5 is the highest of them. T solved it. No one remarked that the first term needed to be defined apart from the pattern.

1 comments:

Anonymous said...

It's binary!!!
The number of 1's (or ons) needed to display the placeholder!)

1 = 0001 = 1
2 = 0010 = 1
3 = 0011 = 2
4 = 0100 = 1
...
11 = 1011 = 3

The sequence is:
1121223122 -> 3, 2, 3, 3, 4, 1, 2, 2, 3,

T!

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