The following are the first fifty terms of a sequence: 1, 4, 9, 16, 25, 36, 49, 64, 81, 1, 2, 5, 10, 17, 26, 37, 50, 65, 82, 4, 5, 8, 13, 20, 29, 40, 53, 68, 85, 9, 10, 13, 18, 25, 34, 45, 58, 73, 90, 16, 17, 20, 25, 32, 41, 52, 65, 80, 97, 25, . . . 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. If it hasn't been solved by Thursday, I'll give a hint. 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 (3, 11, 21, 33, 47, 63, 81, 101, 123, 147, . . .) was defined explicitly a(n)=n2+5n-3. KWTsui solved it.
Tuesday, July 21, 2009
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6 comments:
Hey Peter, this is Jill -- see? I knew we'd see each other again! But I won't be submitting a description of the pattern of this sequence. I'm allergic to Math or things that closely resemble Math . . .
I see you're pretty adament about letting everyone know that you don't like math.
I think that I saw you on the CBC campus yesterday. Indeed, I was wrong.
Did you start reading my blog before we met, or is this a recent development? Either way, thank you for your readership.
Peter Eddy is a HUGE jerkface.
That is all.
The n-th term of the sequence is the sum of the square of the digits of n. For example, for the 15th term,
a(15) = 1^2 + 5^2 = 26
@Anonymous: Good thing Jesus gave up his life to pay for Peter's jerkfaceness (Mk 10:45)! :P
Jerkface or not, Jeremy sounded like you!
Recent development. But now I'm like so YOU'RE the Peter Eddy that writes long comments on Jer's blog. Haha.
And maybe you did see me at CBC -- I was visiting on Tuesday at lunchtime to meet the T.O. metro team.
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