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jcaron2 · Feb 28, 2011, 09:08 AM

Not for the faint of heart:

Find the next number in the sequence:

1, 1, 3, 3, 15, 15, 33, 68, 100, 109, 199, 210, 282, 399, 497, 527,

Unknown008 · Feb 28, 2011, 09:29 AM

Hm... nice one. I'll give that a try :)

jcaron2 · Feb 28, 2011, 08:09 PM

Let me know if you want a hint. ;)

jcaron2 · Mar 1, 2011, 04:14 PM

All right, since it's been a while, I'll give you a hint. If you don't want the hint, don't read on.

If the numbers in the sequence are $a_1, \;a_2,\; a_3,\; ...$ notice that the difference between $a_n$ and $a_{n+1}$ is always divisible by n.

ebaines · Mar 1, 2011, 07:28 PM

I get 767 for the next number in the series. But I must admit - I needed the hint!

Unknown008 · Mar 2, 2011, 12:22 AM

Okay, I'm getting nowhere with all the jumping of numbers. :(

Could the next term be 937 by any chance? I don't think so...

jcaron2 · Mar 2, 2011, 05:54 AM

This one's definitely pretty challenging. Not 937 I'm afraid. I admire your perseverance though!

Unknown008 · Mar 2, 2011, 06:32 AM

I hope you understand that I was referring to the term after 767... Maybe 801?

The number of multiples after each 'peak' was 0, -1, 1, -2, so I'm thinking it's now 2(17)

ebaines · Mar 2, 2011, 07:01 AM

The term after 767 is indeed 801! I don't understand your explanation however.

Unknown008 · Mar 2, 2011, 07:07 AM

Lol, I hit the sack while I was blinded :p

I don't know really... taking the differences, we see the multiples of 'n' as being in that order:

0, 1, 0, 3, 0, 3, 5, 4, 1, 10, 1, 6, 9, 7, 2, 15,

First peak is at 1, where it doesn't get any higher, and then, we get 0. (-0)
Second peak at 3, then we get 0 (0)
Third peak at 5, then we get 4 (-1)
4th peak at 10, then we get 1 (1)
5th peak at 9, then we get 7 (-2)
I take the 6th peak as 15, and I guess (2)

2(17) + 767 gives 801

In the pattern I described above, we see -0, 0, -1, 1, -2, 2 but that doesn't help me know the rest of the sequence :o

ebaines · Mar 2, 2011, 07:42 AM

OK - I see what you're doing. I'm not familiar with the term "I hit the sack while I was blinded" - but I'm guessing it's similar to a phrase we use here in the US: "even a blind squirrel finds some nuts once in a while." So you get the right answer, but for the wrong reason.

Spoiler Alert:

In looking at the sequence of divisors as you have it:

0, 1, 0, 3, 0, 3, 5, 4, 1, 10, 1, 6, 9, 7, 2, 15

Note that the nth term here is always less than n. Which got me to thinking about divisors and remainders - note how each of these terms is actually the remainder of the nth term in the sequence 1,1,3,3,15,15,. divided by n. You can probably get it now.

Unknown008 · Mar 2, 2011, 08:09 AM

Well, I just made that expression up :p :o

And I just noticed I made a mistake. I put 10 instead of 9 for the first 10.

But thanks, now I got it. With this one, even the next term takes some time to be found.

Really nice one Josh :)

jcaron2 · Mar 7, 2011, 04:02 PM

LOL! Yeah, that would make it quite a bit harder to figure out the sequence. :)