



Senior Member


Feb 28, 2011, 09:08 AM


Challenging number sequence
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,



Uber Member


Feb 28, 2011, 09:29 AM


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



Senior Member


Feb 28, 2011, 08:09 PM


Let me know if you want a hint. ;)



Senior Member


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 notice that the difference between and is always divisible by n.



Expert


Mar 1, 2011, 07:28 PM


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



Uber Member


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...



Senior Member


Mar 2, 2011, 05:54 AM


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



Uber Member


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)



Expert


Mar 2, 2011, 07:01 AM


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



Uber Member


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



Expert


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.



Uber Member


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 :)



Senior Member


Mar 7, 2011, 04:02 PM


Comment on Unknown008's post
LOL! Yeah, that would make it quite a bit harder to figure out the sequence. :)


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