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,
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,
Hm... nice one. I'll give that a try :)
Let me know if you want a hint. ;)
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.
I get 767 for the next number in the series. But I must admit - I needed the hint!
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...
This one's definitely pretty challenging. Not 937 I'm afraid. I admire your perseverance though!
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)
The term after 767 is indeed 801! I don't understand your explanation however.
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
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.
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 :)
LOL! Yeah, that would make it quite a bit harder to figure out the sequence. :)
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