Question

Mathematical Circles

Answer 1

Ok we just determined that the differences have to be absolute values cause otherwise if we have

\[a_{n+2} = a_{n+1}-a_n\]
\[1 = \sum_{k=0}^{29} a_{n}= \sum_{k=0}^{29} a_{n+2} = \sum_{k=0}^{29} a_{n+1}-a_n =\sum_{k=0}^{29} a_{n} -\sum_{k=0}^{29} a_{n} =0\]
Which is a contradiction.

Answer 2

Let's try this with three people for a beginning. \(a_1, a_2, a_3\) where\[a_1 = |a_2 - a_3|\]\[a_2 = |a_3 - a_1|\]\[a_3 = |a_1 - a_2| \]\[a_1 + a_2+a_3=1\]Now I kinda see how squaring can get us somewhere if not all the way through.\[a_1^2 = a_2^2 + a_3^2 - 2a_2a_3\]\[a_2^2 = a_1^2 + a_3^2 - 2a_1a_3\]\[a_3^2 = a_1^2 + a_2^2 - 2a_1a_2\]That gives us this thing:\[2(a_1a_2 + a_2a_3+a_3a_1) = a_1^2+a_2^2 + a_3^2 \]Probably some inequality?

Answer 3

lol 6 puppies divides 30 puppies around a circle.

Since \(a_{n+2} =|a_{n+1} - a_n |\) for all n, then that means all the \(a_n\) are positive. So:

\[1= \sum_{n=0}^{29} a_n = \sum_{n=0}^{29} |a_{n+1} - a_n |\]

So for all the \(a_n\) we have:

\[0 \le a_n < 1\]

and I think we can't have repeats, I guess that's my next thing to look into.

Answer 4

Yeah I like your idea I'll try looking into squares and smaller cases of 3 people as well seems simpler. Maybe I'll do 4 people cause it's an even number.

Answer 5

Sorta looking at this, taking it two steps deep \[a_{n+4}=|a_{n+3}-a_{n+2}| = ||a_{n+2}-a_{n+1}|-|a_{n+1}-a_n||\]

Answer 6

Is it possible that the sequence is:

0,1,1,0,1,1,0,...

Answer 7

arg no that will never all add up to 1, it's already more than 1.

Answer 8

BUT what if we take that, count the number of 1s that would be in it, and then divide by the total? :O

Answer 9

Ahahaha! Yes!

0,1,1, repeat this cycle to 30, and you see that \(\frac{2}{3}\) of them are 1s. So let's call the terms \(b_n\) and if you add them all together you get:

\[\sum_{n=0}^{29} b_n = 30*\frac{2}{3} = 20\]

So, divide the equation by 20 to get

\[\sum_{n=0}^{29} \frac{b_n}{20} = 1\]

\[0,\frac{1}{20},\frac{1}{20},0,\frac{1}{20},\frac{1}{20},...\]

Answer 10

I think this isn't the only answer, since any cyclic subsequence that divides 30 will also be able to be fit into this as well. Like suppose instead of the sequence having period 3 like mine did, we find one with period 10 or 6, then that can also be normalized in the same way. I don't even know if the one I found is the only one of period 3 that works either.

Answer 11

In general if you find a periodic subsequence of period \(T\) that divides 30, then:

\[N=\sum_{n=0}^{T-1}a_n\]
\[1= \sum_{n=0}^{29} \frac{T*a_n}{N*30}\]

Answer 12

YOU'RE A GENIUS! A MADMAN!

Answer 13

If you asked me to lay on the floor I'd do it, cause I'm an absolute madman!