Gifted Issues Discussion homepage Forums General Discussion How would you solve this?

This was DS8's "train of thought":-

He first listed out the info up to #23 and saw the pattern which was:

No. of people <=> Best Seat
1 <=> 1
2 <=> 1
3 <=> 3
4 <=> 1
5 <=> 3
6 <=> 5
7 <=> 7
8 <=> 1
9 <=> 3
10 <=> 5
11 <=> 7
12 <=> 9
13 <=> 11
14 <=> 13
15 <=> 15
16 <=> 1
17 <=> 3
18 <=> 5
19 <=> 7
20 <=> 9
21 <=> 11
22 <=> 13
23 <=> 15
: :
: :


He noticed that when the # of pple is 4, 8 16 .... the best seat would be 1 (the first seat). He then associated them with the power of 2s.

He also noticed that the increment for each cycle is by 2. Eg. [1] [1,3] [1,3,5,7] [1,3,5,7,9,11,13,15 ....] This is the part when he thought that there should be a {times 2} in his formula.

Since he didn't know how to "write out" a formula, I told him to describe to me. He started with :-

1) "Find the difference between the no. of pple (n) and the closest power of 2 which is less than n"

2) Then multiply by 2

3) Add 1 (this "1" according to him is to add back the 1st seat)

After making him explained to me like 100x , he wrote down the formula... (He can't wait to go play his lego!)

1+(n-2^) x 2
(where the power of 2 is denoted by 2^)

n = 1, 2, 3, 4, 5, 6, 7, 8, ......
nth term = [1][1,3][1,3,5,7][1,3,5,7,9,...][1,....]

The [1,3,5...] sequence would normally have this formula: 2n-1

Now, the sequence restarts when a binary number is reached.

eg for the 3rd interval
n=4,5,6,7
nth term=[1,3,5,7]

What needs to be done is to convert this n so that it starts at 1, and the formula 2n-1 can be used. To do that, we'll need to subtract n by the closest binary number that's smaller than n and add 1.

In this 3rd interval, the binary number is 4, so apply the formula (n-2^)+1

n=4,5,6,7 becomes
n=1,2,3,4 after we subtract n by 4 and add 1
we can now apply the earlier formula 2n-1 to get the nth term that we wanted.

So what we're doing is.
2((n-2^)+1)-1
--> 2(n-2^)+2-1
--> 2(n-2^)+1