Vivek
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James, I accidentally noted the pattern of consecutive numbers in the sequences. And checked whether the same pattern could work for any N and it did ðŸ™‚ I think the proof is not profound.
Creating puzzles like this, opens up lots of interesting stuffs. Keep doing that. Will have an eye on your site… ðŸ™‚

I too love to learn number theory deeply, James. I got some introduction to number theory from the Discrete Math book by Kenneth Rosen. Also “A friendly introduction to number theory” by Joseph Silverman, is one nice book.

James, Got it ðŸ˜€ The proof is easy and the sequence can be generated with simple rules for every N. Before We noted the cases for N=P and N=P+1. Initially we said P is the prime. But It can be any odd number as sum of first n natural number is divided by n, if n is odd. So, If we can get the sequence for the odd number, then from that the sequence of the next even number can be generated.
Now The rules to create the sequence for odd numbers is as follows :
Suppose N=2m+1.
Then, a1=2m+1, a2=1.
a(2k)=m+k , for all 2<=k<=m
a(2k+1)=k+1, for all 1<=k<=m
This is it. This will generate our sequence. It is easy to note that a2 divides a1.
Also, It is easy to prove that
S(2k)=(m+k)(k+1), for all 1<=k<=m
S(2k+1)=(m+k+1)(k+1) , for all 1<=k<=m
From this, We can see a(2k+1) divides S(2k) and a(2k+2) divides S(2k+1).
The easy way to form the sequence is as follows:
Consider the case when N=13.
We start with 13,1,2 subsequently we write consecutive numbers 3,4,5 and so on till (13+1)/2 that is 7 with single gap between the terms. After that winding up to the first gap, we continue filling the gaps with consecutive numbers 8,9,10 and so on, till all the gaps are filled.
Step 1: 13 , 1 , 2 , x , 3 , x , 4 , x , 5 , x , 6 , x , 7
Step 2: 13 , 1 , 2 , 8 , 3 , 9 , 4 , 10 , 5 , 11 , 6 , 12 , 7
That is it :D.
To form the sequence for 14, We just replace 13 , 1 with 14 and append 13 , 1 at the end.
14 , 2 , 8 , 3 , 9 , 4 , 10 , 5 , 11 , 6 , 12 , 7 , 13 , 1
Another way to form the same sequence:
a1=2m+1
a2=1
a3=2
ak = S(k1)/a(k1) , for all 4<=k<=2m+1

I always wonder at the creativity level of the puzzle creators, James !! Yeah, This one is really a good puzzle. I too wonder whether the solution is possible for all N. It would be awesome If we could come up with some proof.

Thank you ðŸ™‚ Hope you too had awesome days. In this one, Used slanted parabolas to get that heart shape. Later came to know that there are precise nice mathematical heart curves. ðŸ˜›

Interesting (When worlds collide?) it is. I too think about it. I am not sure whether I miss something. Wondering about this : Conduct the fair lottery. If only all the individuals in the group win, That group will be selected. Otherwise the entire group will be rejected. If none of the group wins, Conduct the lottery again ?

Usually the standard explanation goes like this : Consider for example , X1+X2+X3=5. The solution can be considered equivalent to 5 *s separated by 2 s . For example (1,2,2) corresponds to ***** . In the general case, M *s and N1 s and that gives ((M+N1) choose (N1))

Before your “pretty solution clue”, I got that through lots of equations , changing 1/2 to sin 30 and the like. haha..This is cool! ðŸ™‚