Multiple Positions

Previously, I generated a sequence using the following recurrence relation and I made a few conjectures about it.

2i + (2(i-1) - 3) if i is even 
or 2i + (2i-1 + 3) if i is odd

I also wrote an algorithm based on this recurrence relation to generate the first few such integers up to some limit. The first 20 terms of the sequence with initial term i = 3 are:

Obviously, each term in this sequence is divisible by 1 or more primes.

Claim: If a term in this sequence at position b is divisible by a prime p then the term at position
b + (p-1)  is divisible by p.

This claim holds for all other recurrence relations of the form

2i + (2(i-1) - k) if i is even 
or 2i + (2i-1 + k) if i is odd

where k is any odd integer

It is easy to see it because it is a result of the fundamental theorem of algebra in that every integer can be represented as a product of primes.

For example, the first 20 terms of the sequence generated with the recurrence relation

2i + (2(i-1) - 7) if i is even 
or 2i + (2i-1 + 7) if i is odd


The first term 17 is a prime at position 0, and the term at position 0 + (17 - 1) = 16 is divisible by 17.
In fact 1572857/17 = 92521.

Similarly, 55 = 11* 5 is at position 1, so the term in the sequence at position 1 + (11 - 1) = 11 is divisible by 11 and the term at 1 + (5 - 1) = 5 is divisible by 5.

I'm willing to bet that the 90th term in the sequence is divisible by 89 and the 201st term is divisible by 199.

377 = 13*29 is at position 4 in the sequence and the term in the sequence that's at position 4+(13-1) = 16 is 1572857, which is divisible by 13. In fact, the term at position 4+(29-1) = 32 is 103079215097 and it is divisible by 29 among other integers.

And so on.

One of the claims I made for the sequence where k = 3 is that every single element in it is divisible by 3. It is also easy to see why this is so using the claim above since the first 2 terms of the sequence at position 0 and 1 respectively are divisible by 3. Namely, if the initial i = 3 then clearly the first term of the sequence at position 0 is 23 + (22 - 3) = 15, which is divisible by 3 so the term at position 0+(3-1) = 2 will also be divisible by 3. Therefore, every second term will be divisible by 3. To calculate the term at position 1, I plug in i = 4 so 24+ (23 - 3) = 21, which is also divisible by 3 so every third element of the sequence will also be divisible by 3. 

Similarly, if k is equal to any other odd multiple of 3, then the sequence generated using the relation

2i + (2(i-1) - k) if i is even 
or 2i + (2i-1 + k) if i is odd

is composed of terms such that every term is divisible by 3.

I also made another claim related to the sequence when k = 3. I claimed that for each term n where

n = 2i + (2(i-1) - 3) if i is even 
or n = 2i + (2i-1 + 3) if i is odd

the order of 2 mod n is i+(i-2) 

I haven't been able to see why this is the case yet but it holds for the first 50 terms of the sequence, and beyond as far as I can see.

Below is a graphic I made of different sequences generated for different k with initial i = 4.

The interesting thing about this is that only the sequence for k = 1 exists in the OEIS.org database of sequences but it is generated using a different relation, and the sequence itself is an amalgamation of sequences concerning powers of 3.

Neither of the other sequences I presented here have found their way into the OEIS.org database of integer sequences.