2

^{i}+ (2^{(i-1)}- 3) if i is even
or 2

^{i}+ (2^{i-1}+ 3) if i is oddI 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

2

^{i}+ (2^{(i-1)}- k) if i is even
or 2

where k is any odd integer

^{i}+ (2^{i-1}+ k) if i is oddwhere 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

2

^{i}+ (2^{(i-1)}- 7) if i is even
or 2

^{i}+ (2^{i-1}+ 7) if i is oddThe 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 2

^{3}+ (2

^{2}- 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 2

^{4}+ (2

^{3}- 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

2

^{i}+ (2^{(i-1)}- k) if i is even
or 2

^{i}+ (2^{i-1}+ k) if i is oddis 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 = 2

^{i}+ (2^{(i-1)}- 3) if i is even
or n = 2

^{i}+ (2^{i-1}+ 3) if i is odd**the order of 2 mod n is i+(i-2)**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.