Prime Number Sieve using LCM Function How to prove the following conjectures ?
Definition :
Let $b_n=b_{n-2}+\operatorname{lcm}(n-1 , b_{n-2})$ with $b_1=2$ , $b_2=2$ and $n>2$ .
Let $a_n=b_{n+2}/b_n-1$
Conjectures :

*

*Every term of this sequence $a_i$ is either prime or $1$ .


*Every odd prime number is member of this sequence .


*Every new prime in sequence is a next prime from the largest prime already listed .
Maxima implementation of sieve :
/* input number n */
load(functs);
n:200;
b1:2;
b2:2;
max:2;
k:3;
i:1;
while max<=n do (
    if i=1 then(
        print(max),
        i:0
        ),
    b3:b1+lcm(k-1,b1), 
    a:b3/b1-1, 
    k:k+1, 
    b1:b2, 
    b2:b3, 
    if max<a then (
       max:a, 
       i:1
       )
    );

Implementation of the sieve in PARI/GP can be found here.
 A: This is sequence A135506 in OEIS, and as far as I know, no proof to conjecture 1 has been published. Benoit Cloitre announced one in this paper, but I don't think it's been published.
Conjecture 2 is actually very simple to proof, I did so in my bachelor's thesis. Unfortunately, this is only in German, I can write it up in English if you're interested.
I think conjecture 3 is pretty much a direct consequence from the fact that $a_p=p$ for every prime, and this is the first time $p$ can occur in the sequence.
Edit: I actually realised that your definition of the quotient is slightly different since it goes two steps back. I'll check the argument - it shouldn't make much of a difference.
Edit: This is the full argument for conjectures 2 & 3. First we need the general relation between gcd $(a,b)$ and lcm $[a,b]$:
$$a\cdot b = (a,b)\cdot[a,b].$$
Then we note that the lowest common multiple $[n-1,b_{n-2}]$ is in particular a multiple of $b_{n-2}$, say $kb_{n-2}$ with $1\le k\le n-1$. Hence we have
$$b_n=b_{n-2}(k+1),$$
so in every step the term $b_n$ gets a new factor between $2$ and $n$ which means in particular that all prime factors of $b_n$ are less or equal to $n$.
Now we rearrange $a_n$ with the above observation to 
$$a_n=\frac{n+1}{(n+1,b_n)}.$$
Let $p$ be a prime. Then $(p,b_{p-1})=1$ since all prime factors of $b_{p-1}$ are strictly smaller than $p$. But then 
$$a_{p-1}=\frac{p}{(p,b_{p-1})}=p$$
as claimed in conjecture 2.
Further, we have obviously $a_n\leq n+1$ for all $n$, so the first index for which the prime $p$ can appear in the sequence is $p-1$ which immediately implies conjecture 3.
As mentioned before, conjecture 1 seems quite a bit more difficult and I'd be interested in a proof myself.
PS: It took me a couple of pages to prove these facts when I first worked on them 6 years ago, but it's clearly much simpler now that I thought about it again.
Edit: I wrote a little longer blog post about this sequence.
