Finding the numbers $n$ such that $2n+5$ is composite. Let $n$ be a positive integer greater than zero. I write 
$$a_n =
\begin{cases}
1 , &\text{ if } n=0 \\
1 , &\text{ if } n=1 \\
n(n-1), & \text{ if $2n-1$ is prime} \\
3-n, & \text{ otherwise}
\end{cases}$$
The sequence goes like this $$1,1,2,6,12,-2,30,42,-5,72,90,-8,132,-10,-11,\ldots$$ I would like to prove the following two claims.

claim 1 : If $a_n>0$ and
  ${a_n \above 1.5 pt 3} \notin \mathbb{Q}$ then $\sqrt{4a_n+1}$ is prime. 

The table below illustrates what I am seeing: 
\begin{array}{| l | l | l | l }
\hline
n & a_n  & {a_n \above 1.5 pt 3} & \sqrt{4a_n+1}\\ \hline
0 & 1 &  .333333.. & 2.2360679.. \\ 
1 & 1 &  .333333.. & 3 \\ 
2 & 2 &  .666666..  & 3   \\ 
3 & 6 &  2   & 5   \\ 
4 & 12 &  4   & 7   \\ 
6 & 30 & 10   & 11    \\ 
7 & 42 & 14  & 13     \\ 
9 & 72 & 24  & 17 &   \\ 
10 & 90 & 30   & 19  \\ 
12 & 132 & 44   & 23  \\ 
15 & 210 & 70   & 29  \\ 
16 & 240 & 80  & 31  \\ 
19 & 342 & 114   & 37 \\ 
21 & 420 & 140   & 41 \\ 
22 & 462 & 154   & 43 \\ 
    \hline
    \end{array}

claim 2: If $a_n<0$ then $2a_n+5$ is composite

 A: $n=13$ is a counter example for claim 2.
A: Ignore $n = 0, 1$ since they're kind of irrelevant.  Then $a_n \leq 0$ for all $n$ unless $2n - 1$ is prime, by the definition.  In that case $a_n = n(n - 1)$. So, 
$$4a_n + 1 = 4(n^2 - n) + 1 = (2n-1)^2$$
So, then $\sqrt{4a_n + 1} = 2n - 1$, which is prime. 
A: Just to clarify I have that both claims are trivially true and are explicit in the definition of $a_n$. 
For claim 1 we have from @callus: 

proof of claim 1: $a_n \leq 0$ for all $n$ unless $2n - 1$ is prime, by the definition.  In that case $a_n = n(n - 1)$. So, $4a_n +
> 1 = 4(n^2 - n) + 1 = (2n-1)^2$. So  $\sqrt{4a_n + 1} = 2n - 1$,
  which is prime.

For claim 2 we have from @Thomas Andrews

proof of claim 2: Given that if $a_n<0$ then $a_n=3−n$ and $2a_n+5=11−2n$, It is hard to see why this is composite. However,
  $5−2a_n=2n−1$ is known to be composite, which means $5+2|a_n|$ is
  composite.

A: Claim 1 is vaccuously true, since given any $n$, $a_{n}\in\mathbb{Z}$.
It also looks like n=14 is a counterexample to claim 2.
Proof: 
$3-14=-11$ and $2(-11)+5=-17$ has positive divisors $1$ and $17$ only.
