Ladder primes, a special class of prime numbers

Question

Introduction

In a discussion with a friend we found an interesting problem. I could not identify it in the vast collection of so called super primes. Hence, and for reasons that become clear below, I coined the term ladder primes. If this problem is known please tell me, you can then most probably ignore this post.

Consider this sequence of relations

$$3+1 = 4 = 2*2$$ $$5+1 = 6 = 2*3$$ $$7+1 = 8 = 2*4$$

$$11+1 = 12 = 2*6 $$

$$13+1 = 14 = 2*7$$ $$17+1 = 18 = 2*9$$
and so on

We start with a prime number $p$, then add $1$ and divide by $2$. If the result is again a prime number, then we will say that $p$ is a ladder prime of order $1$, if not, $p$ is called ladder prime of order $0$. In the sequence above we see that $3, 5, 13$ have order $1,$ and $7, 11,$ and $17$ have order $0$.

Definitions

Definition 1: Order

A ladder prime (number) $p$ of order $k = 0, 1, 2, \ldots$ is defined as a prime for which $p + k = (k+1) q$ with some prime $q$.

The reader can readily find examples of ladder primes of order $1$.

But we can go one step further and look for a prime which has simultaneously order $1$ and order $2$.

The smallest is 13. In fact,

$13 + 1 = 2q -> q = 7$ ok
$13 + 2 = 3q -> q = 5$ ok
$13 + 3 = 4q -> q = 4$ not ok, no prime

Other examples are $37,157,541,877,1201,1381$. These all have with $13$ in common that the highest order is $2$.

The leads us to the next definition.

Definition 2: Degree

The degree $n$ of a ladder prime $p$ is defined as the highest successive order of $p$, i.e. $p$ has simultaneously order $1, 2, \ldots, n$ but does not have order $n+1$.

$p + 1 = 2$ * Prime
$p + 2 = 3$ * Prime
$p + 3 = 4$ * Prime
...
$p + n = (n+1)$ * Prime

but

$p + (n+1) != (n+2)$ * Prime

Remark: The term "ladder prime" stems from this "ladder" of equations to be satisfied by the prime number is a question.

Examples

(each times at least the 4 smallest of the kind)

Order 2
$13,37,157,541,877,1201,1381$

Order 3
$12721,16921,19441,24481$

Order 4
$19441, 266401, 423481, 539401$

Order 5
$5516281, 16831081, 18164161, 29743561$

Order 6
$5516281, 18164161, 51755761, 175472641$

Order 7
No ladder prime found up to the $10^7$th prime ($179424673$).

Questions

Q$1$: What is the number $c(p,k)$ of ladder primes of order $k$ up to a given ladder prime $p$?

Q$2$: Are there infinitely many ladder primes of any given order up to $n = 6$?

Q$3$: Are there ladder primes of order $7$, and if so, how many?

Q$4$: Same as Q$1$ but for the degree $d(p,k)$

Answer 1

Suppose $p$ is an odd prime, and suppose $\frac{p+1}{2}$ and $\frac{p+2}{3}$ are prime.

Then $p$ is congruent to $1$ modulo $6$, so we may write $p=6k+1$ for some integer $k$.

So the question of whether or not there are infinitely many primes $p$ such that $\frac{p+1}{2}$ and $\frac{p+2}{3}$ are prime is equivalent to asking whether or not there are infinitely many $k$ such that $$ 6k+1, 3k+1, \text{ and }\, 2k+1 $$ are all prime. This is an open question, conjectured to be true by Dickson's conjecture.

The conjecture applies similarly for your higher order cases.

Added: As for the density of such numbers, take a look at the Bateman-Horn conjecture.