Question about the Collatz Conjecture, Nicomachus's Triangle, and more Using $(3p+1)/2$ starting with $p = 44102911$, we find an ordered set of $8$ primes.  By computer, we find that this is the only ordered set of $8$ primes $< 300000000$ primes.
The primes:
$\{44102911, 66154367, 99231551, 148847327, 223270991, 334906487, 502359731, 753539597\}$
When we take the differences and remove the common factor, $344554$, we have:
$\{64, 96, 144, 216, 324, 486, 729\}$
As you can see, this is the $7th$ row of Nicomachus's Triangle.
A036561

Question: What would it take to show that a count of $8$ is the upper bounds for this ordered set?
Terry Tao has a recent blog about this.
Collatz on arxiv and another 
Collatz on arxiv.
Set of $7$ primes
$\{89599, 134399, 201599, 302399, 453599, 680399, 1020599\}$
When we take the differences and remove the common factor, $1400$, we have:
$\{32, 48, 72, 108, 162, 243\}$
As you can see, this is the $6th$ row of the triangle.
 A: The numbers $p_i$ in your list are of the form $p_i=344554\cdot2^{7-i}3^i-1,$ for $i=0,1,\ldots,7$, so there is nothing mysterious about the appearance of a row from Nicomachus' triangle, as
$$
\frac{p_{i+1}-p_i}{344554}=3^{i+1}2^{6-i}-3^i2^{7-i}=3^i2^{6-i}(3-2)=3^i2^{6-i}.
$$ 
Of course, it is rare that they would all primes, but as Gerry Myerson explained, there is no obvious reason, why we couldn't find even longer sequences of primes given that the residue class of $-1$ is a fixed point for the mapping $x\mapsto(3x+1)/2$ modulo $q$, $q$ any odd prime.
As an example of a constraint to your search for such sequences, implicit in Gerry's answer, let us consider the following. If you search for such a sequence of length ten, then all the primes in that sequence must be congruent to $-1$ modulo $11$. This is because the mapping $x\mapsto(3x+1)/2$ permutes the other residue classes modulo $11$ as a 10-cycle:
$$
6\mapsto 4\mapsto 1\mapsto 2\mapsto 9\mapsto 3\mapsto 5\mapsto8\mapsto 7\mapsto 0(\mapsto 6),
$$
so any integer $x$ that is not $\equiv-1\pmod{11}$ will become divisible by eleven after at most nine iterations.

Trying to address the question on the emergence of Nicomachus' triangle. Assume that we start a Collatz run from the number $a_0=m\cdot 2^k-1$, where $k>0$ and $m$ is an odd integer. Observe that every odd positive integer $a_0$ can be written in this way. As $a_0$ is odd, the next Collatz step gives $3a_0+1=m\cdot2^k\cdot3-2$. This is an even number, so the next step yields $a_1=m\cdot2^{k-1}\cdot3-1.$ If $k>1$, then we can repeat the same reasoning, and two further steps of Collatz gives us
$a_2=(3a_1+1)/2=m\cdot2^{k-2}\cdot3^2-1.$ The logic will be broken after $2k$ Collatz steps, because $a_k=m\cdot2^03^k-1$ is an even number, so the two types of steps stop alternating at this point.
Assume that the numbers $a_i$ are primes in the range $i=0,1,\ldots,\ell$. Obviously then $\ell<k$. Let us consider the differences 
$\Delta_i=a_{i}-a_{i-1}$, for $i=1,2,\ldots,\ell$. Then
$$
\Delta_i=(m\cdot2^{k-i}3^i-1)-(m\cdot2^{k-i+1}3^{i-1}-1)=m\cdot2^{k-i}3^{i-1}(3-2)
=m\cdot2^{k-i}3^{i-1}.
$$
We immediately see that the greatest common divisor of the numbers $\Delta_i,i=1,2,\ldots,\ell$ is $d=m\cdot2^{k-\ell}.$ This is because, we always have $3\Delta_i=2\Delta_{i+1}$. Thus the numbers
$$
\frac{\Delta_i}{d}=\frac{m\cdot2^{k-i}3^{i-1}}{m\cdot2^{k-\ell}}=2^{\ell-i}3^{i-1}
$$
form the $\ell^{\text{th}}$ row Nicomachus' triangle with $i$ ranging from $1$ to $\ell$.
A: I would expect you'd get arbitrarily long sequences of primes by iterating $(3p+1)/2$. You're asking whether some finite set of linear expressions can be simultaneously prime. The general belief is that it can be, if there is no simple congruential reason why it can't. For small primes $q$, just make sure $p+1$ is a multiple of $q$. For sufficiently large primes $q$, there should be no problem. 
