Discrete vs Continuous approach
To give some background on the rationality of these arguments, we refer to the concepts of a discrete probabilistic model and a continuous probabilistic model.
Essentially, the model I have used relies on intervals of the form $\alpha_{p,k}$ which can be thought of as cycles of residues modulo $p$. The cycles themselves are treated as discrete packages, whereby we know that $p_n \#$ is the end of the cycles associated with prime numbers upto $p_n$, so we can build this discrete probabilistic model of totatives to the $n$th primorial number. Thereby the rest of this approach relies on discrete probability theory. That is, it relies on being able to identify the amount of intervals of the form $\alpha_{p,k}$ in intervals smaller than $P_n$.
If we don't, we are in effect treating this like a continuous probabilistic model, requiring the calculation of variance to compensate for the differences between the actual discrete probabilistic situation and mathematical tradition, where I am in particularly referring to the probabilities of residue conditions in smaller intervals smaller than $P_n$. Although the variance approach is valid, for now it is simpler to ignore it.
Traditionally, the OP question has been approached with a disregard of using the cycles I mentioned as discrete packages, producing results which rely on variance and are attached to general integers as opposed to numbers relating to $p_n\#$. The reason for this change by me, is that we can directly relate prime numbers to primorial numbers; where all prime numbers less than some $p_{n+1}^2$ are those in the interval $\lbrace p_n, .., (p_{n+1}^2 -1) \rbrace$ which are totatives to the $n$th primorial, essentially the aim of my approach is to model the of distribution prime numbers directly from primorial numbers.
$\Lambda(n,2)$
Consider the interval $P_n=\lbrace0,..., p_n\# -1\rbrace$.
Let $L[P_n, q]$ be the interval $P_n$ viewed as a union of $\alpha_q$ intervals, with $q$ prime $\leq p_n$ andLet $|L[P_n, q]|$$L|S(n,a), q|$ be the number of $\alpha_q$ intervals that make upof the interval $P_n$. Now letform $1/a \times P_n$ be any interval that$\alpha_{q,k}$ whose union is defined defined byequal to an interval $(n,a)$, where$s \in S(n,a)$. Restrict $a$ isto integers which are only divisible by each of itsit's prime factors only once, and $a \leq p_n\#$. It follows that immediately from a conjecture stated above.
$L|S(n,1), q| =p_n\# /q$ where $q$ is prime number less than or equal to $p_n$
Consider an interval $1/a \times P_n$. In terms of lights,Also we havecan derive the following identity.result;
$L|S(n,a), q| = 1/a \times L|S(n,1), q|$ iff $gcd(a,q)=1$.
Remember $|L[1/a \times P_n, q]| = 1/a|L[p_n, q]|$ iff$P_n = \lbrace 0, ..., (p_n \#) -1 \rbrace$ is the only interval in $gcd(a,q)=1$$S(n,1)$. This is becauseSo the conjecture above results from the fact that $p_n \#$ divided by$L|S(n,1), q| = p_n \# / q $ and if $q: gcd(a,q)=1 $$gcd(a,q)=1 $ then$ L|S(n,1), q|$ is still divisible by $a$, but giving us a discrete amount of intervals of the form $\alpha_{q,k}$ whose union is some $s \in S(n,a)$. However if $gcd(a,q)>1$, then $p_n \#$ divided by $q$L|S(n,1), q| is not divisible by $a$.
There are half as many $\alpha_3$ intervals inof the form $1/2 \times P_n$$\alpha_{3,k}$ whose union is an interval $s \in S(n,2)$, as there isare of inintervals of the form $\alpha_{3,k}$ whose union is the interval $P_n$, in fact there.
There are half as many $\alpha_q$ intervals inof the form $1/2 \times P_n$$\alpha_{q,k}$ whose union is an interval $s \in S(n,2)$, as there is of inintervals of the form $\alpha_{q,k}$ whose union is the interval $P_n$ wherefor all $2<q\leq p_n$.
In evaluating $\Lambda(n,2)$ remember that $L|S(n,a), q| = 1/a \times L|S(n,1), q|$ iff $gcd(a,q)=1$ therefore $(L|S(n,2),q|)/(L|S(n,1), q|) = 1/2 $ for all $q : gcd(2,q)=1$. Therefore the statement is true for all $q \leq p_n$ except $q=2$.
So consider $L|S(n,2), 2|$. Remember $L|S(n,1),2| = $p_n# /2$.
Note that $p_n# /2$ is odd therefore $(p_n# /2) -1$ is even. Therefore $L|S(n,2), 2]| \geq ((p_n# /2) -1)/2 $.
Consider $L[1/2 \times P_n, 2]$. In the interval $P_n$ there is $p_n\# /2$ amount of $\alpha_2$ intervals, i.e $L[P_n, 2]=p_n\# /2$. (This can be generalized to say that $L[P_n, q]=p_n\# /q$ where $q$ is prime $\leq p_n$.) Note, $p_n\# /2$ is odd therefore $(p_n\# /2) -1$ is even$L|S(n,2), 2]| = b \times L|S(n,1),2]|$. Therefore there are at least $((p_n\# /2) -1)/2$ amount of $\alpha_2$ intervals inThen $1/2 \times P_n$. I$b \times (p_n\# /2) \geq (((p_n\# /2) -1)/2)$.e And so $|L[1/2 \times P_n, 2]| \geq ((p_n\# /2) -1)/2 $$b = (((p_n\# /2) -1)/2) / (p_n \# /2) = (p_n \# /2) -1) / (p_n \#)$.
Consider $|L[1/2 \times P_n, 2]| = b \times |L[P_n,2]|$, then $b \geq (((p_n\# /2) -1)/2) / (p_n\# /2) = ((p_n\# /2) -1) / (p_n \#)$. Note that $1/2 \geq( (p_n\# /2) -1) / (p_n \#)$$1/2 \geq( (p_n\# /2) -1) / (p_n \#) = b$. Therefore $|L[1/2 \times P_n, q]| \geq b \times |L[P_n, q]|$$L|S(n,2), q| \geq b \times L|P_n, q|$ for all $q$ which are prime numbers less than or equal to $\leq p_n$$p_n$.
Therefore $(n,2) \geq $b \times (n,1)$$\Lambda(n,2) \geq b \times \Lambda(n,1)$.
I.e $(n,2) \geq ((p_n\# /2) -1) / (p_n \#) \times \prod_{i=1}^n p_i -1$$\Lambda(n,2) \geq ((p_n\# /2) -1) / (p_n \#) \times \prod_{i=1}^n p_i -1$. fromAnd following the demir inequality mentioned in section (n, p_{n-1}#), we get thatthat;
$(n,2) \geq ((p_n\# /2) -1)/p_n$$\Lambda(n,2) \geq ((p_n\# /2) -1)/p_n$
