Constructing number between zero and one by concatenating digits from square root of primes

Question

Let $a_{n}=$ the first $p_{n}$th digits to the right of decimal point of the square root of the $n$th prime.

Example:

$\sqrt{2}=1.414213562...$

So, $a_{1}=41$

$\sqrt{3}=1.73205080...$

So, $a_{2}=732$

$\sqrt{5}=2.236067977499789696409173668731276...$

So, $a_{3}=23606$

$\sqrt{7}=2.64575131106459059050161575363926...$

So, $a_{4}=6457513$

$\sqrt{11}=3.316624790355399849114932736...$

So, $a_{5}=31662479035$

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The number I want to construct:

$0.a_{1}a_{2}a_{3}a_{4}a_{5}...=0.4173223606645751331662479035...$

This number exists by monotone bounded sequence theorem applied to the sequence of partial sums of the series forming the number.

Could I say that this constant is transcendental?

Update at $12$:$21$AM of May/$31$/$2024$:

I have added the following summation formula for the number:

$0.a_{1}a_{2}a_{3}a_{4}a_{5}...=\sum_{n=1}^{\infty} a_{n}10^{- \sum_{j=1}^{n} p^{j}}=\sum_{n=1}^{\infty} \lfloor10^{p_{n}}(\sqrt{p_{n}}-\lfloor \sqrt{p_{n}} \rfloor) \rfloor 10^{- \sum_{j=1}^{n} p^{j}} $

Answer 1

I'd like to propose a probabilistic argument for the following:

$M=0.a_{1}a_{2}a_{3}...$ is impossible to be periodic, which would mean it is irrational.

Assume $M$ is periodic, so $M$ has period length equaling $m \in \mathbb{N}$.

Let $k_{1}k_{2}...k_{m}$ be the repeating block of digits in $M$.

By definition of $a_{n}$, there exists sufficiently large $s$ such that $a_{s} \geq2m>m$.

This means that we would have the block $k_{1}k_{2}...k_{m}$ is contained somewhere in $a_{n}$ for all square roots of primes $p_{n}$ at least one time when $n \geq s$.

I think this is unlikely to happen.