So I was messing with polynomials and I encountered the following equation: $$26214x^3 - 27761x^2 - 71019x - 21667 = 0.$$ Solving for $x$ using the cubic formula, I got three solutions (as expected, pursuant to the FTOA, namely, the Fundamental Theorem of Algebra).
Let's call the equation $p(x)$ and I will denote by $p(x)_n$ the $n^{\text{th}}$ root of $p(x)$.
$p(x)_{1,2}<0$ but $p(x)_3>0$. In fact, $$p(x)_3 = 2.3571379391713739171440\ldots$$ Notice that we begin with the first four primes $2,3,5,7$ and then we go to $1,3,7,9$. Also notice that the next four primes after $7$ are $11,13,17,19$. That's when I realised that $p(x)_3$ has its decimal places being the last digit of primes, apart from $4,4,0$.
Question:
Let $d_n$ be the last digit of the $n^{\text{th}}$ prime, then is the decimal $2.d_2d_3d_4\ldots$ transcendental? Does it have a formula? Can a formula be constructed?
Thank you in advance.
Transcendence beats me, but I have a proof for irrationality:
To begin, notice that other than the primes $2$ or $5$, no prime can end in any of $0$, $2$, $4$, $5$, $6$ or $8$, since this would imply divisibility by $2$ or $5$. Therefore we see that for all primes other than those two, they must end with either $1$, $3$, $7$, or $9$.
Shiu proved (https://londmathsoc.onlinelibrary.wiley.com/doi/pdf/10.1112/S0024610799007863) that if $a$ and $q$ are coprime integers, there exist arbitrary long strings of prime congruent to $a \bmod q$. For our purpose, this implies that in our number, $2.a_1a_2...$, there exist arbitrary long strings of $1$s, $3$s, $7$s, and $9$s. This is sufficient to prove it cannot be rational.