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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.

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  • $\begingroup$ @Tyler6 hahah thank you for he edit. Me thinking about prime numbers has confused myself :) $\endgroup$
    – Mr Pie
    May 10 '18 at 3:04
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    $\begingroup$ No worries, we’ve all been there ahah $\endgroup$
    – Tyler6
    May 10 '18 at 3:05
  • $\begingroup$ Here's the continued fraction for $0.23571379\dots$: oeis.org/A071775 $\endgroup$ May 20 '18 at 8:54
  • $\begingroup$ @GerryMyerson thanks for that, but both decimals are not quite the same :) $\endgroup$
    – Mr Pie
    May 20 '18 at 8:56
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    $\begingroup$ I don't know what you mean. We're talking about the decimal made from the last digit of the primes, right? That's what the OEIS reference is about, too. $\endgroup$ May 20 '18 at 9:05
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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.

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    $\begingroup$ I appreciate your efforts. I found it rather trivial it was irrational since there are infinitely many primes, so infinitely many decimal places. And, given some formulas by Hardy and Littlewood, we know that there are not infinitely many primes continuing with a certain digit from in a row, so the decimal would not be terminating. Nonetheless, $(+1)$ :) $\endgroup$
    – Mr Pie
    May 10 '18 at 3:44
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    $\begingroup$ @NilotpalKantiSinha the Copeland Erdös Constant is the concatenation of all primes, this is the concatenation of only their last digits $\endgroup$
    – Tyler6
    May 10 '18 at 6:29
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    $\begingroup$ "I found it rather trivial it was irrational since there are infinitely many primes, so infinitely many decimal places." By that logic, $1/3=.333\dots$ is irrational, since it has infinitely many decimal places. $\endgroup$ May 20 '18 at 8:49
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    $\begingroup$ @user, OK. In the first place, Hardy & Littlewood didn't prove, or even "essentially" prove, anything of the kind: they made a conjecture, a conjecture which remains unproved to this day. And if the conjecture is correct, then a prime ending in a digit $d$ is followed by a prime ending in $d$ less than one-fourth of the time. But how much less than one-fourth? The difference goes to zero as you look at more and more primes. Continued next comment.... $\endgroup$ Jun 15 '18 at 6:37
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    $\begingroup$ ....But never mind – that has nothing to do with periodicity. The primes could end in 1, 1, 3, 1, 7, 1, 9, 3, 3, 7, 3, 9, 7, 7, 9, 9 repeating that pattern forever, and each digit would be followed by each other digit exactly one-fourth of the time. $\endgroup$ Jun 15 '18 at 6:37

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