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Given $n\in[0,1]$ with base-b expansion $0.n_1n_2n_3\dots$, define $\Delta_b(n)$ to be the number with the following base-b expansion:

$\huge{ 0.\underbrace{n_1}_{1^{st}\text{ block}}\overbrace{n_1n_2}^{2^{nd}\text{ block}}\underbrace{n_1n_2n_3}_{3^{rd}\text{ block}}\cdots \overbrace{n_1n_2\dots n_i}^{i^{th}\text{ block}}\cdots }$

This is clearly irrational unless $n = \frac{i}{b-1}$ for $i \in \mathbb{Z}$. Under what contexts can it also be shown to be transcendental? Specifically is it transcendental:

  1. When n is transcendental?
  2. When n is transcendental with irrationality measure > 2?
  3. When n is Liouville?
  4. When n is rational?

It feels like Roth's theorem may handle the last one and maybe the one before, but I'm not sure.

[Note that while this is similar to Is $ 0.112123123412345123456\dots $ algebraic or transcendental?, neither implies the other, since the 10th block of that number is $12345678910$ while the 10th block of $\Delta_{10}(C_{10})$ is just $1234567891$ (and the 11th block $12345678910$).]

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Uri Granta
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    Note that you want to exclude $n = \frac i{b-1} = 0.iiii\ldots$, not specifically $\frac i9$. – AlexR Mar 13 '15 at 12:06
  • Are you, in a given base $b$, assuming the sequences in the blocks are cycling from $1$ to $b-1$ then starting again at $1$ and so on? For example in base $4$ the blocks would be $1,12,123,1231,12312,$ etc? [If so in base $2$ your number is $.111...=1$ but in higher bases is at least irrational.] – coffeemath Mar 13 '15 at 12:11
  • @coffeemath: no. There is no cycling in the individual blocks. Hence the blocks for $\Delta_4(C_4)$ would be 1, 12, 123, 1231, 12310, 123101, 1231011, etc. – Uri Granta Mar 13 '15 at 12:18

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