Existence and a construction of an efficiently computable real number with predetermined values of symbols in all of its representations in base $b$ Assuming that $b \geq 2$ and $n \geq 1$, let $f(x, b, n)$ denote the numeric value of an $n$-th symbol of the fractional part of the base-$b$ representation of a real number $x$ (here $n \geq 1$ implies that the first symbol of the fractional part has index $1$). For example, $$\begin{array}{l}
f(\pi, 2, 1) = 0,\\
f(\pi, 2, 2) = 0,\\
f(\pi, 2, 3) = 1,\\
f(\pi, 3, 4) = 2,\\
f(\pi, 5, 7) = 1,\\
f(\pi, 10, 5) = 9,\\
f(\pi, 12, 10) = 11,\\
f(\pi, 16, 4) = 15.
\end{array}$$
Does there exist an irrational real number $v$ such that $f(v, 2t+1, 2t+1) = t$ for any natural number $t \geq 1$? If yes, what can be an efficiently computable example of such a number? Here “efficiently computable” means that there is an efficient algorithm which, given any pair of $b$ and $n$, allows to compute the corresponding value of $f(v, b, n)$.
 A: First, $f(v,b,n)$ can be calculated by
$$\lfloor v\cdot b^n\rfloor \bmod b$$
which can be easily verified using
$$v=\sum_{n=-\infty}^{+\infty}f(v,b,n)\cdot b^{-n}$$
There must be infinitely many $v$ that meets $$f(v,2k+1,2k+1)=k\quad, \forall k\in\mathbb N$$
but one of those is the one that can be represented as follows:
$$v=\frac{a_1}{3^3}+\frac{a_2}{5^5}+\frac{a_3}{7^7}+\cdots=\sum_{k=1}^{\infty}\frac{a_k}{(2k+1)^{2k+1}}\quad (0\le a_k<2k+1)\tag1$$
Then, it is possible to find all $a_k$ sequentially starting from $k=1$ by incrementing $k$ by $1$, because

Lemma-$1$. Changing $a_j\,(j>k)$ does not affect $f(v,2i+1,2i+1)\,(i\le k)$.

Or, equivalently

Lemma-$2$. Changing $a_j\,(j>k)$ does not affect $f(v,2k+1,2k+1)$.

that can be proven by
$$\sum_{j=k+1}^{\infty}\frac{a_j}{(2j+1)^{2j+1}}
\le\sum_{j=k+1}^{\infty}\frac{2j}{(2j+1)^{2j+1}}
<\sum_{j=2k+3}^{\infty}\frac{j-1}{j^j}$$
$$<\sum_{j=2k+3}^{\infty}\frac{1}{j^{j-1}}
<\sum_{j=2k+3}^{\infty}\frac{1}{(2k+3)^{j-1}}
=\frac1{(2k+3)^{2k+2}}\sum_{j=0}^{\infty}\left(\frac{1}{2k+3}\right)^j$$
$$=\frac1{(2k+2)(2k+3)^{2k+1}}
<\frac1{(2k+1)^{2k+1}}$$
The Lemma-$1$. implies that when we define
$$v_n=\sum_{k=1}^{n}\frac{a_k}{(2k+1)^{2k+1}}$$
Then
$$f(v_{\infty},2n+1,2n+1)=f(v_n,2n+1,2n+1)$$
where $v_{\infty}$ is same to $v$ as defined in the Eq.$(1)$.
Now, the procedure to find all digits in $v$ is:
First, find
$$v_1=\frac{a_1}{3^3}$$
It's obvious that
$$a_1 = f(v,3,3) = 1$$
Then
$$v_2=v_1+\frac{a_2}{5^5}=\frac{1}{3^3}+\frac{a_2}{5^5}$$
$$f(v,5,5)=f(v_2,5,5)=\lfloor v_2\cdot 5^5\rfloor \bmod 5$$
$$=\left\lfloor\frac{3125}{27}+a_2\right\rfloor\bmod 5=(115+a_2)\bmod 5=a_2=2$$
Or
$$a_2=\left(f(v_2,5,5)+\lfloor v_1\cdot 5^5\rfloor\right)\bmod5=\left(2+\left\lfloor\frac{3125}{27}\right\rfloor\right)\bmod5=2$$
$$\vdots$$
$$a_{n+1}=\left(f(v_n,{2n+1},{2n+1})+\lfloor v_n\cdot (2n+1)^{2n+1}\rfloor\right)\bmod (2(n+1)+1)$$
$$\vdots$$
, which pretty much defines the procedure. Using this procedure, I could calculate (with MS Excel):
$$a_1=1,\,a_2=2,\,a_3=0,\,a_4=3,\,a_5=7,\,a_6=1,\,\cdots$$
$$v=0.037677045\cdots$$
(*) This way, we can also find $v$ for any
$$0\le f(v,2k+1,2k+1)<2k+1$$
not only
$$f(v,2k+1,2k+1)=k$$
as defined in the question.
