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Let's define the following function: $F(n)$ is equal to the minimal natural number (integer) $x$ such that ${10^x}$ is greater than $n$ (assuming that $n$ is a positive natural number). Then let's define the next function (that takes two arguments): $G(x,y)$ evaluates to the $(F(x) - y + 1)$-th decimal digit of $x$ if $(F(x) - y + 1) \ge 1$ ; otherwise, it evaluates to $0$ (also assuming that $x$ and $y$ are positive integers).

Consider a set of numbers {$N_1$, $N_2$, $N_3$, ...} such that each $i$-th decimal digit of the decimal expansion of $N_k$ is equal to $G({p_i},k + 1)$, assuming that each $N_k$ starts with a zero followed by the floating point, and where $p_1 = 2$, $p_2 = 3$, $p_3 = 5$, ... (prime numbers).

To put it simply, in the decimal expansion of $N_1$ each $i$-th decimal digit is equal to the $2$nd-to-last decimal digit of $p_i$; in the decimal expansion of $N_2$, each $i$-th decimal digit is equal to the $3$rd-to-last decimal digit of $p_i$; in the decimal expansion of $N_3$, each $i$-th decimal digit is equal to the $4$th-to-last decimal digit of $p_i$ etc., that is, in the decimal expansion of $N_k$, each $i$-th decimal digit is equal to the $(k+1)$th-to-last decimal digit of $p_i$. But if we want to extract the $A$-th-to-last digit of number $B$ and $A$ is greater than the total number of digits in the decimal representation of $B$, we assume that the $A$-th-to-last digit of $B$ is $0$ (for example, the $123$-th-to-last digit of $987654$ is $0$, as well as its $12$-th-to-last digit).

For example, if we compute the first $1000000$ digits of $N_1$, that is, the sequence that corresponds to the set {$2$, $3$, $5$, $7$, $11$, $13$, $17$, ..., $15485849$, $15485857$, $15485863$}, we will get the sequence that starts with

0.00001111223344455667778890000123...  

and ends with

...93344568012556781346780034456  

The main question: can we assume that each such number contains all possible finite sequences of decimal digits? If yes (or no), how to prove this? If yes, are these numbers normal (and is it possible to prove this)? Even for three-digit sequences in $N_1$, there are a lot of examples that cannot be found in the first $1000000$ digits of this number, e.g. $208$, $209$, $210$, $198$, $298$, $598$ etc.

The additional question: how to prove that any $N_k$ is irrational and transcendental?

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    $\begingroup$ What is the point of the useless constraint $F(x)\ge y$? Do you mean the $(F(x)-y)$th decimal digit? Why don't you just use straightforward language like "the second-to-last digit of $x$" instead of defining it in this obtuse way? Why do you call it obvious that these numbers are not normal? $\endgroup$ – Erick Wong Apr 26 '17 at 8:11
  • $\begingroup$ I can't see why $N_1$ starts with 0.00001111... . Maybe you miswrote $10x$ to $10^x$? $\endgroup$ – didgogns Apr 26 '17 at 12:56
  • $\begingroup$ @ErickWong: I tried to provide a more formal definition. I edited the question and added a simplified definition. $\endgroup$ – lyrically wicked Apr 27 '17 at 5:13
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    $\begingroup$ @didgogns: the second-to-last digit of the elements of {2,3,5,7} is 0, hence 4 zeros at the beginning of $N_1$; the second-to-last digit of the elements of {11,13,17,19} is 1, hence 4 ones after 4 zeros; ...[continue]...; the second-to-last digits of the elements of {15485849, 15485857, 15485863} are 4, 5 and 6, so the first 1000000 decimal digits of $N_1$ end with "456". $\endgroup$ – lyrically wicked Apr 27 '17 at 5:21
  • $\begingroup$ @lyricallywicked Thank you for the edit. Informal definitions are often much more useful. If you have an overly technical definition and don't give any motivation, then not only is it harder for readers to follow, but if you make even a slight mistake or typo in your definition then the result will mean something completely different from what you intend. Case in point: you wrote $F(x)-2$ but you clearly meant something else: no one could be sure of this if all you gave was a wordless definition. $\endgroup$ – Erick Wong Apr 27 '17 at 5:47

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