Abstract

# Abstract

I investigate the use of the notation $$d;n$$, meaning repeat the digit $$d$$, $$n$$ times. The convergence of rational expressions and the primality of larger sequences are explored. A way of ruling some non-trivial numbers out in $$\mathrm{log}_{10}(digits)$$ is found for prime searches. The relationship of numbers written in such notation and thier prime decompositions is investigated.

# Investigation

Let $$d;n$$ mean repeat the digit $$d$$, $$n$$ times. Let $$|$$ be a digitwise concatenation, such that $$2|3;4|6=233336$$ This allows constants to be created for example $C=\lim_{n\to\infty} \frac{1|3;n|8}{2|8;n|7}$

For a given sequence it is not entirely clear if convergence will be achieved. We may take a partial series. For $$n=24$$ we have $$C_{24}=0.461538461538461538461538653254437869822485207100604251251706...$$ which appears to repeat at first with a $$6$$ digit sequence $$.461538$$, repeating $$4$$ times. FOr $$n=40$$ we have $$C_{40}=0.461538461538461538461538461538461538461557633136094674556213...$$ which repeats 6.5 times. If we assume convergence we could then argue that $C_\infty = 0.\overline{461538} =\frac{6}{13}$

This doesn’t sound, too ridiculous when one considers that $$0.|9;\infty=1$$ However one can notice that $\lim_{n\to\infty} \frac{1|3;n|8}{2|8;n|7}=\lim_{n\to\infty} \frac{1|3;n}{2|8;n}$

In fact the end digits wil be very small, so as we go to infinity, this information will probably be lost.

$\lim_{n\to\infty} \frac{1|3;n}{0|8;n} = 1.5\overline{0} =\frac{12}{8} \\ \lim_{n\to\infty} \frac{1|3;n}{1|8;n} = 0.\overline{7058823529411764} =\frac{12}{17} \\ \lim_{n\to\infty} \frac{1|3;n}{2|8;n} = 0.\overline{461538} =\frac{12}{26} \\ \lim_{n\to\infty} \frac{1|3;n}{3|8;n} = 0.3\overline{428571} =\frac{12}{35} \\ \lim_{n\to\infty} \frac{1|3;n}{4|8;n} = 0.\overline{27} =\frac{12}{44} \\$

Which would at least imply $\lim_{n\to\infty} \frac{1|3;n}{d|8;n}=\frac{12}{(d|8) - d}$

Then we have $\lim_{n\to\infty} \frac{1|2;n}{0|8;n} = 1.375\overline{0} =\frac{11}{8} \\ \lim_{n\to\infty} \frac{1|2;n}{1|8;n} =0.\overline{6470588235294117} = \frac{11}{17} \\$

Which would at then imply $\lim_{n\to\infty} \frac{1|d_1;n}{d_2|8;n}=\frac{(1|d_1) - 1}{(d_2|8) - d_2}$

$\lim_{n\to\infty} \frac{0|2;n}{1|8;n} =0.\overline{1176470588235294} = \frac{2}{17} \\ \lim_{n\to\infty} \frac{1|2;n}{1|8;n} =0.\overline{6470588235294117} = \frac{11}{17} \\ \lim_{n\to\infty} \frac{2|2;n}{1|8;n} =1.\overline{1764705882352941} = \frac{20}{17} \\ \lim_{n\to\infty} \frac{3|2;n}{1|8;n} =1.\overline{7058823529411764} = \frac{29}{17} \\$

Which would at then imply $\lim_{n\to\infty} \frac{d_1|d_2;n}{d_3|8;n}=\frac{(d_1|d_2) - d_1}{(d_3|8) - d_3}$

And then perhaps $\lim_{n\to\infty} \frac{d_1|d_2;n}{d_3|d_4;n}=\frac{(d_1|d_2) - d_1}{(d_3|d_4) - d_3}$

Hence if the beginning digits are made to be $$0$$. We have $\frac{a;\infty}{b;\infty}=\frac{a}{b}$

Which can only be true, if the digits are single digits, i.e $$d\in[0,9]$$ Thus for longer digit phrases for example $10;5 \ne 1010101010 \\ 10;5 = 111110$

However such a phrase allows for insights such as $10;5=(1;5)|0 \\ 100;5=(1;5)|00\\$

Or more generally $(d|0;n);m =(d;m)|(0;n), \;\; d\in[0,9]$

Likewise we have $(0;n|d);m=(0;n)|(d;m), \;\; d\in[0,9]$

For the pipe and semi-comma operations we have identity $0|a = a \\ a;1 = a \\ a;0 = 1 or 1???$

Interesting expressions such as $10101;1=10101 \\ 10101;2=111111 \\ 10101;3=1121211 \\ 10101;4=11222211 \\ 10101;5=112232211 \\$

Drive a cellular automaton? Pascals triangle generator?