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.


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?