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?

# Thing

$\lim_{n\to\infty}\frac{9|8;n}{7|6;n}=\frac{89}{69}$

Try 339/108, for $$\pi$$ like test... Have $\frac{(d_1|d_2)-d_1}{(d_3|d_4)-d_3}=\frac{339}{108} \to \frac{(4|03) - 4}{(1|09)-1}$

Consider $03;5 = 033333 \\ 09;5 = 099999 \\$

attempt to see error convergence with added digits$\frac{403}{109}-\pi= 0.555655052832225110161209827729671427729436105212050142327807...\\ \frac{4033}{1099}-\pi= 0.528107073434774550436355706802389745466160900897869611236156...\\ \frac{40333}{10999}-\pi=0.525377070930617707987033860106259457833924336419057284761940...\\$

# Pascals

Try to generate pascals triangle or similar. $010 \\ 0110 \\ 01210 \\ 013310 \\ 0146410 \\$

The first line is $$010;1$$.
The second line is $$010;2$$.
The third line is $$(010;2);2$$.
The n^th line is $$010;2_1;2_2;2_3...;2_{n-1}$$
We can attempt similar triangles with other starting strings.

1 1;3= 111 111;3=

111 111 111 = 12321

12321;3=

12321 12321 12321 = 1367631

1367631;3

1:3:6:7:6:3:1 :1:3:6:7:6:3:1 :1:3:6:7:6:3:1

1:4:10:16:19:16:10:4:1

$1 \\ 1:1:1 \\ 1:2:3:2:1 \\ 1:3:6:7:7:6:3:1 \\ 1:4:10:16:19:16:10:4:1$

12

12 12 =132

132 132 ==1452

1452 1452 = 15972

$12 \\ 132 \\ 1452 \\ 15972 \\$

1001

1001 1001 = 11011

11011 11011 = 121121

121121 121121 = 1332331

1332331 1332331 = 14655641

$1001 \\ 11011 \\ 121121 \\ 1332331 \\ 14655641$