Here are two somewhat strange sums using the shifted decimal forms of the powers of $3.$

$\begin{equation*}\begin{array}{ccccccc} &1&&&&&& \\ &&3&&&&& \\ &&&9&&&&\\ &&&2&7&&&\\ &&&&8&1&&\\ &&&&2&4&3&\\ &&&&&7&2&9\\ &-&-&-&-&-&-&-\\ &1&4&2&8&5&7&\cdots \end{array}\end{equation*}\ \ \ \ \ \ $ $\begin{equation*}\begin{array}{ccccccccc} &&&&&&&&1& \\ &&&&&&&3&& \\ &&&&&&9&&&\\ &&&&2&7&&&&\\ &&&8&1&&&&\\ &2&4&3&&&&&\\ 7&2&9&&&&&&\\ -&-&-&-&-&-&-&-&-\\ \cdots&2&4&1&3&7&9&3&1 \end{array}\end{equation*}$

The one on the left turns out to repeat the pattern $142857\ 142857\cdots$ if every power of $3$ is included. So putting a decimal point in front We get $\frac{1}{7}.$ This is easy to establish:

If $$x=\frac1{10}+\frac{3}{100}+\frac{9}{1000}+\cdots$$ then $$3x=\frac3{10}+\frac{9}{100}+\frac{27}{1000}+\cdots=10x-1$$ So $7x=1.$

The second sum shifts to the right. If all the powers of $3$ are used is there a periodic pattern and, if so, what does the repeating decimal with that pattern equal? I will give two somewhat unsatisfactory explanations why it is

$$\frac1{29}=0.\mathbf{0344827586206896551724137931}0344827586206896551724137931\cdots$$

This is a phenomenon that occurs for every integer sequence given by a linear recurrence relation. Most famously $$\frac1{89}=0.\mathbf{01123595505617977528089887640449438202247191}011235\cdots$$

and

$$\frac{10}{109}=0.\mathbf{091743119266055\cdots238532110}0917431\cdots$$

For the shifting off to the left we obtain the repeating period from left to right in a familiar order. On the right we work backwards from the "end."

Below are two calculations. What is a better way to explain what is going on in the first one? The second is a candidate, but not a pleasing one.

Approach 1: **10-adic integers**. The $2$-adic and $5$-adic integers are integral domains. Their direct sum is the $10$-adic integers (not an integral domain) which can be thought of as possibly infinite decimal integers $\sum_0^{\infty}a_i10^i$ with the $a_i\in \{0,1,2,\cdots,9\}.$ They are well understood. I'll skimp on justifications for the following. First I claim that $\cdots99999.=-1$ (proof: look what happens when you add $1$.) Of course $0.999\cdots=1$ so $\cdots9999.9999\cdots=0.$ Many infinite decimal integers have no rational value, however I claim that if $q$ is a rational with periodic decimal $q=0.abcdabcdabcd\cdots$ (length $4$ chosen for illustration) then the $10$-adic integer $\cdots abcdabcdabcd.$ is equal to $-q.$ To see this, multiply $\cdots abcdabcdabcd.abcdabcd\cdots$ by $\frac1q$ turning it into $\cdots9999.9999\cdots=0.$ Hence the integer part is the additive inverse of the fractional part. Now I have a clear way to start my period at the end. The $10$-adic integer $$y=\cdots 0344827586206896551724137931\mathbf{0344827586206896551724137931}.$$ satisfies $30y=y-1$ hence $y=\frac{-1}{29}$ So the thing I want is the additive inverse $\frac1{29}.$ That is a tidy calculation and ends up with the desired result. It could be more fully justified, but seems like the wrong way to go at it.

Approach 2:

It seems possible that the thing we want should be $z=\frac{1/3}{10}+\frac{1/9}{100}+\frac{1/27}{1000}+\cdots.$ And there $30z=1+z.$ So yes $z=\frac1{29}$ and the method worked. But is it justified? Why is it clear that adding $0.0\mathbf{3}33\cdots+0.00\mathbf{1}11\cdots+.000\mathbf{037}037\cdots +\cdots$ corresponds to the sifted sum on the right?

To use this approach and get $\frac{10}{109}$ for the right shifted Fibonacci decimals recall that the full sequence is $\cdots 13,-8,5,-3,2,-1,1\ | \ ,0,1,1,2,3,5,\cdots$ So

$$z=\frac{1}{10}+\frac{-1}{100}+\frac{2}{1000}+\frac{-3}{10000}+\cdots$$ satisfies $$10z=1+\frac{-1}{10}+\frac{2}{100}+\frac{-3}{1000}+\frac{5}{10000}+\cdots$$ and $$10z+z=1+\frac{0}{10}+\frac{1}{100}+\frac{-1}{1000}+\frac{2}{10000}+\cdots$$

Thus $11z=1+\frac{z}{10}$ and $110z=10+z.$