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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 $110z=10+z.$

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    $\begingroup$ Just to make sure I understand the premise of the question: you take the 10-adic sum of shifted powers of 3, which give you some periodic expansion. Now we take this period and ask what the value is when we consider a decimal with expansion having the same periodic part. Is that correct? $\endgroup$ – Wojowu Feb 26 '18 at 10:13
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    $\begingroup$ Hmm, when you put it that way it seems less unusual. Think of the motivating question as "it is desired to find the rational whose decimal period encodes the decimal numerals for powers of $3$ (with overlaps, adding and carries.) $\cdots931$ " And the actual question is "why does the first or second approach work?" $\endgroup$ – Aaron Meyerowitz Feb 26 '18 at 10:57
  • $\begingroup$ You should not be embedding ordinary real decimals into the 10-adics; the relevant sums do not converge. (In particular, I can't make any sense of adding the 10-adic $\overline9.0 = -1$ to the real $0.\overline9 = 1$.) I can't really tell if you're just using this notation for motivation, or not using the full algebraic structure; or maybe you're working in some structure like the quotient of $\mathbb Z_{10} \oplus \mathbb R$ by the anti-diagonal embedding of $\mathbb Z$. $\endgroup$ – LSpice Aug 2 '18 at 13:33
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As to the second sum, we may write the sum of the first $28n$ rows as $$\sum_{k=0}^{28n-1}30^k={30^{28n}-1\over 29}={3^{28n}-1\over 29}\cdot 10^{28n}+{10^{28}-1\over 29}\cdot{10^{28n}-1\over 10^{28}-1}=$$ $$N\cdot 10^{28n}+344827586206896551724137931\sum_{k=0}^{n-1}10^{28k}$$ an integer whose decimal expansion, from right to left, is $0344827586206896551724137931$ repeated $n$ times, followed by the expansion of some integer $N$. And, of course, if we do the limit for $n\to \infty$ in the 10-adic distance, we get the 10-adic expansion of $-1/29$.

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  • $\begingroup$ One could as well ask about $x=\sum_0^NF_k10^k.$ Of course $F_k=\frac{\tau^n+(-1/ \tau)^n}{\sqrt{5}}$ so one can sum geometric progressions. How to know from that to break at $109n-1$ is not obvious to me. Maybe something about $F_k \mod 10^k$ at $109.$ And $110x=x-10$ which tells us something.. $\endgroup$ – Aaron Meyerowitz Feb 28 '18 at 18:23

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