---
title: "Polynomial Counting 3: The third degree"
format:
  html:
    html-math-method: katex
date: "2021-02-06"
date-modified: "2025-02-08"
jupyter: python3
categories:
  - algebra
  - phinary
  - binary
  - python
---

```{python}
#| echo: false

from IPython.display import Markdown
from sympy import latex
from tabulate import tabulate
import matplotlib.pyplot as plt

color = lambda c, text: f"<span style='color: {c}'>{text}</span>"
```

This post assumes you have read the [first](../1) post, which introduces generalized positional counting, and the [second](../2), which restricts the focus and specifies the general aim.

Thus far, the systems have been based on quadratic polynomials and two-term linear recurrences,
  or more directly, carries of width two.
Numbers larger than one introduce certain complications to carries.
With that in mind, let's go back to things closer to phinary and Fibonacci.


Tribonacci and Beyond
---------------------

The *Tribonacci numbers* ([OEIS A000073](http://oeis.org/A000073)) come from elongating the Fibonacci recurrence to 3 terms.
In other words, this is the carry/recurrence $\langle 1,1,1|$.
The Tribonacci numbers are seeded with two zeros followed by a single one; thus, the sequence begins $0, 0, 1, 1, 2, 4, 7, 13,\dots$
The *Tribonacci constant* is the limiting ratio of these numbers (i.e., the positive real root of $x^3 - x^2 - x - 1$) with approximate value
  $T = 1.8393\dots$

The number two bounds both this number and the terms in the recurrence, so its expansion can be deduced:

$$
\textcolor{red}{2} = 1.\textcolor{red}{111}_T = \textcolor{blue}{1.11}1_T = \textcolor{blue}{1}0.001_T
$$

Naturally, the string "111" will be illegal in both systems.
The Tribonacci expansions of the integers are as follows:

:::: {.row .text-center width="60%"}
::: {#canonical-tribonacci-table .column}

| *n* (Decimal) | *n* (base $T$) | *n* (Tribonacci) |
|--------------:|:--------------:|------------------|
|  0            |    0           |             0    |
|  1            |    1           |             1    |
|  2            |   10.001       |            10    |
|  3            |   11.001       |            11    |
|  4            |  100.100011    |           100    |
|  5            |  101.100011    |           101    |
|  6            |  110.101011    |           110    |
|  7            | 1000.101011    |          1000    |
|  8            | 1001.101011    |          1001    |
|  9            | 1010.110100011 |          1010    |
|  10           | 1100.010100011 |          1011    |

:::
::::

Since "111" is a rarer string than "11", the integral base more closely resembles standard binary.

Extensions of the Fibonacci numbers, which have an $n$-term recurrence
  (seeded with $n - 1$ zeroes followed by a one) are called "n-nacci" or "n-bonacci" numbers.
If we discard the seeding terms, then for larger and larger *n*, this sequence of sequences appears
  to approach the powers of two.

$$
\begin{matrix}
  \text{Fibonacci: }  &   &   & 0,& 1,& 1,& 2,& 3,& 5,& 8\dots \\
  \text{Tribonacci: } &   & 0,& 0,& 1,& 1,& 2,& 4,& 7,& 13\dots \\
  \text{Tetranacci: } & 0,& 0,& 0,& 1,& 1,& 2,& 4,& 8,& 15\dots \\
  \vdots \\
  \text{Binary: } & (\dots,& 0,& 0,& 1,)& 1,& 2,& 4,& 8,& 16\dots
\end{matrix}
$$

In the limit, the rightmost and leftmost "1"s in the carry $\langle 1, 1, ..., 1, 1 |$ are
  infinitely far away from each other.
This is similar to binary, in which $2 = 1.1111\dots_2$.
This argument is clearer if we directly manipulate the carry as a polynomial $p_n(x)$:

$$
\begin{align*}
p_n &= x^n - x^{n-1} - … - x - 1 \\
-p_n &= -x^n + x^{n-1} + … + x + 1 \\
x^n - p_n &= x^{n-1} + … + x + 1 \\
x - p_n x^{-n+1} &= 1 + x^{-1} + … x^{-n+2} + x^{-n+1} \\[8pt]
\text{Let } n &\rightarrow \infty \\
x &= 1 + x^{-1} + x^{-2} + x^{-3} + …
\end{align*}
$$

Since both *φ* and $T$ are greater than 1, we can assume that $x > 1$, which causes the
  $x^{-n+1}$ term to vanish in the limit.
Unfortunately, this means $p_n$ should diverge, invalidating the above argument.

Ignoring this, the last line can be manipulated as a power series:

$$
\begin{align*}
\frac{1}{1 - x} &= 1 + x + x^2 + x^3 + \dots \\
\frac{1}{1 - (1/x)} &= 1 + x^{-1} + x^{-2} + x^{-3} + \dots \\
x &= \frac{1}{1- (1/x)} \\
x - 1 &= 1 \\
x &= 2
\end{align*}
$$

Which is to imply that, in a non-rigorous sense and by partially assuming the conclusion,
  that the $n$-nacci constants approach two.


Golder than Gold
----------------

All carries entirely made up of "1"s correspond to the $n$-nacci constants.
While this would appear to exhaust every sequence without going to negative numbers,
  it ignores the potential of carries with a "0".

Starting simple, Narayana's cows sequence ([OEIS A000930](http://oeis.org/A000930)) corresponds to
  the recurrence $\langle 1,0,1|$.
It may be seeded with either three 1's, or in the same way as above, with a sequence of 0's followed by a 1.
The limiting ratio $\psi \approx 1.4656\dots$ is called the supergolden ratio.
The introduction of "0"s turns out to have big implications, since the concatenation trick to devise
  the expansion of two no longer works.
However, hope is not lost.

$$
\begin{gather*}
  1\textcolor{red}{1}000_\psi
  = 10\textcolor{red}{101}_\psi
  = \textcolor{blue}{101}01_\psi
  = \textcolor{blue}{1}00001_\psi \\
  \textcolor{red}{2} = 1.\textcolor{red}{101}_\psi
  = \textcolor{blue}{1.1}01_\psi
  = \textcolor{blue}{1}0.001\textcolor{blue}{1}_\psi
  = 10.00\textcolor{purple}{11}_\psi
  = 10.0\textcolor{purple}{1}0000\textcolor{purple}{1}_\psi
\end{gather*}
$$

Incredibly, the carry contains not only an explicit rule for "101", but implicit rules "2" and "11".
This sort of makes sense: since $\psi < \varphi$, adjacent place values are "too dense",
  and therefore we can rewrite the string "11".
On the other hand, attempting to manipulate "1001" and "10001" results in circuiting back to the original string.

$$
\begin{split}
  100\textcolor{red}{1}000000_\psi
  &= 1000\textcolor{red}{101}000_\psi
    = 1000\textcolor{orange}{1}01000_\psi
    = 10000\textcolor{orange}{1}1\textcolor{orange}{1}00_\psi \\
  &= 10000\textcolor{blue}{11}100_\psi
    = 1000\textcolor{blue}{1}0010\textcolor{blue}{1}_\psi
    = 1000100\textcolor{red}{101}_\psi \\
  &= 100010\textcolor{red}{1}000_\psi
    = 1000\textcolor{orange}{101}000_\psi
    = 100\textcolor{orange}1000000_\psi \\ \\
  \textcolor{red}{1}0001000_\psi
  &= 0\textcolor{red}{101}1000_\psi
    = 010\textcolor{blue}{11}000_\psi
    = 01\textcolor{blue}{1}0000\textcolor{blue}{1}_\psi \\
  &= 0\textcolor{purple}{11}00001_\psi
    = \textcolor{purple}{1}0000\textcolor{purple}{1}01_\psi \\
  &= 10000\textcolor{red}{101}_\psi = 1000\textcolor{red}{1}000_\psi
\end{split}
$$

I call these strings of "0"s sandwiched between "1"s *spacings*.
The smallest spacing, with a width of zero, is the Fibonacci recurrence.
If we forbid both the width-1 and width-0 spacings from appearing in supergolden expansions,
  we obtain the following list:

:::: {.row .text-center width="60%"}
::: {#canonical-supergolden-table .column}

<style>
#canonical-supergolden-table td > em
{
  color: red !important;
  font-style: normal !important;
}
</style>

| *n* (Decimal) | *n* (base $\psi$)   |
|--------------:|:-------------------:|
|   0           |       0             |
|   1           |       1             |
|   2           |      10.0100001     |
|   3           |    *100.0110001*    |
|   3           |     100.1000100001  |
|  *4*          |    *110.0001100001* |
|   4           |    1000.1000100001  |
|  *5*          |   *1010.0001100001* |
|   5           |   10000.0010001     |
|   6           |   10001.0010001     |
|  *7*          |  *10010.0110002*    |
|  *7*          |  *10010.1000012*    |
|   7           |  100000.0001000001  |
|   8           |  100001.0001000001  |
|   9           |  100100.0000001001  |
|   10          | 1000000.0000001001  |

:::
::::

In the above table, some intermediate steps are shown in red,
  but the last entry for each integer is canonical.
Not only does the number of digits grow radically faster than in phinary (or even binary),
  but there are many more intermediate steps.

For space reasons, I do not show the integral dual.
Slowly-growing sequences, which have largely uninteresting integral systems, dominate the rest of this post.
Therefore, the remainder of this post will focus solely on fractional systems.


Radiant Plastic
---------------

A similar degree 3 recurrence is $\langle 0,1,1|$.
Its root corresponds to the plastic ratio $\rho \approx 1.3247\dots$.
A number of sequences share this recurrence; when seeded with $0, 0, 1$ as before,
  the best match is the Padovan sequence ([OEIS A000931](http://oeis.org/A000931)),
  which begins with an additional one.

Since $\rho < \psi < \varphi$, we should expect that spacings of at least width two are illegal.
The system turns out to be even stricter than that: spacings up to and including width three are expandable.

$$
\begin{array}{c|c}
  \text{Width} & \textcolor{red}{0} & \textcolor{green}{1} & 2 & \textcolor{blue}{3} \\ \hline
  & 0\textcolor{red}{011}_\rho
  & 010\textcolor{red}{1}000_\rho
  & 0\textcolor{red}{1}00\textcolor{red}{1}00000_\rho
  & 0\textcolor{red}{1}0001_\rho
  \\
  & \textcolor{red}{1}000_\rho
  & 0100\textcolor{red}{011}_\rho
  & 00\textcolor{red}{011}\textcolor{red}{011}00_\rho
  & 00\textcolor{red}{011}1_\rho
  \\
  && 0\textcolor{blue}{10001}1_\rho
  & 000\textcolor{blue}{11011}00_\rho
  & 000\textcolor{orange}{111}_\rho
  \\
  && \textcolor{blue}{1}000001_\rho
  & 00\textcolor{blue}{1}0101000_\rho
  & 00\textcolor{orange}{1}100_\rho
  \\
  &&
  & 0010\textcolor{green}{101}000_\rho
  & 0\textcolor{red}{011}00_\rho
  \\
  &&
  & 001\textcolor{green}{1000001}_\rho
  & \textcolor{red}{1}00000_\rho
  \\
  &&
  & 0\textcolor{red}{011}000001_\rho
  \\
  &&
  & \textcolor{red}{1}000000001_\rho
  \\\\
  \hline
  & s_0  = \rho^2
  & s_1 = \rho s_{5}
  & s_2 = \rho s_{8}
  & s_3 = \rho
\end{array}
$$

The final line is shorthand for the spacing rules:

- The zero-spacing is equal to a one in the place value two higher than it
- The one-spacing is equal to an five-spacing which begins one place value higher than it
- The two-spacing is equal to an eight-spacing which begins one place value higher than it
- The three-spacing is equal to a one in the place value one higher than it

With these rules, we can write a canonical expansion expansion for 2:

$$
\begin{align*}
  \textcolor{red}{2}
  &= 1.\textcolor{red}{011}
    = \textcolor{blue}{1.011}
    = \textcolor{blue}{1}\overbrace{0.0100000}^{8 \text{ digits}}\textcolor{blue}{1} \\
  &= \textcolor{purple}{10.01}000001
    = \textcolor{purple}{1}00.000000\textcolor{purple}{1}1 \\
  &= 100.00000\textcolor{red}{011}
    = 100.0000\textcolor{red}{1}
\end{align*}
$$

Since $\rho < \sqrt 2$, the largest place value in the expansion of two is $\rho^2$, which
  distinguishes it from previous systems.

Similarly to how the width-two and width-three spacings are allowed in the supergolden ratio base,
  we realize that the width-four spacing cannot be expanded further in the plastic ratio base:

$$
\begin{align*}
  \textcolor{red}{1}0000\textcolor{red}{1}000
  &= 0\textcolor{red}{011}00\textcolor{red}{011}
    = 001\textcolor{blue}{10001}1
    = 00\textcolor{blue}{2}000001 \\
  &= \textcolor{blue}{1}000000\textcolor{blue}{1}1
    = 100001000
\end{align*}
$$

With these implicit rules derived for spacings of width smaller than three,
  the plastic expansions of the integers up to ten are as follows:

:::: {.row .text-center width="70%"}
::: {#canonical-plastic-table .column}

<style>
#canonical-plastic-table td > em
{
  color: red !important;
  font-style: normal !important;
}
</style>

| *n* (Decimal) | *n* (base $\rho$)            |
|--------------:|:----------------------------:|
|  0            |         0                    |
|  1            |         1                    |
|  2            |       100.00001              |
| *3*           |      *101.00001*             |
| *3*           |     *1000.00101*             |
|  3            |      1000.01000001           |
| *4*           |    *10000.0101000000001*     |
|  4            |     10000.1000001000001      |
| *5*           |    *10100.0000001000001*     |
|  5            |    100000.1000001000001      |
| *6*           |   *100001.1000001000001*     |
| *6*           |   *100100.0000001000001*     |
| *6*           |  *1000000.0010001000001*     |
|  6            |   1000000.0100000000001      |
| *7*           |   1000001.0100000000001      |
|  7            |   1000010.0000100000001      |
| *8*           |  *1001000.0000100000001*     |
| *8*           | *10000000.0100100000001*     |
| *8*           | *10000000.1000000001001*     |
|  8            |  10000000.100000001000000001 |
|  9            |  10000100.000000001000000001 |
|  10           |  10001000.001000001000000001 |

:::
::::

Clearly, this base is incredibly sensitive.
A number as small as 8 has a fractional part as small as $\rho^{-18}$ in its expansion.


Skipped Spacings
----------------

When discussing the expandable spacings in the supergolden and plastic bases, they jumped from
  width one to width three.
Did we forgot width 2?
The strings "1001" and "10001" are associated to the carries
  $\langle 1, 0, 0, 1|$ and $\langle 1, 0, 0, 0, 1|$.

| Spacing Width | Carry                                      | Integral Sequence                        | Root                      |
|---------------|--------------------------------------------|------------------------------------------|---------------------------|
| 0             | $\langle 1,1|$                             | Fibonacci numbers, Lucas numbers         | Golden Ratio, $\varphi$   |
| 1             | $\langle 1,0,1|$                           | Narayana's cows sequence                 | Supergolden Ratio, $\psi$ |
| 2             | $\langle 1,0,0,1|$                         | [OEIS A003269](https://oeis.org/A003269) | Unnamed $(\upsilon?)$     |
| 3             | $\langle 1,0,0,0,1| \simeq \langle 0,1,1|$ | Padovan sequence, Perrin sequence        | Plastic Ratio, $\rho$     |

I chose the symbol $\upsilon$ since it came from the second half of the Greek alphabet,
  like the others.

While the width-two spacing is a irreducible polynomial like its predecessors,
  the width-three one can be factored

```{python}
from sympy.abc import x

def spacing(width):
  return x**(width + 2) - x**(width + 1) - 1

spacing(3).factor()
```

This also means that despite $\rho$ and $\psi$ being cubic roots,
  $\upsilon$ is irreducibly a quartic root, despite being between them in terms of spacing width.

The right polynomial may be familiar from the previous post as $\langle 1, -1 |$,
  which only has complex roots.
It happens to be the sixth cyclotomic polynomial ($\Phi_6$), and allows the width-three spacing
  to be equivalent to the simpler plastic ratio carry.

As previously stated, the cyclotomic polynomials cause big trouble for carries.
Other than factoring, there appears to be no way to derive $\langle 0,1,1|$ from $\langle 1,0,0,0,1|$.
At best, we can observe the following:

$$
\begin{matrix}
  \langle 1,0,0,0,1| && \langle 0,1,1| \\ \hline
  11111 &\iff& 11111 \\
  101110 && 1000111 \\
  1001100 &\iff& 1001100 \\
  10001000 && 1100000 \\
  100000000 &\iff& 100000000 \\
\end{matrix}
$$


### Chopped Circles

The appearance of a cyclotomic factor is not unique to the width-three spacing.
Each of the smaller-width rules in the supergolden and plastic bases can be reexamined as polynomials.
After converting and factoring, their cyclotomic factors become clear:

$$
\begin{gather*}
  11000_\psi = 100001_\psi  \\
  1\bar{1}\bar{1}001_\psi = 0 \\
  \psi^5 - \psi^4 - \psi^3 + 1 \\
  (\psi - 1)(\psi + 1)(\psi^3 - \psi^2 - 1) \\
  \Phi_1 \Phi_2 \langle 1,0,1|
  \\ \\
  \begin{gather*}
    101000_\rho = 1000001_\rho &
    100100000_\rho = 1000000001_\rho \\
    1\bar{1}0\bar{1}001_\rho = 0 &
    1\bar{1}00\bar{1}00001_\rho = 0 \\
    \rho^6 - \rho^5 - \rho^3 + 1 &
    \rho^9 - \rho^8 - \rho^5 + 1 \\
    (\rho - 1)(\rho^2 + 1)(\rho^3 - \rho - 1) &
    \dots(\rho^2 - \rho + 1)(\rho^3 - \rho - 1) \\
    \Phi_1 \Phi_4 \langle 0,1,1| &
    \Phi_1 \Phi_2 \Phi_4 \Phi_6 \langle 0,1,1|
  \end{gather*}
\end{gather*}
$$

Though I am uncertain without a proof, it seems that cyclotomic polynomials
  play a role in spacing out "1"s.
This is to say that spacings have a "fundamental" irreducible polynomial.
By multiplying certain cyclotomic polynomials by the fundamental, (all?) smaller spacings
  can produce spacings the size of the fundamental or less.

Naturally, I attempted to write a program to compute lesser spacings from an implicit rule.
Generally, this entailed assembling all lesser spacings, then expanding the rightmost 1
  if unable to find any spacing, else replacing and continuing.
Unfortunately, since it operates in lockstep, it gets stuck easily, and I had little success.
My Haskell code can be found [here](zero_spacing.hs).

### Inherently Factorable Spacings

The width-three spacing is not fundamental, since it can be factored.
We can collect reducible spacings into a table:

```{python}
is_irreducible = lambda p: len(p.factor().as_ordered_factors()) == 1
display = lambda x: color("green", "Yes") if x else color("red", "No")

Markdown(tabulate(
  zip(*(
      [["*n*", "Irreducible?"]]
      + [[ str(i), display(is_irreducible(spacing(i))) ] for i in range(20)]
  ))
))
```


In the range given, this table appears to repeat every six terms.
We can examine each of these factors directly:

$$
\begin{align*}
  x^5 - x^4 - 1 &= (x^2 - x + 1)
    (x^3 - x - 1) \\
    &= 1\bar{1}1_x * 10\bar{1}\bar{1}_x \\
  x^{11} - x^{10} - 1 &= (x^2 - x + 1)
    (x^9 - x^7 - x^6 + x^4 + x^3 - x - 1) \\
    &= 1\bar{1}1_x * 10\bar{1}\bar{1}0110\bar{1}\bar{1}_x \\
  x^{17} - x^{16} - 1 &= (x^2 - x + 1)
    (x^{15} - x^{13} - \dots + x^4 + x^3 - x - 1) \\
    &= 1\bar{1}1_x * 10\bar{1}\bar{1}0110\bar{1}\bar{1}0110\bar{1}\bar{1}_x \\
  5, 11, 17, \dots &= 6n + 5
\end{align*}
$$

Each of these polynomials has $\Phi_6$ as a common factor.
The other factor appears to be a truncation of the repeating string "$\underline{0110\bar{1}\bar{1}}$".
^[Since I already use the overbar for negative digits, I gather repeating terms using an underline as a vinculum.]
Interpreted as a power series, this is the reciprocal of $-\Phi_6$ ([OEIS A010892](https://oeis.org/A010892)).

As the carry approaches infinite width, the $x$ terms go to 0 for $x < 1$.
This shows that like the $n$-nacci constants approach two, the spacing constants approach one.


Repeating Expansions
--------------------

Let's not get too eager here and instead return to base $\upsilon$.
It should be possible to expand the strings "11" and "101", since it is the fundamental spacing of width 2.
However, we immediately run into an issue:

$$
\begin{array}{c|c}
  \text{Width} & \textcolor{green}{0} & \textcolor{blue}{1} & \textcolor{red}{2} \\ \hline
  & 01\textcolor{red}{1}00000_\upsilon
  & 010\textcolor{red}{1}0000_\upsilon
  & 0\textcolor{red}{1001}_\upsilon
  \\
  & 010\textcolor{red}{1001}0_\upsilon
  & 0100\textcolor{red}{1001}_\upsilon
  & \textcolor{red}{1}0000_\upsilon
  \\
  & 0\textcolor{blue}{101}0010_\upsilon
  & 0\textcolor{orange}{1001}001_\upsilon
  \\
  & \textcolor{blue}{1}000001\textcolor{blue}{1}_\upsilon
  & \textcolor{orange}{1}0000001_\upsilon
  \\
  & 100000\textcolor{green}{11}_\upsilon \\
  & \vdots \\ \hline
  & s_0 = \ ? & s_1 = \rho s_6 & s_2 = \rho
\end{array}
$$

The width-0 expansion is recursive.
If we continue to apply its rule, we generate the string "100001" repeating.
Fortunately, each repeating unit is a width-four spacing, which is allowed by the carry.

Repeating expansions imply a representation by geometric series:

$$
\begin{gather*}
  1.\underbrace{\underline{0\dots1}}_{n}{}_x = 1 + x^{-n} + x^{-2n} + x^{-3n} + \dots
  = \frac{1}{1 - (1/x)^n} = \frac{x^n}{x^n - 1}
  \\
  10 = 1.\underbrace{\underline{0\dots1}}_{n}{}_x ~\iff~ x = \frac{x^n}{x^n - 1} \\
  1 = \frac{x^{n-1}}{x^n - 1} \\ \\
  x^{n-1} = x^n - 1
\end{gather*}
$$

The final expression is just the definition of a carry, so we can immediately write:

$$
\begin{align*}
  10_\varphi &= 1.\underline{01}_\varphi & &
  \text{Carry } \langle 1,1| \\
  10_\psi &= 1.\underline{001}_\psi & &
  \text{Carry } \langle 1,0,1| \\
  10_\upsilon &= 1.\underline{0001}_\upsilon & &
  \text{Carry } \langle 1,0,0,1| \\
  10_{\rho} &= 1.\underline{00001}_\rho & &
  \text{Carry } \langle 1,0,0,0,1| = \langle 0,1,1|
\end{align*}
$$

This neatly ties repeating spacings in with carries.
^[Recall that when we naively computed ten in base phi, we got "10100.0100101010101010101".
After a certain point, this expansion alternates between 0 and 1. Assuming that this is true repetition
and applying $10_\varphi = 1.\underline{01}_\varphi$, one obtains "10100.0101", which is canonical. ]

But we didn't want "10" as a repeating expansion, we wanted "11".
Attempting to justify with the same strategy:

$$
\begin{align*}
  0.11_\upsilon &= 1.\underline{00001}_\upsilon & \text{Carry } \langle 1,0,0,1|
  \\
  \frac{1}{x} + \frac{1}{x^2} &= \frac{x^n}{x^n - 1} \\
  x + 1 &= \frac{x^{n+2}}{x^n - 1}
  \\[8pt]
  x^{n+1} + x^n - x - 1 &= x^{n+2} \\
  x^{n+2} - x^{n+1} - x^n + x + 1 &= 0 \\
  x^n(x^2 - x - 1) + x + 1 &= 0
\end{align*}
$$

```{python}
repeating_pair = lambda n: (x**n * (x**2 - x - 1) + x + 1).expand().factor()

Markdown(tabulate(
  zip(*(
      [["*n*", "Irreducible?"]]
      + [[ str(i), display(is_irreducible(repeating_pair(i))) ] for i in range(1, 10)]
  ))
))
```

For $n$ = 5, the polynomial factors as the product of $\langle 1, 0, 0, 1|$ and $\langle 0, 1, 1|$.
The second factor should come as no surprise since we know directly that $0.11_\rho = 10_\rho$
  and we already evaluated $10_{\rho} = 1.\underline{00001}_\rho$

Checking more $n$, this polynomial actually seems to be the only reducible one for quite a while.

```{python}
#| freeze: true

all(is_irreducible(repeating_pair(i)) for i in range(10, 100))
```

This bestows some level of intrigue upon the repeating expansion of "11".
Since repeating expansions are required for base $\upsilon$ I will elect to not show the expansion of the integers.


### Weird Modulus Patterns

This can be generalized slightly to match any spacing with any repeating spacing:

$$
\begin{gather*}
  0.\underbrace{10\dots01}_m = 1.\underbrace{\underline{0\dots01}}_n
  ~\Leftrightarrow~
  \frac{1}{x} + \frac{1}{x^m} = \frac{x^n}{x^n - 1}
  \\
  x^{m-1} + 1 = \frac{x^{n+m}}{x^n - 1} \\
  \\
  x^{n+m-1} + x^n - x^{m-1} - 1 = x^{n+m}\\
  x^{n+m} - x^{n+m-1} - x^n + x^{m-1} + 1 = 0 \\
  (x^n)(x^m - x^{m-1} - 1) + x^{m-1} + 1 = 0 \\
\end{gather*}
$$

If the polynomial has factors, then fractional systems where those factors are carries have
  repeating expansions of width $n$ for a spacing of width $m$.
Graphing the tuples of reducible polynomials produces a strange pattern
  apparently arranged in lines (going from the lower left to the upper right):

```{python}
general_repeats = lambda m, n: (x**n)*(x**m - x**(m-1) - 1) + x**(m-1) + 1

plt.xticks(range(20))
plt.yticks(range(20))
plt.xlabel("$m - 1$")
plt.ylabel("$n - 1$")
plt.imshow([
  [ is_irreducible(general_repeats(m + 1, n + 1)) for m in range(20) ]
  for n in range(20)
])
```

The leftmost purple square is the exception, which corresponds to the earlier $(m, n) = (2, 5)$.
Every purple square here is a reducible polynomial, the first few of whose factors are:


| $(m, n)$ | Sum | Polynomial Factors
|----------|-----|-------------------
| (2, 5)   |   7 | $(x^3 - x - 1) (x^4 - x^3 - 1)$
| (3, 11)  |  14 | $(x^4 - x^3 + x^2 - x + 1)$ <br> $(x^{10} - x^8 - x^7 - x^6 - x^5 + x^3 + x^2 + x + 1)$
| (7, 7)   |     | $(x^4 - x^3 + x^2 - x + 1)$ <br> $(x^{10} - x^8 - x^5 + x + 1)$
| (13, 1)  |     | $(x^4 - x^3 + x^2 - x + 1)$ <br> $(x^{10} + x^7 - x^5 - x^2 + 1)$
| (7, 17)  |  24 | $(x^4 - x^3 + x^2 - x + 1)$ <br> $(x^{20} - x^{18} - x^{15} - x^{12} + x^{10} + x^7 - x^5 + x + 1)$
| (13, 11) |     | $(x^4 - x^3 + x^2 - x + 1)$ <br> $(x^{20} - x^{18} - x^{15} + x^{13} + x^{10} + \dots + x + 1)$
| (17, 7)  |     | $(x^4 - x^3 + x^2 - x + 1)$ <br> $(x^{20} - x^{18} - x^{15} + x^{13} + x^{12} + \dots + x + 1)$
| (17, 17) |     | $(x^4 - x^3 + x^2 - x + 1)$ <br> $(x^{30} - x^{28} - x^{25} + x^{23} + x^{20} - x^{18} - \dots + x + 1)$

Only the first entry has factor that is a carry associated to a spacing.
Every other entry appears to have $(x^4 - x^3 + x^2 - x + 1)$ as a common factor,
  which is the 10th cyclotomic polynomial.

These solutions also have sums which increment in tens, which correspond to the
  vertical and horizontal lines of purple squares in the image.
Both $m$ and $n$ solutions work in a similar way: the symmetric solution (7, 7)
  has similar solutions (7, 17), (17, 7), and (17, 17).
This pattern also holds up in the "next" terms (7, 27), (17, 27), (27, 27), (27, 17), and (27, 7).
It appears as though there is a congruence of $m + n\ (\text{mod } mn)$,
  where m and n come from the exception (2, 5).
This pattern is similar to the one in which one in every six spacings can be factored.


Closing
-------

In summary, the Fibonacci recurrence can be generalized in at least two two different ways.
Introducing "0"s into the carry is more touch-and-go than would otherwise seem,
  though they produce (with much effort) valid bases.

The [next post](../4) will explore a more distant cousin of the Fibonacci numbers,
  and other strange power series which arise from its discussion.