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Polynomial Counting: The third degree
======================================

This post assumes you have read the [first]() post, which introduces generalized positional counting, and the [second](), 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 2. With the complications from numbers larger than 1 addressed in the previous post, let's go back to 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 1; 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 $T = 1.8393…$. The value 2 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:

```{}
	Tribonacci Constant     Tribonacci
	(Base T)                 Numbers
	     ______
	3210 123456              7 4 2 1
0	   0                           0
1	   1                           1
2	  10.001                     1 0
3	  11.001                     1 1
4	 100.100011                1 0 0
5	 101.100011                1 0 1
6	 110.101011                1 1 0
7	1000.101011              1 0 0 0
8	1001.101011              1 0 0 1
9	1010.110100011           1 0 1 0
10	1100.010100011           1 0 1 1
```

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 numbers. If we discard the seeding terms, then for larger and larger *n*, this sequence of sequences appears to approach the powers of 2.

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

In the limit, the rightmost and leftmost "1"s in the carry are infinitely far away. 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:

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

Since both $\phi$ 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 equation. Despite this, it can still be manipulated as an equation in 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 2.


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

All carries entirely made up of 1s 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 no longer works. However, hope is not lost.

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

The carry contains implicit rules for not only "101", but also "11". This sort of makes sense: since $\psi < \phi$, 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 0, 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

```{}
	Supergolden Ratio
	        __________
	6543210 123456789A
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 $\rho\ (\approx 1.3247…)$. 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 1.

Since $\rho < \psi < \phi$, we should expect that spacings of at least width 2 are illegal. The system turns out to be even stricter than that: widths up to and including 3 being expandable.

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

With these spacings, 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 2 is $\rho^2$, distinguishing it from previous systems.

Similarly to how the width-2 and width-3 spacings are allowed in the supergolden ratio base, we realize that the width-4 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 about width-3 spacings and below, the plastic expansions of the integers up to 10 are as follows:

```{}
	Plastic Ratio
	         __________________
	76543210 123456789ABCDEFGHI
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 1 to width 3, implying 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, $\phi$ |
| 1             | $\langle 1,0,1|$ | Narayana's cows sequence | Supergolden Ratio, $\psi$ |
| 2             | $\langle 1,0,0,1|$ |                          | Unnamed $(\upsilon?)$ |
| 3             | $\langle 1,0,0,0,1| = \langle 0,1,1|$ | Padovan sequence, Perrin sequence | Plastic Ratio, $\rho$ |

The symbol $\upsilon$ comes from the second half of the Greek alphabet, like the others. There do not appear to be any sequences in the OEIS with this recurrence signature. While the width-2 spacing is a irreducible polynomial like its predecessors, the width-3 one can be factored

$$
\begin{align*}
\langle 1,0,0,0, 1| = x^5 - x^4 - 1 &= (x^2 - x + 1)(x^3 - x - 1) \\
&= \Phi_6 \langle 0, 1, 1|
\end{align*}
$$

The right polynomial is familiar from the previous post as the one with only complex roots. It happens to be the sixth cyclotomic polynomial, and allows the width-3 spacing to be equivalent to the simpler plastic ratio carry. 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.

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-3 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]().


### Inherently Factorable Spacings

The width-3 spacing is not fundamental, since it can be factored. Using [WolframAlpha](https://www.wolframalpha.com/input/?i=Table%5B+IrreduciblePolynomialQ%5B+x%5En+-+x%5E%28n-1%29+-+1%5D%2C+%7Bn%2C+1%2C+20%7D+%5D), I was able to collect reducible spacings into a table. In the range given, it appears to repeat every six terms.

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

Each of these polynomials has $\Phi_6$ as a common factor. The other factor appears to be a truncation of the repeating string "$\underline{0110\bar{1}\bar{1}}$". Note: 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 $-\Phi_6$. As the carry approaches infinite width, the higher terms go to 0 for $x < 1$:


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

Let's not get too eager here and 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 4 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:

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

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

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

For *n* = 5, the polynomial has $\langle 1, 0, 0, 1|$. This polynomial has another factor, $\langle 0,1,1|$, which should come as no surprise since we know directly that $0.11_\rho = 10_\rho$. Quickly checking [WolframAlpha](http://Table[ IrreduciblePolynomialQ [x^n(x^2 - x - 1) + x + 1], {n, 1, 20} ]) for *n* from 0 to 20 shows that this is the only *n* in that range whose carry polynomial is reducible. 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:

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

Another [table](https://www.wolframalpha.com/input/?i=Table%5B+IrreduciblePolynomialQ+%5Bx%5En%28x%5Em+-+x%5E%28m-1%29+-+1%29+%2B+x%5E%28m-1%29+%2B+1%5D%2C+%7Bn%2C+1%2C+20%7D%2C+%7Bm%2C+1%2C+20%7D%5D) can be generated by searching for $(m, n)$ for which this polynomial is reducible. If the polynomial has factors, then bases 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):

![]()

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 | $(x^4 - x^3 + x^2 - x + 1)$ $(x^{10} - x^8 - x^7 - x^6 - x^5 + x^3 + x^2 + x + 1)$
| (7, 7)   |     | $(x^4 - x^3 + x^2 - x + 1)$ $(x^{10} - x^8 - x^5 + x + 1)$
| (13, 1)  |     | $(x^4 - x^3 + x^2 - x + 1)$ $(x^{10} + x^7 - x^5 - x^2 + 1)$
| (7, 17)  |  24 | $(x^4 - x^3 + x^2 - x + 1)$ $(x^{20} - x^{18} - x^{15} - x^{12} + x^{10} + x^7 - x^5 + x + 1)$
| (13, 11) |     | $(x^4 - x^3 + x^2 - x + 1)$ $ $(x^{20} - x^{18} - x^{15} + x^{13} + x^{10} + \dots + x + 1)$
| (17, 7)  |     | $(x^4 - x^3 + x^2 - x + 1)$ $(x^{20} - x^{18} - x^{15} + x^{13} + x^{12} + \dots + x + 1)$
| (17, 17) |     | $(x^4 - x^3 + x^2 - x + 1)$ $(x^{30} - x^{28} - x^{25} + x^{23} + x^{20} - x^{18} - \dots + x + 1)$

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