# The Behavior of a Three-Term Hofstadter-Like Recurrence with Linear Initial
Conditions

Nathan Fox111Department of Quantitative Sciences, Canisius University,
Buffalo, New York<EMAIL_ADDRESS>
###### Abstract

In this paper, we study the three-term nested recurrence relation
$B(n)=B(n-B(n-1))+B(n-B(n-2))+B(n-B(n-3))$ subject to initial conditions where
the first $N$ terms are the integers $1$ through $N$. This recurrence is the
three-term analog of Hofstadter’s famous $Q$-recurrence
$Q(n)=Q(n-Q(n-1))+Q(n-Q(n-2))$. Nested recurrences are highly sensitive to
their initial conditions. Some initial conditions lead to finite sequences,
others lead to predictable sequences, and yet others lead to sequences that
appear to be chaotic and infinite. A corresponding study to this one was
previously carried out on the $Q$-recurrence. As with that work, we consider
two families of sequences, one where terms with nonpositive indices are
undefined and a second where terms with nonpositive indices are defined to be
zero. We find similar results here as with the $Q$-recurrence, as we can
completely characterize the sequences for sufficiently large $N$. The results
here are, in a sense, simpler, as our sequences are all finite for
sufficiently large $N$.

##  1 Introduction

The Hofstadter $Q$-recurrence [geb]

$Q(n)=Q(n-Q(n-1))+Q(n-Q(n-2))$

has been the subject of numerous studies. Hofstadter’s original $Q$-sequence
starts with initial conditions $Q(1)=Q(2)=1$. This sequence has tantalizing
properties that have thus far evaded proof, though they have been the subject
of statistical studies [pinn, spotbased]. As such, most studies of the
$Q$-recurrence analyze sequences generated by other initial conditions
[golomb, rusk, symbhof, genrusk, gengol]. One recent approach involves
studying a family of initial conditions described by a parameter [hof1thruN].

In this paper, we apply the parametrized initial condition approach to a
different recurrence, the three-term Hofstadter-like recurrence

$B(n)=B(n-B(n-1))+B(n-B(n-2))+B(n-B(n-3)).$

This recurrence is known to generate a well-behaved sequence [slowtrihof] when
given initial conditions $B(1)=1$, $B(2)=2$, $B(3)=3$, $B(4)=4$, and $B(5)=5$.
For convenience, we refer to this well-behaved sequence as the _$B$
-sequence_. Aside from that one article, this recurrence has not been widely
reported on. This is presumably because most natural initial conditions lead
to finite sequences. If a sequence $B^{*}(n)$ generated by the $B$-recurrence
ever has $B^{*}(n-1)\geq n$ or $B^{*}(n-1)\leq 0$, then $B^{*}(n)$ would be
undefined. When this sort of behavior occurs, we say that the sequence _dies_
after $n-1$ terms, or that it dies at index $n$. This assumes, as is standard,
that the first term defined by the initial conditions is $B^{*}(1)$. In this
paper, we consider such initial conditions, but we also consider infinite
initial conditions that define values for $B^{*}(n)$ when $n\leq 0$. In this
realm, $B^{*}(n-1)\geq n$ does not lead to sequence death, but $B^{*}(n-1)\leq
0$ still does. So, to avoid some confusion later in the paper, when a sequence
dies because $B^{*}(n-1)\leq 0$, we say that the sequence _ends_.

###  1.1 Notation

Going forward, the only recurrence relation we discuss is the $B$-recurrence,
but we study it with many different initial conditions. We introduce analogous
notation to [hof1thruN]. The notation $B(n)$ refers to the $n$th term of the
$B$-sequence itself. The notation $B^{*}(n)$ refers to a generic sequence that
satisfies the $B$-recurrence. For any other specific sequence satisfying the
$B$-recurrence, we use $B$ with a subscript that we define for that particular
sequence.

We use angle brackets to denote our initial conditions. For example,
$\left<1,2,3,4,5\right>$ is shorthand for the initial conditions for the
$B$-sequence. Sometimes, we wish to define $B^{*}(n)=0$ for all $n\leq 0$ in
order to prevent our sequences from dying too soon. This convention is noted
with a symbol $\bar{0}$ followed by a semicolon at the start of the initial
conditions. For example, $\left<\bar{0};1,1,1\right>$ is shorthand for
$B^{*}(n)=0$ for $n\leq 0$, $B^{*}(1)=1$, $B^{*}(2)=1$, and $B^{*}(3)=1$.

###  1.2 Structure of this Paper

This paper’s structure mirrors that of the analogous study on the
$Q$-recurrence [hof1thruN]. In Section 2, we characterize the sequences
generated by the $B$-recurrence via family of initial conditions of the form
$\left<1,2,3,\ldots,N\right>$. Then, in Section 3, we study the more general
initial conditions $\left<\bar{0};1,2,3,\ldots,N\right>$. Finally, we suggest
some future research directions in Section 4.

##  2 Behavior of the $B$ Recurrence with Linear Initial Conditions

In this section, we consider sequences obtained from the $B$-recurrence and
initial conditions of the form $\left<1,2,3,\ldots,N\right>$ for some integer
$N\geq 3$. Henceforth, we denote this sequence for a given value of $N$ by
$B_{N}$.

We have the following result, which characterizes the behaviors of almost all
of these sequences.

######  Theorem 1.

For $N=3$, $N=4$, or $N\geq 10$, the sequence $B_{N}$ dies. Furthermore, if
$N\geq 14$, the sequence has exactly $N+24$ terms.

###### Proof.

Computing terms, we obtain that $B_{3}(4)=6$, $B_{4}(5)=6$,
$B_{10}(1015)=1036$, $B_{11}(117)=120$, $B_{12}(45)=47$, and $B_{13}(73)=82$.
So, these sequences all die [oeis]*A373230, A373231, A373232, A373233.

Now, we treat $N$ as a symbolic parameter and apply the process from
[hof1thruN]. This allows us to compute $24$ terms following the initial
conditions. These $24$ terms are:

$\displaystyle 6,\,$ $\displaystyle
N+1,\,N+2,\,N+3,\,9,\,N+4,\,N+5,\,N+6,\,12,\,N+7,\,N+8,\,N+9,\,15,\,N+10,\,$
$\displaystyle N+11,\,17,\,N+13,\,18,\,N+13,\,N+15,\,N+16,\,22,\,21,\,2N+11.$

See Appendix A for explicit computations of these terms, along with a bound on
the values of $N$ for which that computation and all previous computations are
valid. In particular, note that the calculations are valid for $N\geq 9$.

The last term we have is $B(N+24)=2N+11$. We try to compute $B_{N}(N+25)$:

$\displaystyle B_{N}(N+25)$
$\displaystyle=B_{N}(N+25-B_{N}(N+24))+B_{N}(N+25-B_{N}(N+23))$
$\displaystyle\phantom{=}+B_{N}(N+25-B_{N}(N+22))$
$\displaystyle=B_{N}(N+25-\left(2N+11\right))+B_{N}(N+25-21)+B_{N}(N+25-22)$
$\displaystyle=B_{N}\left(-N+14\right)+B_{N}\left(N+4\right)+B_{N}\left(N+3\right).$

If $N\geq 14$, then $-N+14\leq 0$, so $B_{N}(-N+14)$ is undefined and the
sequence dies, as required.

∎

Theorem 1 says that $B_{N}$ dies for all but five values $N$. The sequences
$B_{5}$ and $B_{6}$ are identical; both are the $B$-sequence [slowtrihof].
Sequences $B_{7}$, $B_{8}$, and $B_{9}$ are more akin to Hofstadter’s Q
sequence. Like Hofstadter’s, it is unclear whether any of these sequences die.
All last for at least $30$ million terms [oeis]*A373227, A373228, A373229.
Plots of the first hundred thousand terms of each of these sequences are shown
in Figure 1.

Figure 1: Plots of the first 100,000 terms of $B_{7}$ (left), $B_{8}$
(center), and $B_{9}$ (right)

##  3 Linear Initial Conditions with Extra Zeroes

The sequences in Section 2 almost all die. As in [hof1thruN], we now consider
what happens if we prevent them from dying quickly by defining their values to
be zero at nonpositive integers. For an integer $N\geq 3$, let $B_{\bar{N}}$
denote the sequence obtained from the $B$-recurrence with initial conditions
$\left<\bar{0};1,2,3,\ldots,N\right>$.

In [hof1thruN], it was seen that the corresponding behavior for the Hofstadter
$Q$-recurrences depends on the congruence class of $N$ modulo $5$. Three of
those cases lead to the end of the sequence, one leads to a semi-predictable
pattern that seems to go on forever, and the fifth case leads to a dependence
on the congruence class of $N$ modulo $25$ and, thereafter, potentially higher
powers of $5$. For the sequences $B_{\bar{N}}$, the dependence is instead on
the congruence class of $N$ modulo $7$. Here, all seven cases lead to the end
of the sequence without needing to consider any cases involving higher powers
of $7$. But, all cases require fairly large values of $N$ to be valid. We have
the following main result.

######  Theorem 2.

Let $N\geq 72$ be a natural number. Then the following period-$7$ pattern
begins at index $N+67$ in $B_{\bar{N}}$, where $k$ denotes a positive integer:

$\begin{cases}B_{\bar{N}}(N+7k)=7k+2\\\ B_{\bar{N}}(N+7k+1)=N+7k+2\\\
B_{\bar{N}}(N+7k+2)=N+7k+4\\\ B_{\bar{N}}(N+7k+3)=7\\\
B_{\bar{N}}(N+7k+4)=2N+2k+45\\\ B_{\bar{N}}(N+7k+5)=2N+k-7\\\
B_{\bar{N}}(N+7k+6)=N-2.\end{cases}$

This pattern lasts through index $2N+\nu$, where

$\nu=\begin{cases}-1&\text{if }N\equiv 0\pmod{7}\\\ -2&\text{if }N\equiv
1\pmod{7}\\\ -2&\text{if }N\equiv 2\pmod{7}\\\ -2&\text{if }N\equiv
3\pmod{7}\\\ 2&\text{if }N\equiv 4\pmod{7}\\\ 1&\text{if }N\equiv 5\pmod{7}\\\
0&\text{if }N\equiv 6\pmod{7}.\end{cases}$

After this,

  * •

If $N\equiv 0\pmod{7}$ and $N\geq 196$, then $B_{\bar{N}}$ ends after $2N+27$
terms.

  * •

If $N\equiv 1\pmod{7}$ and $N\geq 2087$, then $B_{\bar{N}}$ ends after
$2N+254$ terms.

  * •

If $N\equiv 2\pmod{7}$ and $N\geq 3201$, then $B_{\bar{N}}$ ends after
$2N+524$ terms.

  * •

If $N\equiv 3\pmod{7}$ and $N\geq 4315$, then $B_{\bar{N}}$ ends after
$2N+560$ terms.

  * •

If $N\equiv 4\pmod{7}$ and $N\geq 200$, then $B_{\bar{N}}$ ends after $2N+20$
terms.

  * •

If $N\equiv 5\pmod{7}$ and $N\geq 32478$, then $B_{\bar{N}}$ ends after
$2N+4547$ terms.

  * •

If $N\equiv 6\pmod{7}$ and $N\geq 118$, then $B_{\bar{N}}$ ends after $2N+9$
terms.

The proof of Theorem 2 uses the following lemma.

######  Lemma 1.

Let $K\geq 7$, $c\geq 1$ and $0\leq\gamma\leq 6$. Then, let $\lambda$ and
$\mu$ be integers such that

$\lambda\geq\begin{cases}-2c+2&\text{if }\gamma=0\\\ -2c+1&\text{if
}\gamma=1\\\ -2c+4&\text{if }\gamma=2\\\ -2c+3&\text{if }\gamma=3\\\
-2c+2&\text{if }\gamma=4\\\ -2c+1&\text{if }\gamma=5\\\ -2c&\text{if
}\gamma=6\end{cases}\hskip 14.45377pt\text{and}\hskip
14.45377pt\mu\geq\begin{cases}-c&\text{if }\gamma=0\\\ -c&\text{if
}\gamma=1\\\ -c+3&\text{if }\gamma=2\\\ -c+2&\text{if }\gamma=3\\\
-c+1&\text{if }\gamma=4\\\ -c&\text{if }\gamma=5\\\ -c-1&\text{if
}\gamma=6.\end{cases}$

Define $L=K-7c-\gamma$ and $M=K+L+5$. Then, for arbitrary integers
$a_{1},a_{2},\ldots,a_{L}$, denote the sequence resulting from the
$B$-recurrence and the initial conditions

$\left<\bar{0};1,2,\ldots,K,6,a_{1},a_{2},\ldots,a_{L},2K+\lambda-2,2K+\mu-1,K-2\right>$

by $B_{C}$. The sequence $B_{C}$ follows the following pattern from
$B_{C}(M-3)$ through $B_{C}(2K+\nu)$

$\begin{cases}B_{C}(M+7k)=L+7k+7\\\ B_{C}(M+7k+1)=M+7k+2\\\
B_{C}(M+7k+2)=M+7k+4\\\ B_{C}(M+7k+3)=7\\\ B_{C}(M+7k+4)=2K+2k+\lambda\\\
B_{C}(M+7k+5)=2K+k+\mu\\\ B_{C}(M+7k+6)=K-2\end{cases}$

where

$\nu=\begin{cases}-2&\text{if }\gamma=0\\\ -2&\text{if }\gamma=1\\\ 2&\text{if
}\gamma=2\\\ 1&\text{if }\gamma=3\\\ 0&\text{if }\gamma=4\\\ -1&\text{if
}\gamma=5\\\ -2&\text{if }\gamma=6.\end{cases}$

###### Proof.

The proof is by induction on the index. The base cases are $B_{C}(M-3)$
through $B_{C}(M-1)$, which are part of the initial conditions. Now, suppose
$M\leq n\leq 2K+\nu$, and suppose that $B_{C}(n^{\prime})$ is what we want it
to be for all $M-3\leq n^{\prime}<n$.

There are seven cases to consider:

$n-M\equiv 0\pmod{7}$:

In this case, $n=M+7k$ for some $k$. Applying the $B$-recurrence, we have

$\displaystyle B_{C}(M+7k)$ $\displaystyle=B_{C}(M+7k-B_{C}(M+7k-1))$
$\displaystyle\phantom{=}+B_{C}(M+7k-B_{C}(M+7k-2))$
$\displaystyle\phantom{=}+B_{C}(M+7k-B_{C}(M+7k-3))$
$\displaystyle=B_{C}(M+7k-\left(K-2\right))+B_{C}(M+7k-\left(2K+k-1+\mu\right))$
$\displaystyle\phantom{=}+B_{C}(M+7k-\left(2K+2k-2+\lambda\right))$
$\displaystyle=B_{C}(L+7k+7)+B_{C}(-K+L+6k+6-\mu)$
$\displaystyle\phantom{=}+B_{C}(-K+L+5k+7-\lambda).$

We know that $n\leq 2K+\nu$. Since $n=M+7k$, we have $n-M=7k\leq
2K+\nu-M=K-L-5+\nu=7c+\gamma-5+\nu$. Observe that

$\gamma-5+\nu=\begin{cases}-7&\gamma=0\\\ -6&\gamma=1\\\
-1&\text{otherwise.}\end{cases}$

Since $7k\equiv 0\pmod{7}$, we actually have $7k\leq 7c-7$, since
$\left\lfloor\frac{\gamma-5+\nu}{7}\right\rfloor$ always equals $-1$. In
particular, this means that $L+7k+7\leq L+7c=K-\gamma\leq K$. As a result,
$B_{C}(L+7k+7)=L+7k+7$.

We also have

$-K+L+6k+6-\mu=-7c-\gamma+6k+6-\mu\leq-7c-\gamma+\frac{6}{7}(7c-7)+6-\mu=-c-\gamma-\mu.$

Observe that

$-c-\gamma-\mu=\begin{cases}0&\gamma=0\\\ -1&\gamma=1\\\
-5&\text{otherwise,}\end{cases}$

which implies that $-c-\gamma-\mu\leq 0$. In turn, this means that
$-K+L+6k+6-\mu\leq 0$, implying that $B_{C}(-K+L+6k+6-\mu)=0$.

Similarly, we have that

$-K+L+5k+7-\lambda=-7c-\gamma+5k+7-\lambda\leq-7c-\gamma+\frac{5}{7}(7c-7)+7-\lambda=-2c-\gamma+2-\lambda.$

Observe that

$-2c-\gamma+2-\lambda=\begin{cases}0&\gamma=0\text{ or }\gamma=1\\\
-4&\text{otherwise,}\end{cases}$

which implies that $-2c-\gamma+2-\lambda\leq 0$. In turn, this means that
$-K+L+5k+7-\lambda\leq 0$, implying that $B_{C}(-K+L+5k+7-\lambda)=0$. So,
$B_{C}(M+7k)=L+7k+7$, as required.

$n-M\equiv 1\pmod{7}$:

In this case, $n=M+7k+1$ for some $k$. Applying the $B$-recurrence, we have

$\displaystyle B_{C}(M+7k+1)$ $\displaystyle=B_{C}(M+7k+1-B_{C}(M+7k))$
$\displaystyle\phantom{=}+B_{C}(M+7k+1-B_{C}(M+7k-1))$
$\displaystyle\phantom{=}+B_{C}(M+7k+1-B_{C}(M+7k-2))$
$\displaystyle=B_{C}(M+7k+1-\left(L+7k+7\right))+B_{C}(M+7k+1-\left(K-2\right))$
$\displaystyle\phantom{=}+B_{C}(M+7k+1-\left(2K+k-1+\mu\right))$
$\displaystyle=B_{C}(K-1)+B_{C}(L+7k+8)+B_{C}(-K+L+6k+7-\mu).$

We know that $B_{C}(K-1)=K-1$. We also know that $n\leq 2K+\nu$. Since
$n=M+7k+1$, we have $n-M-1=7k\leq 2K+\nu-M-1=K-L-6+\nu=7c+\gamma-6+\nu$.
Observe that

$\gamma-6+\nu=\begin{cases}-8&\gamma=0\\\ -7&\gamma=1\\\
-2&\text{otherwise.}\end{cases}$

Since $7k\equiv 0\pmod{7}$, we actually have

$7k\leq\begin{cases}7c-14&\gamma=0\\\ 7c-7&\text{otherwise.}\end{cases}$

In particular, this means that if $\gamma=0$ then $L+7k+8\leq L+7c-6=K-6<K$,
and if $\gamma\neq 0$ then $L+7k+8\leq L+7c+1=K-\gamma+1\leq K$. As a result,
$B_{C}(L+7k+8)=L+7k+8$.

We also have $-K+L+6k+7-\mu=-7c-\gamma+6k+7-\mu$. When $\gamma=0$,
$-7c-\gamma+6k+7-\mu\leq-7c-\gamma+\frac{6}{7}(7c-14)+7-\mu=-c-\gamma-5-\mu=-5<0.$
When $\gamma\neq 0$,
$-7c-\gamma+6k+7-\mu\leq-7c-\gamma+\frac{6}{7}(7c-7)+7-\mu=-c-\gamma+1-\mu.$
Observe that

$-c-\gamma+1-\mu=\begin{cases}1&\gamma=0\\\ 0&\gamma=1\\\
-4&\text{otherwise.}\end{cases}$

Putting all cases together imply that $-c-\gamma+1-\mu\leq 0$ when $\gamma\neq
0$. In turn, this all means that $-K+L+6k+7-\mu\leq 0$, implying that
$B_{C}(-K+L+6k+7-\mu)=0$. So,
$B_{C}(M+7k+1)=\left(K-1\right)+\left(L+7k+8\right)=K+L+7k+7=M+7k+2$, as
required.

$n-M\equiv 2\pmod{7}$:

In this case, $n=M+7k+2$ for some $k$. Applying the $B$-recurrence, we have

$\displaystyle B_{C}(M+7k+2)$ $\displaystyle=B_{C}(M+7k+2-B_{C}(M+7k+1))$
$\displaystyle\phantom{=}+B_{C}(M+7k+2-B_{C}(M+7k))$
$\displaystyle\phantom{=}+B_{C}(M+7k+2-B_{C}(M+7k-1))$
$\displaystyle=B_{C}(M+7k+2-\left(M+7k+2\right))$
$\displaystyle\phantom{=}+B_{C}(M+7k+2-\left(L+7k+7\right))+B_{C}(M+7k+2-\left(K-2\right))$
$\displaystyle=B_{C}(0)+B_{C}(K)+B_{C}(L+7k+9).$

We know that $B_{C}(0)=0$ and that $B_{C}(K)=K$. We also know that $n\leq
2K+\nu$. Since $n=M+7k+2$, we have $n-M-2=7k\leq
2K+\nu-M-2=K-L-7+\nu=7c+\gamma-7+\nu$. Observe that

$\gamma-7+\nu=\begin{cases}-9&\gamma=0\\\ -8&\gamma=1\\\
-3&\text{otherwise.}\end{cases}$

Since $7k\equiv 0\pmod{7}$, we actually have

$7k\leq\begin{cases}7c-14&\gamma=0\text{ or }\gamma=1\\\
7c-7&\text{otherwise.}\end{cases}$

In particular, this means that if $\gamma\leq 1$ then $L+7k+9\leq
L+7c-4=K-\gamma-4<K$, and if $\gamma\geq 2$ then $L+7k+9\leq
L+7c+2=K-\gamma+2\leq K$. As a result, $B_{C}(L+7k+9)=L+7k+9$. So,
$B_{C}(M+7k+2)=K+L+7k+9=M+7k+4$, as required.

$n-M\equiv 3\pmod{7}$:

In this case, $n=M+7k+3$ for some $k$. Applying the $B$-recurrence, we have

$\displaystyle B_{C}(M+7k+3)$ $\displaystyle=B_{C}(M+7k+3-B_{C}(M+7k+2))$
$\displaystyle\phantom{=}+B_{C}(M+7k+3-B_{C}(M+7k+1))$
$\displaystyle\phantom{=}+B_{C}(M+7k+3-B_{C}(M+7k))$
$\displaystyle=B_{C}(M+7k+3-\left(M+7k+4\right))$
$\displaystyle\phantom{=}+B_{C}(M+7k+3-\left(M+7k+2\right))$
$\displaystyle\phantom{=}+B_{C}(M+7k+3-\left(L+7k+7\right))$
$\displaystyle=B_{C}(-1)+B_{C}(1)+B_{C}(K+1)=0+1+6=7,$

as required.

$n-M\equiv 4\pmod{7}$:

In this case, $n=M+7k+4$ for some $k$. Applying the $B$-recurrence, we have

$\displaystyle B_{C}(M+7k+4)$ $\displaystyle=B_{C}(M+7k+4-B_{C}(M+7k+3))$
$\displaystyle\phantom{=}+B_{C}(M+7k+4-B_{C}(M+7k+2))$
$\displaystyle\phantom{=}+B_{C}(M+7k+4-B_{C}(M+7k+1))$
$\displaystyle=B_{C}(M+7k+4-7)+B_{C}(M+7k+4-\left(M+7k+4\right))$
$\displaystyle\phantom{=}+B_{C}(M+7k+4-\left(M+7k+2\right))$
$\displaystyle=B_{C}(M+7k-3)+B_{C}(0)+B_{C}(2).$

We know that $B_{C}(0)=0$ and that $B_{C}(2)=2$. By induction, we have
$B_{C}(M+7k-3)=2K+2k+\lambda-2$. So, $B_{C}(M+7k+4)=2K+2k+\lambda$, as
required.

$n-M\equiv 5\pmod{7}$:

In this case, $n=M+7k+5$ for some $k$. Applying the $B$-recurrence, we have

$\displaystyle B_{C}(M+7k+5)$ $\displaystyle=B_{C}(M+7k+5-B_{C}(M+7k+4))$
$\displaystyle\phantom{=}+B_{C}(M+7k+5-B_{C}(M+7k+3))$
$\displaystyle\phantom{=}+B_{C}(M+7k+5-B_{C}(M+7k+2))$
$\displaystyle=B_{C}(M+7k+5-\left(2K+2k+\lambda\right))+B_{C}(M+7k+5-7)$
$\displaystyle\phantom{=}+B_{C}(M+7k+5-\left(M+7k+4\right))$
$\displaystyle=B_{C}(-K+L+5k+10-\lambda)+B_{C}(M+7k-2)+B_{C}(1).$

We know that $B_{C}(1)=1$. By induction, we have $B_{C}(M+7k-2)=2K+k+\mu-1$.
We also know that $n\leq 2K+\nu$. Since $n=M+7k+5$, we have $n-M-5=7k\leq
2K+\nu-M-5=K-L-10+\nu=7c+\gamma-10+\nu$. Observe that

$\gamma-10+\nu=\begin{cases}-12&\gamma=0\\\ -11&\gamma=1\\\
-6&\text{otherwise.}\end{cases}$

Since $7k\equiv 0\pmod{7}$, we actually have

$7k\leq\begin{cases}7c-14&\gamma=0\text{ or }\gamma=1\\\
7c-7&\text{otherwise.}\end{cases}$

From here, we have $-K+L+5k+10-\lambda=-7c-\gamma+5k+10-\lambda$. When
$\gamma\leq 1$,
$-7c-\gamma+5k+10-\lambda\leq-7c-\gamma+\frac{5}{7}(7c-14)+10-\lambda=-2c-\gamma-\lambda<0.$
When $\gamma\geq 2$,
$-7c-\gamma+5k+10-\lambda\leq-7c-\gamma+\frac{5}{7}(7c-7)+10-\lambda=-2c-\gamma+5-\lambda.$
Observe that $-2c-\gamma+5-\lambda=-1$ when $\gamma\geq 2$, which is less than
$0$. In turn, this all means that $-K+L+5k+10-\lambda\leq 0$ regardless of
value of $\gamma$, implying that $B_{C}(-K+L+5k+10-\lambda)=0$. Therefore
$B_{C}(M+7k+5)=2K+k+\mu$, as required.

$n-M\equiv 6\pmod{7}$:

In this case, $n=M+7k+6$ for some $k$. Applying the $B$-recurrence, we have

$\displaystyle B_{C}(M+7k+6)$ $\displaystyle=B_{C}(M+7k+6-B_{C}(M+7k+5))$
$\displaystyle\phantom{=}+B_{C}(M+7k+6-B_{C}(M+7k+4))$
$\displaystyle\phantom{=}+B_{C}(M+7k+6-B_{C}(M+7k+3))$
$\displaystyle=B_{C}(M+7k+6-\left(2K+k+\mu\right))$
$\displaystyle\phantom{=}+B_{C}(M+7k+6-\left(2K+2k+\lambda\right))+B_{C}(M+7k+6-7)$
$\displaystyle=B_{C}(-K+L+6k+11-\mu)+B_{C}(-K+L+5k+11-\lambda)$
$\displaystyle\phantom{=}+B_{C}(M+7k-1).$

By induction, we have $B_{C}(M+7k-1)=K-2$. We also know that $n\leq 2K+\nu$.
Since $n=M+7k+6$, we have $n-M-6=7k\leq
2K+\nu-M-6=K-L-11+\nu=7c+\gamma-11+\nu$. Observe that

$\gamma-11+\nu=\begin{cases}-13&\gamma=0\\\ -12&\gamma=1\\\
-7&\text{otherwise.}\end{cases}$

Since $7k\equiv 0\pmod{7}$, we actually have

$7k\leq\begin{cases}7c-14&\gamma=0\text{ or }\gamma=1\\\
7c-7&\text{otherwise.}\end{cases}$

From here, we have $-K+L+6k+11-\mu=-7c-\gamma+6k+11-\mu$. When $\gamma\leq 1$,
$-7c-\gamma+6k+11-\mu\leq-7c-\gamma+\frac{6}{7}(7c-14)+11-\mu=-c-\gamma-1-\mu<0.$
When $\gamma\geq 2$,
$-7c-\gamma+6k+11-\mu\leq-7c-\gamma+\frac{6}{7}(7c-7)+11-\mu=-c-\gamma+5-\mu.$
Observe that $-c-\gamma+5-\mu=0$ when $\gamma\geq 2$. In turn, this all means
that $-K+L+6k+11-\mu\leq 0$ regardless of value of $\gamma$, implying that
$B_{C}(-K+L+6k+11-\mu)=0$. Similarly, we have
$-K+L+5k+11-\lambda=-7c-\gamma+5k+11-\lambda$. When $\gamma\leq 1$,
$-7c-\gamma+5k+11-\lambda\leq-7c-\gamma+\frac{5}{7}(7c-14)+11-\lambda=-2c-\gamma+1-\lambda<0.$
When $\gamma\geq 2$,
$-7c-\gamma+5k+11-\lambda\leq-7c-\gamma+\frac{5}{7}(7c-7)+11-\lambda=-2c-\gamma+6-\lambda.$
Observe that $-2c-\gamma+6-\lambda=0$ when $\gamma\geq 2$. In turn, this all
means that $-K+L+5k+11-\lambda\leq 0$ regardless of value of $\gamma$,
implying that $B_{C}(-K+L+5k+11-\lambda)=0$. Therefore, $B_{C}(M+7k+6)=K-2$,
as required.

∎

We now prove Theorem 2.

###### Proof.

We refer the reader to Appendix A for terms $B_{\bar{N}}(1)$ through
$B_{\bar{N}}(N+28)$. Those calculations, which are for $B_{N}$, also apply to
$B_{\bar{N}}$. From there, Appendix B calculates and lists terms
$B_{\bar{N}}(N+25)$ through $B_{\bar{N}}(N+69)$. Those calculations are all
valid provided $N\geq 67$. Observe that the values $K=N$,
$c=\left\lfloor\frac{N-65}{7}\right\rfloor$, $\gamma=\left(N-65\right)\bmod
7$, $\lambda=65$, and $\mu=3$ satisfy the conditions of Lemma 1 provided that
$N\geq 72$ (so that $c\geq 1$), and the first $N+69$ terms of $B_{\bar{N}}$
can be used as initial conditions as per that lemma. Keeping in mind that
these choices of parameters mean that $L=65$ and $M=N+70$, Lemma 1 implies we
have the following pattern for $B_{\bar{N}}$ for some time:

$\begin{cases}B_{\bar{N}}(N+70+7k^{\prime})=65+7k^{\prime}+7=7k^{\prime}+72\\\
B_{\bar{N}}(N+70+7k^{\prime}+1)=N+70+7k^{\prime}+2=N+7k^{\prime}+72\\\
B_{\bar{N}}(N+70+7k^{\prime}+2)=N+70+7k^{\prime}+4=N+7k^{\prime}+74\\\
B_{\bar{N}}(N+70+7k^{\prime}+3)=7\\\
B_{\bar{N}}(N+70+7k^{\prime}+4)=2N+2k^{\prime}+65\\\
B_{\bar{N}}(N+70+7k^{\prime}+5)=2N+k^{\prime}+3\\\
B_{\bar{N}}(N+70+7k^{\prime}+6)=N-2.\end{cases}$

Re-indexing so that $k=k^{\prime}+10$ allows us to rewrite this pattern as

$\begin{cases}B_{\bar{N}}(N+7k)=7\left(k-10\right)+72=7k+2\\\
B_{\bar{N}}(N+7k+1)=N+7\left(k-10\right)+72=N+7k+2\\\
B_{\bar{N}}(N+7k+2)=N+7\left(k-10\right)+74=N+7k+4\\\ B_{\bar{N}}(N+7k+3)=7\\\
B_{\bar{N}}(N+7k+4)=2N+2\left(k-10\right)+65=2N+2k+45\\\
B_{\bar{N}}(N+7k+5)=2N+\left(k-10\right)+3=2N+k-7\\\
B_{\bar{N}}(N+7k+6)=N-2,\end{cases}$

which is the desired pattern. Lemma 1 guarantees that this pattern persists
through index $2N+\nu$, where $\nu$ depends on $\gamma$, which, in turn,
depends on $N$. Here, we have

$\nu=\begin{cases}-1&\text{if }N\equiv 0\pmod{7}\\\ -2&\text{if }N\equiv
1\pmod{7}\\\ -2&\text{if }N\equiv 2\pmod{7}\\\ -2&\text{if }N\equiv
3\pmod{7}\\\ 2&\text{if }N\equiv 4\pmod{7}\\\ 1&\text{if }N\equiv 5\pmod{7}\\\
0&\text{if }N\equiv 6\pmod{7},\end{cases}$

as required.

We now prove the remainder of Theorem 2, regarding the ways in which
$B_{\bar{N}}$ can end. For each possibility of $N\bmod 7$, we can compute
terms of $B_{\bar{N}}$ from index $2N+\nu+1$ onward, using the now known
values of $B_{\bar{N}}$ for all smaller indices. These computations are akin
to those in the appendices, and like those we track how large $N$ needs to be
for the computations to be valid. The result is a lengthy and tedious list of
terms and bounds, but the claimed end conditions in the statement of the
theorem are all validated. The full length of the computations can be found on
GitHub222https://github.com/nhf216/B-recurrence-data, with one file for each
value of $N\bmod 7$. ∎

The sequences corresponding to the minimum values of $N$ for each congruence
class in Theorem 2 are all available in OEIS [oeis]*A373234, A373235, A373236,
A373237, A373238, A274058, A373239.

###  3.1 The Remaining Values of $N$

Theorem 2 characterizes the behavior of $B_{\bar{N}}$ for all but $6079$
values of $N\geq 3$. These sequences can be studied individually by generating
the sequences and observing the terms. This study is carried out in [thesis];
what follows is a summary of those findings. All of these sequences end before
$150$ million terms except when

$\displaystyle N\in\\{4,$ $\displaystyle
5,6,7,8,9,10,11,12,13,14,15,18,81,182,193,429,822,1892,2789,3442,7292,$
$\displaystyle 20830,23511,25163\\}.$

Of these $B_{\bar{5}}$ and $B_{\bar{6}}$ are the $B$-sequence, so they last
forever. For

$N\in\left\\{4,7,8,9,10,11,12,13,14,15,18\right\\},$

$B_{\bar{N}}$ appears to last forever but exhibits chaotic behavior, akin to
the sequences in Figure 1. To discuss the remaining values, we need four more
results like Lemma 1. All come from [thesis] and are stated without proof. All
proofs are straightforward but tedious inductive arguments, much like the
proof of Lemma 1. Here, unlike in Lemma 1, the bounds on the parameters are
generally not optimized.

######  Lemma 2.

Let $K\geq 1$ and $M\geq K+5$ be integers. Then, for arbitrary integers
$a_{1},a_{2},\ldots,a_{K}$, denote the sequence resulting from the
$B$-recurrence and the initial conditions

$\left<\bar{0};a_{1},a_{2},\ldots,a_{K},2,M,2\right>$

by $B_{D}$. The sequence $B_{D}$ follows the following pattern starting from
$B_{D}(K+1)$ (and lasting forever):

$\begin{cases}B_{D}(K+2k)=2^{k-1}\cdot M\\\ B_{D}(K+2k+1)=2.\end{cases}$

######  Lemma 3.

Let $K\geq 3$, and $\mu\geq 1$ be integers. Then, for arbitrary integers
$a_{4},a_{5},\ldots,a_{K}$, denote the sequence resulting from the
$B$-recurrence and the initial conditions

$\left<\bar{0};1,2,3,a_{4},a_{5},\ldots,a_{K},K+\mu,3,K+3,K+\mu+1,5\right>$

by $B_{E}$. The sequence $B_{E}$ follows the following pattern from
$B_{E}(K+1)$ through
$B_{E}\\!\left(K+\left\lfloor\frac{5\mu-15}{2}\right\rfloor\right)$:

$\begin{cases}B_{E}(K+5k)=5\\\ B_{E}(K+5k+1)=K+3k+\mu\\\ B_{E}(K+5k+2)=3\\\
B_{E}(K+5k+3)=K+5k+3\\\ B_{E}(K+5k+4)=K+3k+\mu+1.\end{cases}$

######  Lemma 4.

Let $K$, $\lambda$, $\mu_{1}$, $\mu_{2}$, and $\gamma$ be positive integers
with $\lambda>31+K$, $\mu_{1}>\lambda$, $\mu_{2}>\lambda$, and
$\gamma>\lambda$. Then, for arbitrary integers $a_{1},a_{2},\ldots,a_{K}$,
denote the sequence resulting from the $B$-recurrence and the initial
conditions

$\left<\bar{0};a_{1},a_{2},\ldots,a_{K},\lambda,7,\mu_{2},16,\mu_{2},16,\mu_{1},\lambda,7,\mu_{2},16,2\mu_{2},16,\mu_{2},25,\gamma,\lambda,7\right>$

by $B_{T}$. The sequence $B_{T}$ follows the following pattern from
$B_{T}(K+1)$ through $B_{T}(\lambda)$:

$\begin{cases}B_{T}(K+16k)=\mu_{1}\cdot 2^{k-1}+\gamma-\mu_{1}\\\
B_{T}(K+16k+1)=\lambda\\\ B_{T}(K+16k+2)=7\\\ B_{T}(K+16k+3)=\mu_{2}\cdot
2^{k}\\\ B_{T}(K+16k+4)=16\\\ B_{T}(K+16k+5)=\mu_{2}\cdot 2^{k}\\\
B_{T}(K+16k+6)=16\\\ B_{T}(K+16k+7)=\mu_{1}\cdot
2^{k}\end{cases}\quad\begin{cases}B_{T}(K+16k+8)=\lambda\\\
B_{T}(K+16k+9)=7\\\ B_{T}(K+16k+10)=\mu_{2}\cdot 2^{k}\\\
B_{T}(K+16k+11)=16\\\ B_{T}(K+16k+12)=\mu_{2}\cdot 2^{k}\\\
B_{T}(K+16k+13)=16\\\ B_{T}(K+16k+14)=\mu_{2}\cdot 2^{k}\\\
B_{T}(K+16k+15)=25.\end{cases}$

######  Lemma 5.

Let $K$, $\lambda$, $\mu_{1}$, $\mu_{2}$, $\gamma_{1}$, $\gamma_{2}$, and
$\gamma_{3}$ be positive integers with $\lambda>31+K$, $\mu_{1}>\lambda$,
$\mu_{2}>\lambda$, $\gamma_{1}>\lambda$, $\gamma_{2}>\lambda$, and
$\gamma_{3}>\lambda$. Then, for arbitrary integers $a_{1},a_{2},\ldots,a_{K}$,
denote the sequence resulting from the $B$-recurrence and the initial
conditions

$\left<\bar{0};a_{1},a_{2},\ldots,a_{K},16,\mu_{2},7,\gamma_{2},\lambda,16,\lambda,16,\mu_{1},10,\gamma_{3},\mu_{2},7,\lambda,16,\gamma_{1}\right>$

by $B_{U}$. The sequence $B_{U}$ follows the following pattern from
$B_{U}(K+1)$ through $B_{U}(\lambda)$:

$\begin{cases}B_{U}(K+16k)=\mu_{1}\cdot 2^{k}+\gamma_{1}-2\mu_{1}\\\
B_{U}(K+16k+1)=16\\\ B_{U}(K+16k+2)=\mu_{2}\cdot 2^{k}\\\ B_{U}(K+16k+3)=7\\\
B_{U}(K+16k+4)=7k+\gamma_{2}\\\ B_{U}(K+16k+5)=\lambda\\\ B_{U}(K+16k+6)=16\\\
B_{U}(K+16k+7)=\lambda\end{cases}\quad\begin{cases}B_{U}(K+16k+8)=16\\\
B_{U}(K+16k+9)=\mu_{1}\cdot 2^{k}\\\ B_{U}(K+16k+10)=10\\\
B_{U}(K+16k+11)=16k+\gamma_{3}\\\ B_{U}(K+16k+12)=\mu_{2}\cdot 2^{k}\\\
B_{U}(K+16k+13)=7\\\ B_{U}(K+16k+14)=\lambda\\\
B_{U}(K+16k+15)=16.\end{cases}$

For $N\in\left\\{81,182,429,822,1892,2789,7292,23511,25163\right\\}$, Lemma 2
eventually applies, so these sequences continue forever. Both
$B_{\overline{193}}$ and $B_{\overline{3442}}$ also continue forever.
Infinitely many prefixes of these sequences satisfy the hypotheses of Lemma 3.
In other words, these sequences consist of infinitely many chunks of the sort
described by Lemma 3 with some sporadic terms in between. Each such chunk
lasts approximately six times as long as the previous one. The first ten
million terms of $B_{\overline{193}}$ are shown in Figure 2. The only
remaining sequence is $B_{\overline{20830}}$. This sequence ends, but it has a
total of $84975\cdot 2^{560362}+31$ terms, far too many to compute. First,
some initial terms of $B_{\overline{20830}}$ satisfy the conditions of Lemma
4, so that lemma governs the behavior of the sequence for awhile. Then,
shortly after the chunk of $B_{\overline{20830}}$ described by Lemma 4
concludes, Lemma 5 applies. That lemma then governs the behavior of the
sequence for a long time. That chunk terminates at index $84975\cdot
2^{560362}$, after which the sequence lasts only $31$ additional terms
[oeis]*A283887.

Figure 2: Plots of the first 10,000,000 terms of $B_{\overline{193}}$ with
linear scale (left) and logarithmic scale (right)

##  4 Future Work

Studying nested recurrence relations with symbolic initial conditions of this
type was initiated in [hof1thruN] with Hofstadter’s $Q$-recurrence, and it is
continued here with the three-term analog to that recurrence. The obvious idea
is to continue adding more terms to the recurrence. But, that study was
undertaken in [thesis], and it suggests that chaos reigns supreme for the
recurrences with four or more terms, even for large $N$. One question, then,
would be whether any other symbolic initial conditions lead to predictable
solutions to these recurrences. Another direction would be to study these same
sorts of initial conditions with other recurrences, such as the Conolly
recurrence [con] or the Tanny recurrence [tanny].

The other big research direction suggested by this work is further exploration
and discovery of results like Lemmas 1, 2, 3, 4, and 5. These five lemmas all
describe temporary or permanent solutions to the $B$-recurrence that consist
of interleavings of simpler sequences. These particular results appear in this
paper because they are needed to analyze the sequences $B_{\bar{N}}$ for
various values of $N$. But, they are by no means exhaustive among results of
this type. The first known solution of this type to a nested recurrence is
Golomb’s solution to the $Q$-recurrence [golomb], which uses initial
conditions $\left<3,2,1\right>$. Later, Ruskey gave a solution to the
$Q$-recurrence with exponentially growing subsequences [rusk] via initial
conditions $\left<\bar{0};3,6,5,3,6,8\right>$. A general algorithmic framework
for finding and proving these sorts of solutions was laid out in [symbhof].
All of these solutions last forever; the first known occurrence of temporary
interleaved solutions is in [hof1thruN], in analogous work to this paper on
the $Q$-recurrence. Perhaps such solutions could also be worked into the
framework from [symbhof].

##  Appendix A First $24$ terms following initial conditions of $B_{N}$

Assuming $N\geq 9$, these are the first $24$ terms of $B_{N}$ following the
initial conditions.

$\displaystyle\mathbf{B_{N}(N+1)}$
$\displaystyle=B_{N}(N+1-B_{N}(N))+B_{N}(N+1-B_{N}(N-1))$
$\displaystyle\phantom{=}+B_{N}(N+1-B_{N}(N-2))$
$\displaystyle=B_{N}(N+1-N)+B_{N}(N+1-(N-1))+B_{N}(N+1-(N-2))$
$\displaystyle=B_{N}(1)+B_{N}(2)+B_{N}(3)=1+2+3=\mathbf{6}$
$\displaystyle(N\geq 3)$ $\displaystyle\mathbf{B_{N}(N+2)}$
$\displaystyle=B_{N}(N+2-B_{N}(N+1))+B_{N}(N+2-B_{N}(N))$
$\displaystyle\phantom{=}+B_{N}(N+2-B_{N}(N-1))$
$\displaystyle=B_{N}(N+2-6)+B_{N}(N+2-N)+B_{N}(N+2-(N-1))$
$\displaystyle=B_{N}(N-4)+B_{N}(2)+B_{N}(3)=(N-4)+2+3=\mathbf{N+1}$
$\displaystyle(N\geq 5)$ $\displaystyle\mathbf{B_{N}(N+3)}$
$\displaystyle=B_{N}(N+3-B_{N}(N+2))+B_{N}(N+3-B_{N}(N+1))$
$\displaystyle\phantom{=}+B_{N}(N+3-B_{N}(N))$
$\displaystyle=B_{N}(N+3-(N+1))+B_{N}(N+3-6)+B_{N}(N+3-N)$
$\displaystyle=B_{N}(2)+B_{N}(N-3)+B_{N}(3)=2+(N-3)+3=\mathbf{N+2}$
$\displaystyle(N\geq 4)$ $\displaystyle\mathbf{B_{N}(N+4)}$
$\displaystyle=B_{N}(N+4-B_{N}(N+3))+B_{N}(N+4-B_{N}(N+2))$
$\displaystyle\phantom{=}+B_{N}(N+4-B_{N}(N+1))$
$\displaystyle=B_{N}(N+4-(N+2))+B_{N}(N+4-(N+1))+B_{N}(N+4-6)$
$\displaystyle=B_{N}(2)+B_{N}(3)+B_{N}(N-2)=2+3+(N-2)=\mathbf{N+3}$
$\displaystyle(N\geq 3)$ $\displaystyle\mathbf{B_{N}(N+5)}$
$\displaystyle=B_{N}(N+5-B_{N}(N+4))+B_{N}(N+5-B_{N}(N+3))$
$\displaystyle\phantom{=}+B_{N}(N+5-B_{N}(N+2))$
$\displaystyle=B_{N}(N+5-(N+3))+B_{N}(N+5-(N+2))+B_{N}(N+5-(N+1))$
$\displaystyle=B_{N}(2)+B_{N}(3)+B_{N}(4)=2+3+4=\mathbf{9}$
$\displaystyle(N\geq 4)$ $\displaystyle\mathbf{B_{N}(N+6)}$
$\displaystyle=B_{N}(N+6-B_{N}(N+5))+B_{N}(N+6-B_{N}(N+4))$
$\displaystyle\phantom{=}+B_{N}(N+6-B_{N}(N+3))$
$\displaystyle=B_{N}(N+6-9)+B_{N}(N+6-(N+3))+B_{N}(N+6-(N+2))$
$\displaystyle=B_{N}(N-3)+B_{N}(3)+B_{N}(4)=(N-3)+3+4=\mathbf{N+4}$
$\displaystyle(N\geq 4)$ $\displaystyle\mathbf{B_{N}(N+7)}$
$\displaystyle=B_{N}(N+7-B_{N}(N+6))+B_{N}(N+7-B_{N}(N+5))$
$\displaystyle\phantom{=}+B_{N}(N+7-B_{N}(N+4))$
$\displaystyle=B_{N}(N+7-(N+4))+B_{N}(N+7-9)+B_{N}(N+7-(N+3))$
$\displaystyle=B_{N}(3)+B_{N}(N-2)+B_{N}(4)=3+(N-2)+4=\mathbf{N+5}$
$\displaystyle(N\geq 4)$ $\displaystyle\mathbf{B_{N}(N+8)}$
$\displaystyle=B_{N}(N+8-B_{N}(N+7))+B_{N}(N+8-B_{N}(N+6))$
$\displaystyle\phantom{=}+B_{N}(N+8-B_{N}(N+5))$
$\displaystyle=B_{N}(N+8-(N+5))+B_{N}(N+8-(N+4))+B_{N}(N+8-9)$
$\displaystyle=B_{N}(3)+B_{N}(4)+B_{N}(N-1)=3+4+(N-1)=\mathbf{N+6}$
$\displaystyle(N\geq 4)$ $\displaystyle\mathbf{B_{N}(N+9)}$
$\displaystyle=B_{N}(N+9-B_{N}(N+8))+B_{N}(N+9-B_{N}(N+7))$
$\displaystyle\phantom{=}+B_{N}(N+9-B_{N}(N+6))$
$\displaystyle=B_{N}(N+9-(N+6))+B_{N}(N+9-(N+5))+B_{N}(N+9-(N+4))$
$\displaystyle=B_{N}(3)+B_{N}(4)+B_{N}(5)=3+4+5=\mathbf{12}$
$\displaystyle(N\geq 5)$ $\displaystyle\mathbf{B_{N}(N+10)}$
$\displaystyle=B_{N}(N+10-B_{N}(N+9))+B_{N}(N+10-B_{N}(N+8))$
$\displaystyle\phantom{=}+B_{N}(N+10-B_{N}(N+7))$
$\displaystyle=B_{N}(N+10-12)+B_{N}(N+10-(N+6))+B_{N}(N+10-(N+5))$
$\displaystyle=B_{N}(N-2)+B_{N}(4)+B_{N}(5)=(N-2)+4+5=\mathbf{N+7}$
$\displaystyle(N\geq 5)$ $\displaystyle\mathbf{B_{N}(N+11)}$
$\displaystyle=B_{N}(N+11-B_{N}(N+10))+B_{N}(N+11-B_{N}(N+9))$
$\displaystyle\phantom{=}+B_{N}(N+11-B_{N}(N+8))$
$\displaystyle=B_{N}(N+11-(N+7))+B_{N}(N+11-12)+B_{N}(N+11-(N+6))$
$\displaystyle=B_{N}(4)+B_{N}(N-1)+B_{N}(5)=4+(N-1)+5=\mathbf{N+8}$
$\displaystyle(N\geq 5)$ $\displaystyle\mathbf{B_{N}(N+12)}$
$\displaystyle=B_{N}(N+12-B_{N}(N+11))+B_{N}(N+12-B_{N}(N+10))$
$\displaystyle\phantom{=}+B_{N}(N+12-B_{N}(N+9))$
$\displaystyle=B_{N}(N+12-(N+8))+B_{N}(N+12-(N+7))+B_{N}(N+12-12)$
$\displaystyle=B_{N}(4)+B_{N}(5)+B_{N}(N)=4+5+N=\mathbf{N+9}$
$\displaystyle(N\geq 5)$ $\displaystyle\mathbf{B_{N}(N+13)}$
$\displaystyle=B_{N}(N+13-B_{N}(N+12))+B_{N}(N+13-B_{N}(N+11))$
$\displaystyle\phantom{=}+B_{N}(N+13-B_{N}(N+10))$
$\displaystyle=B_{N}(N+13-(N+9))+B_{N}(N+13-(N+8))+B_{N}(N+13-(N+7))$
$\displaystyle=B_{N}(4)+B_{N}(5)+B_{N}(6)=4+5+6=\mathbf{15}$
$\displaystyle(N\geq 6)$ $\displaystyle\mathbf{B_{N}(N+14)}$
$\displaystyle=B_{N}(N+14-B_{N}(N+13))+B_{N}(N+14-B_{N}(N+12))$
$\displaystyle\phantom{=}+B_{N}(N+14-B_{N}(N+11))$
$\displaystyle=B_{N}(N+14-15)+B_{N}(N+14-(N+9))+B_{N}(N+14-(N+8))$
$\displaystyle=B_{N}(N-1)+B_{N}(5)+B_{N}(6)=(N-1)+5+6=\mathbf{N+10}$
$\displaystyle(N\geq 6)$ $\displaystyle\mathbf{B_{N}(N+15)}$
$\displaystyle=B_{N}(N+15-B_{N}(N+14))+B_{N}(N+15-B_{N}(N+13))$
$\displaystyle\phantom{=}+B_{N}(N+15-B_{N}(N+12))$
$\displaystyle=B_{N}(N+15-(N+10))+B_{N}(N+15-15)+B_{N}(N+15-(N+9))$
$\displaystyle=B_{N}(5)+B_{N}(N)+B_{N}(6)=5+N+6=\mathbf{N+11}$
$\displaystyle(N\geq 6)$ $\displaystyle\mathbf{B_{N}(N+16)}$
$\displaystyle=B_{N}(N+16-B_{N}(N+15))+B_{N}(N+16-B_{N}(N+14))$
$\displaystyle\phantom{=}+B_{N}(N+16-B_{N}(N+13))$
$\displaystyle=B_{N}(N+16-(N+11))+B_{N}(N+16-(N+10))+B_{N}(N+16-15)$
$\displaystyle=B_{N}(5)+B_{N}(6)+B_{N}(N+1)=5+6+6=\mathbf{17}$
$\displaystyle(N\geq 6)$ $\displaystyle\mathbf{B_{N}(N+17)}$
$\displaystyle=B_{N}(N+17-B_{N}(N+16))+B_{N}(N+17-B_{N}(N+15))$
$\displaystyle\phantom{=}+B_{N}(N+17-B_{N}(N+14))$
$\displaystyle=B_{N}(N+17-17)+B_{N}(N+17-(N+11))+B_{N}(N+17-(N+10))$
$\displaystyle=B_{N}(N)+B_{N}(6)+B_{N}(7)=N+6+7=\mathbf{N+13}$
$\displaystyle(N\geq 7)$ $\displaystyle\mathbf{B_{N}(N+18)}$
$\displaystyle=B_{N}(N+18-B_{N}(N+17))+B_{N}(N+18-B_{N}(N+16))$
$\displaystyle\phantom{=}+B_{N}(N+18-B_{N}(N+15))$
$\displaystyle=B_{N}(N+18-(N+13))+B_{N}(N+18-17)+B_{N}(N+18-(N+11))$
$\displaystyle=B_{N}(5)+B_{N}(N+1)+B_{N}(7)=5+6+7=\mathbf{18}$
$\displaystyle(N\geq 7)$ $\displaystyle\mathbf{B_{N}(N+19)}$
$\displaystyle=B_{N}(N+19-B_{N}(N+18))+B_{N}(N+19-B_{N}(N+17))$
$\displaystyle\phantom{=}+B_{N}(N+19-B_{N}(N+16))$
$\displaystyle=B_{N}(N+19-18)+B_{N}(N+19-(N+13))+B_{N}(N+19-17)$
$\displaystyle=B_{N}(N+1)+B_{N}(6)+B_{N}(N+2)=6+6+(N+1)=\mathbf{N+13}$
$\displaystyle(N\geq 6)$ $\displaystyle\mathbf{B_{N}(N+20)}$
$\displaystyle=B_{N}(N+20-B_{N}(N+19))+B_{N}(N+20-B_{N}(N+18))$
$\displaystyle\phantom{=}+B_{N}(N+20-B_{N}(N+17))$
$\displaystyle=B_{N}(N+20-(N+13))+B_{N}(N+20-18)+B_{N}(N+20-(N+13))$
$\displaystyle=B_{N}(7)+B_{N}(N+2)+B_{N}(7)=7+(N+1)+7=\mathbf{N+15}$
$\displaystyle(N\geq 7)$ $\displaystyle\mathbf{B_{N}(N+21)}$
$\displaystyle=B_{N}(N+21-B_{N}(N+20))+B_{N}(N+21-B_{N}(N+19))$
$\displaystyle\phantom{=}+B_{N}(N+21-B_{N}(N+18))$
$\displaystyle=B_{N}(N+21-(N+15))+B_{N}(N+21-(N+13))+B_{N}(N+21-18)$
$\displaystyle=B_{N}(6)+B_{N}(8)+B_{N}(N+3)=6+8+(N+2)=\mathbf{N+16}$
$\displaystyle(N\geq 8)$ $\displaystyle\mathbf{B_{N}(N+22)}$
$\displaystyle=B_{N}(N+22-B_{N}(N+21))+B_{N}(N+22-B_{N}(N+20))$
$\displaystyle\phantom{=}+B_{N}(N+22-B_{N}(N+19))$
$\displaystyle=B_{N}(N+22-(N+16))+B_{N}(N+22-(N+15))$
$\displaystyle\phantom{=}+B_{N}(N+22-(N+13))$
$\displaystyle=B_{N}(6)+B_{N}(7)+B_{N}(9)=6+7+9=\mathbf{22}$
$\displaystyle(N\geq 9)$ $\displaystyle\mathbf{B_{N}(N+23)}$
$\displaystyle=B_{N}(N+23-B_{N}(N+22))+B_{N}(N+23-B_{N}(N+21))$
$\displaystyle\phantom{=}+B_{N}(N+23-B_{N}(N+20))$
$\displaystyle=B_{N}(N+23-22)+B_{N}(N+23-(N+16))+B_{N}(N+23-(N+15))$
$\displaystyle=B_{N}(N+1)+B_{N}(7)+B_{N}(8)=6+7+8=\mathbf{21}$
$\displaystyle(N\geq 8)$ $\displaystyle\mathbf{B_{N}(N+24)}$
$\displaystyle=B_{N}(N+24-B_{N}(N+23))+B_{N}(N+24-B_{N}(N+22))$
$\displaystyle\phantom{=}+B_{N}(N+24-B_{N}(N+21))$
$\displaystyle=B_{N}(N+24-21)+B_{N}(N+24-22)+B_{N}(N+24-(N+16))$
$\displaystyle=B_{N}(N+3)+B_{N}(N+2)+B_{N}(8)=(N+2)+(N+1)+8=\mathbf{2N+11}$
$\displaystyle(N\geq 8)$

##  Appendix B Terms $25$ through $69$ following initial conditions of
$B_{\bar{N}}$

Assuming that $N\geq 67$, these are terms $25$ through $69$ of $B_{\bar{N}}$
following the initial conditions. (Terms $1$ through $24$ agree with $B_{N}$
and can be found in Appendix A.)

$\displaystyle\mathbf{B_{\bar{N}}(N+25)}$
$\displaystyle=B_{\bar{N}}(N+25-B_{\bar{N}}(N+24))+B_{\bar{N}}(N+25-B_{\bar{N}}(N+23))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+25-B_{\bar{N}}(N+22))$
$\displaystyle=B_{\bar{N}}(N+25-(2N+11))+B_{\bar{N}}(N+25-21)+B_{\bar{N}}(N+25-22)$
$\displaystyle=B_{\bar{N}}(-N+14)+B_{\bar{N}}(N+4)+B_{\bar{N}}(N+3)$
$\displaystyle=0+(N+3)+(N+2)=\mathbf{2N+5}$ $\displaystyle(N\geq 14)$
$\displaystyle\mathbf{B_{\bar{N}}(N+26)}$
$\displaystyle=B_{\bar{N}}(N+26-B_{\bar{N}}(N+25))+B_{\bar{N}}(N+26-B_{\bar{N}}(N+24))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+26-B_{\bar{N}}(N+23))$
$\displaystyle=B_{\bar{N}}(N+26-(2N+5))+B_{\bar{N}}(N+26-(2N+11))+B_{\bar{N}}(N+26-21)$
$\displaystyle=B_{\bar{N}}(-N+21)+B_{\bar{N}}(-N+15)+B_{\bar{N}}(N+5)=0+0+9=\mathbf{9}$
$\displaystyle(N\geq 21)$ $\displaystyle\mathbf{B_{\bar{N}}(N+27)}$
$\displaystyle=B_{\bar{N}}(N+27-B_{\bar{N}}(N+26))+B_{\bar{N}}(N+27-B_{\bar{N}}(N+25))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+27-B_{\bar{N}}(N+24))$
$\displaystyle=B_{\bar{N}}(N+27-9)+B_{\bar{N}}(N+27-(2N+5))+B_{\bar{N}}(N+27-(2N+11))$
$\displaystyle=B_{\bar{N}}(N+18)+B_{\bar{N}}(-N+22)+B_{\bar{N}}(-N+16)=18+0+0=\mathbf{18}$
$\displaystyle(N\geq 22)$ $\displaystyle\mathbf{B_{\bar{N}}(N+28)}$
$\displaystyle=B_{\bar{N}}(N+28-B_{\bar{N}}(N+27))+B_{\bar{N}}(N+28-B_{\bar{N}}(N+26))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+28-B_{\bar{N}}(N+25))$
$\displaystyle=B_{\bar{N}}(N+28-18)+B_{\bar{N}}(N+28-9)+B_{\bar{N}}(N+28-(2N+5))$
$\displaystyle=B_{\bar{N}}(N+10)+B_{\bar{N}}(N+19)+B_{\bar{N}}(-N+23)$
$\displaystyle=(N+7)+(N+13)+0=\mathbf{2N+20}$ $\displaystyle(N\geq 23)$
$\displaystyle\mathbf{B_{\bar{N}}(N+29)}$
$\displaystyle=B_{\bar{N}}(N+29-B_{\bar{N}}(N+28))+B_{\bar{N}}(N+29-B_{\bar{N}}(N+27))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+29-B_{\bar{N}}(N+26))$
$\displaystyle=B_{\bar{N}}(N+29-(2N+20))+B_{\bar{N}}(N+29-18)+B_{\bar{N}}(N+29-9)$
$\displaystyle=B_{\bar{N}}(-N+9)+B_{\bar{N}}(N+11)+B_{\bar{N}}(N+20)$
$\displaystyle=0+(N+8)+(N+15)=\mathbf{2N+23}$ $\displaystyle(N\geq 9)$
$\displaystyle\mathbf{B_{\bar{N}}(N+30)}$
$\displaystyle=B_{\bar{N}}(N+30-B_{\bar{N}}(N+29))+B_{\bar{N}}(N+30-B_{\bar{N}}(N+28))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+30-B_{\bar{N}}(N+27))$
$\displaystyle=B_{\bar{N}}(N+30-(2N+23))+B_{\bar{N}}(N+30-(2N+20))+B_{\bar{N}}(N+30-18)$
$\displaystyle=B_{\bar{N}}(-N+7)+B_{\bar{N}}(-N+10)+B_{\bar{N}}(N+12)=0+0+(N+9)=\mathbf{N+9}$
$\displaystyle(N\geq 10)$ $\displaystyle\mathbf{B_{\bar{N}}(N+31)}$
$\displaystyle=B_{\bar{N}}(N+31-B_{\bar{N}}(N+30))+B_{\bar{N}}(N+31-B_{\bar{N}}(N+29))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+31-B_{\bar{N}}(N+28))$
$\displaystyle=B_{\bar{N}}(N+31-(N+9))+B_{\bar{N}}(N+31-(2N+23))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+31-(2N+20))$
$\displaystyle=B_{\bar{N}}(22)+B_{\bar{N}}(-N+8)+B_{\bar{N}}(-N+11)=22+0+0=\mathbf{22}$
$\displaystyle(N\geq 22)$ $\displaystyle\mathbf{B_{\bar{N}}(N+32)}$
$\displaystyle=B_{\bar{N}}(N+32-B_{\bar{N}}(N+31))+B_{\bar{N}}(N+32-B_{\bar{N}}(N+30))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+32-B_{\bar{N}}(N+29))$
$\displaystyle=B_{\bar{N}}(N+32-22)+B_{\bar{N}}(N+32-(N+9))+B_{\bar{N}}(N+32-(2N+23))$
$\displaystyle=B_{\bar{N}}(N+10)+B_{\bar{N}}(23)+B_{\bar{N}}(-N+9)=(N+7)+23+0=\mathbf{N+30}$
$\displaystyle(N\geq 23)$ $\displaystyle\mathbf{B_{\bar{N}}(N+33)}$
$\displaystyle=B_{\bar{N}}(N+33-B_{\bar{N}}(N+32))+B_{\bar{N}}(N+33-B_{\bar{N}}(N+31))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+33-B_{\bar{N}}(N+30))$
$\displaystyle=B_{\bar{N}}(N+33-(N+30))+B_{\bar{N}}(N+33-22)+B_{\bar{N}}(N+33-(N+9))$
$\displaystyle=B_{\bar{N}}(3)+B_{\bar{N}}(N+11)+B_{\bar{N}}(24)=3+(N+8)+24=\mathbf{N+35}$
$\displaystyle(N\geq 24)$ $\displaystyle\mathbf{B_{\bar{N}}(N+34)}$
$\displaystyle=B_{\bar{N}}(N+34-B_{\bar{N}}(N+33))+B_{\bar{N}}(N+34-B_{\bar{N}}(N+32))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+34-B_{\bar{N}}(N+31))$
$\displaystyle=B_{\bar{N}}(N+34-(N+35))+B_{\bar{N}}(N+34-(N+30))+B_{\bar{N}}(N+34-22)$
$\displaystyle=B_{\bar{N}}(-1)+B_{\bar{N}}(4)+B_{\bar{N}}(N+12)=0+4+(N+9)=\mathbf{N+13}$
$\displaystyle(N\geq 4)$ $\displaystyle\mathbf{B_{\bar{N}}(N+35)}$
$\displaystyle=B_{\bar{N}}(N+35-B_{\bar{N}}(N+34))+B_{\bar{N}}(N+35-B_{\bar{N}}(N+33))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+35-B_{\bar{N}}(N+32))$
$\displaystyle=B_{\bar{N}}(N+35-(N+13))+B_{\bar{N}}(N+35-(N+35))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+35-(N+30))$
$\displaystyle=B_{\bar{N}}(22)+B_{\bar{N}}(0)+B_{\bar{N}}(5)=22+0+5=\mathbf{27}$
$\displaystyle(N\geq 22)$ $\displaystyle\mathbf{B_{\bar{N}}(N+36)}$
$\displaystyle=B_{\bar{N}}(N+36-B_{\bar{N}}(N+35))+B_{\bar{N}}(N+36-B_{\bar{N}}(N+34))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+36-B_{\bar{N}}(N+33))$
$\displaystyle=B_{\bar{N}}(N+36-27)+B_{\bar{N}}(N+36-(N+13))+B_{\bar{N}}(N+36-(N+35))$
$\displaystyle=B_{\bar{N}}(N+9)+B_{\bar{N}}(23)+B_{\bar{N}}(1)=12+23+1=\mathbf{36}$
$\displaystyle(N\geq 23)$ $\displaystyle\mathbf{B_{\bar{N}}(N+37)}$
$\displaystyle=B_{\bar{N}}(N+37-B_{\bar{N}}(N+36))+B_{\bar{N}}(N+37-B_{\bar{N}}(N+35))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+37-B_{\bar{N}}(N+34))$
$\displaystyle=B_{\bar{N}}(N+37-36)+B_{\bar{N}}(N+37-27)+B_{\bar{N}}(N+37-(N+13))$
$\displaystyle=B_{\bar{N}}(N+1)+B_{\bar{N}}(N+10)+B_{\bar{N}}(24)=6+(N+7)+24=\mathbf{N+37}$
$\displaystyle(N\geq 24)$ $\displaystyle\mathbf{B_{\bar{N}}(N+38)}$
$\displaystyle=B_{\bar{N}}(N+38-B_{\bar{N}}(N+37))+B_{\bar{N}}(N+38-B_{\bar{N}}(N+36))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+38-B_{\bar{N}}(N+35))$
$\displaystyle=B_{\bar{N}}(N+38-(N+37))+B_{\bar{N}}(N+38-36)+B_{\bar{N}}(N+38-27)$
$\displaystyle=B_{\bar{N}}(1)+B_{\bar{N}}(N+2)+B_{\bar{N}}(N+11)=1+(N+1)+(N+8)=\mathbf{2N+10}$
$\displaystyle(N\geq 1)$ $\displaystyle\mathbf{B_{\bar{N}}(N+39)}$
$\displaystyle=B_{\bar{N}}(N+39-B_{\bar{N}}(N+38))+B_{\bar{N}}(N+39-B_{\bar{N}}(N+37))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+39-B_{\bar{N}}(N+36))$
$\displaystyle=B_{\bar{N}}(N+39-(2N+10))+B_{\bar{N}}(N+39-(N+37))+B_{\bar{N}}(N+39-36)$
$\displaystyle=B_{\bar{N}}(-N+29)+B_{\bar{N}}(2)+B_{\bar{N}}(N+3)=0+2+(N+2)=\mathbf{N+4}$
$\displaystyle(N\geq 29)$ $\displaystyle\mathbf{B_{\bar{N}}(N+40)}$
$\displaystyle=B_{\bar{N}}(N+40-B_{\bar{N}}(N+39))+B_{\bar{N}}(N+40-B_{\bar{N}}(N+38))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+40-B_{\bar{N}}(N+37))$
$\displaystyle=B_{\bar{N}}(N+40-(N+4))+B_{\bar{N}}(N+40-(2N+10))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+40-(N+37))$
$\displaystyle=B_{\bar{N}}(36)+B_{\bar{N}}(-N+30)+B_{\bar{N}}(3)=36+0+3=\mathbf{39}$
$\displaystyle(N\geq 36)$ $\displaystyle\mathbf{B_{\bar{N}}(N+41)}$
$\displaystyle=B_{\bar{N}}(N+41-B_{\bar{N}}(N+40))+B_{\bar{N}}(N+41-B_{\bar{N}}(N+39))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+41-B_{\bar{N}}(N+38))$
$\displaystyle=B_{\bar{N}}(N+41-39)+B_{\bar{N}}(N+41-(N+4))+B_{\bar{N}}(N+41-(2N+10))$
$\displaystyle=B_{\bar{N}}(N+2)+B_{\bar{N}}(37)+B_{\bar{N}}(-N+31)=(N+1)+37+0=\mathbf{N+38}$
$\displaystyle(N\geq 37)$ $\displaystyle\mathbf{B_{\bar{N}}(N+42)}$
$\displaystyle=B_{\bar{N}}(N+42-B_{\bar{N}}(N+41))+B_{\bar{N}}(N+42-B_{\bar{N}}(N+40))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+42-B_{\bar{N}}(N+39))$
$\displaystyle=B_{\bar{N}}(N+42-(N+38))+B_{\bar{N}}(N+42-39)+B_{\bar{N}}(N+42-(N+4))$
$\displaystyle=B_{\bar{N}}(4)+B_{\bar{N}}(N+3)+B_{\bar{N}}(38)=4+(N+2)+38=\mathbf{N+44}$
$\displaystyle(N\geq 38)$ $\displaystyle\mathbf{B_{\bar{N}}(N+43)}$
$\displaystyle=B_{\bar{N}}(N+43-B_{\bar{N}}(N+42))+B_{\bar{N}}(N+43-B_{\bar{N}}(N+41))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+43-B_{\bar{N}}(N+40))$
$\displaystyle=B_{\bar{N}}(N+43-(N+44))+B_{\bar{N}}(N+43-(N+38))+B_{\bar{N}}(N+43-39)$
$\displaystyle=B_{\bar{N}}(-1)+B_{\bar{N}}(5)+B_{\bar{N}}(N+4)=0+5+(N+3)=\mathbf{N+8}$
$\displaystyle(N\geq 5)$ $\displaystyle\mathbf{B_{\bar{N}}(N+44)}$
$\displaystyle=B_{\bar{N}}(N+44-B_{\bar{N}}(N+43))+B_{\bar{N}}(N+44-B_{\bar{N}}(N+42))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+44-B_{\bar{N}}(N+41))$
$\displaystyle=B_{\bar{N}}(N+44-(N+8))+B_{\bar{N}}(N+44-(N+44))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+44-(N+38))$
$\displaystyle=B_{\bar{N}}(36)+B_{\bar{N}}(0)+B_{\bar{N}}(6)=36+0+6=\mathbf{42}$
$\displaystyle(N\geq 36)$ $\displaystyle\mathbf{B_{\bar{N}}(N+45)}$
$\displaystyle=B_{\bar{N}}(N+45-B_{\bar{N}}(N+44))+B_{\bar{N}}(N+45-B_{\bar{N}}(N+43))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+45-B_{\bar{N}}(N+42))$
$\displaystyle=B_{\bar{N}}(N+45-42)+B_{\bar{N}}(N+45-(N+8))+B_{\bar{N}}(N+45-(N+44))$
$\displaystyle=B_{\bar{N}}(N+3)+B_{\bar{N}}(37)+B_{\bar{N}}(1)=(N+2)+37+1=\mathbf{N+40}$
$\displaystyle(N\geq 37)$ $\displaystyle\mathbf{B_{\bar{N}}(N+46)}$
$\displaystyle=B_{\bar{N}}(N+46-B_{\bar{N}}(N+45))+B_{\bar{N}}(N+46-B_{\bar{N}}(N+44))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+46-B_{\bar{N}}(N+43))$
$\displaystyle=B_{\bar{N}}(N+46-(N+40))+B_{\bar{N}}(N+46-42)+B_{\bar{N}}(N+46-(N+8))$
$\displaystyle=B_{\bar{N}}(6)+B_{\bar{N}}(N+4)+B_{\bar{N}}(38)=6+(N+3)+38=\mathbf{N+47}$
$\displaystyle(N\geq 38)$ $\displaystyle\mathbf{B_{\bar{N}}(N+47)}$
$\displaystyle=B_{\bar{N}}(N+47-B_{\bar{N}}(N+46))+B_{\bar{N}}(N+47-B_{\bar{N}}(N+45))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+47-B_{\bar{N}}(N+44))$
$\displaystyle=B_{\bar{N}}(N+47-(N+47))+B_{\bar{N}}(N+47-(N+40))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+47-42)$
$\displaystyle=B_{\bar{N}}(0)+B_{\bar{N}}(7)+B_{\bar{N}}(N+5)=0+7+9=\mathbf{16}$
$\displaystyle(N\geq 7)$ $\displaystyle\mathbf{B_{\bar{N}}(N+48)}$
$\displaystyle=B_{\bar{N}}(N+48-B_{\bar{N}}(N+47))+B_{\bar{N}}(N+48-B_{\bar{N}}(N+46))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+48-B_{\bar{N}}(N+45))$
$\displaystyle=B_{\bar{N}}(N+48-16)+B_{\bar{N}}(N+48-(N+47))+B_{\bar{N}}(N+48-(N+40))$
$\displaystyle=B_{\bar{N}}(N+32)+B_{\bar{N}}(1)+B_{\bar{N}}(8)=(N+30)+1+8=\mathbf{N+39}$
$\displaystyle(N\geq 8)$ $\displaystyle\mathbf{B_{\bar{N}}(N+49)}$
$\displaystyle=B_{\bar{N}}(N+49-B_{\bar{N}}(N+48))+B_{\bar{N}}(N+49-B_{\bar{N}}(N+47))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+49-B_{\bar{N}}(N+46))$
$\displaystyle=B_{\bar{N}}(N+49-(N+39))+B_{\bar{N}}(N+49-16)+B_{\bar{N}}(N+49-(N+47))$
$\displaystyle=B_{\bar{N}}(10)+B_{\bar{N}}(N+33)+B_{\bar{N}}(2)=10+(N+35)+2=\mathbf{N+47}$
$\displaystyle(N\geq 10)$ $\displaystyle\mathbf{B_{\bar{N}}(N+50)}$
$\displaystyle=B_{\bar{N}}(N+50-B_{\bar{N}}(N+49))+B_{\bar{N}}(N+50-B_{\bar{N}}(N+48))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+50-B_{\bar{N}}(N+47))$
$\displaystyle=B_{\bar{N}}(N+50-(N+47))+B_{\bar{N}}(N+50-(N+39))+B_{\bar{N}}(N+50-16)$
$\displaystyle=B_{\bar{N}}(3)+B_{\bar{N}}(11)+B_{\bar{N}}(N+34)=3+11+(N+13)=\mathbf{N+27}$
$\displaystyle(N\geq 11)$ $\displaystyle\mathbf{B_{\bar{N}}(N+51)}$
$\displaystyle=B_{\bar{N}}(N+51-B_{\bar{N}}(N+50))+B_{\bar{N}}(N+51-B_{\bar{N}}(N+49))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+51-B_{\bar{N}}(N+48))$
$\displaystyle=B_{\bar{N}}(N+51-(N+27))+B_{\bar{N}}(N+51-(N+47))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+51-(N+39))$
$\displaystyle=B_{\bar{N}}(24)+B_{\bar{N}}(4)+B_{\bar{N}}(12)=24+4+12=\mathbf{40}$
$\displaystyle(N\geq 24)$ $\displaystyle\mathbf{B_{\bar{N}}(N+52)}$
$\displaystyle=B_{\bar{N}}(N+52-B_{\bar{N}}(N+51))+B_{\bar{N}}(N+52-B_{\bar{N}}(N+50))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+52-B_{\bar{N}}(N+49))$
$\displaystyle=B_{\bar{N}}(N+52-40)+B_{\bar{N}}(N+52-(N+27))+B_{\bar{N}}(N+52-(N+47))$
$\displaystyle=B_{\bar{N}}(N+12)+B_{\bar{N}}(25)+B_{\bar{N}}(5)=(N+9)+25+5=\mathbf{N+39}$
$\displaystyle(N\geq 25)$ $\displaystyle\mathbf{B_{\bar{N}}(N+53)}$
$\displaystyle=B_{\bar{N}}(N+53-B_{\bar{N}}(N+52))+B_{\bar{N}}(N+53-B_{\bar{N}}(N+51))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+53-B_{\bar{N}}(N+50))$
$\displaystyle=B_{\bar{N}}(N+53-(N+39))+B_{\bar{N}}(N+53-40)+B_{\bar{N}}(N+53-(N+27))$
$\displaystyle=B_{\bar{N}}(14)+B_{\bar{N}}(N+13)+B_{\bar{N}}(26)=14+15+26=\mathbf{55}$
$\displaystyle(N\geq 26)$ $\displaystyle\mathbf{B_{\bar{N}}(N+54)}$
$\displaystyle=B_{\bar{N}}(N+54-B_{\bar{N}}(N+53))+B_{\bar{N}}(N+54-B_{\bar{N}}(N+52))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+54-B_{\bar{N}}(N+51))$
$\displaystyle=B_{\bar{N}}(N+54-55)+B_{\bar{N}}(N+54-(N+39))+B_{\bar{N}}(N+54-40)$
$\displaystyle=B_{\bar{N}}(N-1)+B_{\bar{N}}(15)+B_{\bar{N}}(N+14)$
$\displaystyle=(N-1)+15+(N+10)=\mathbf{2N+24}$ $\displaystyle(N\geq 15)$
$\displaystyle\mathbf{B_{\bar{N}}(N+55)}$
$\displaystyle=B_{\bar{N}}(N+55-B_{\bar{N}}(N+54))+B_{\bar{N}}(N+55-B_{\bar{N}}(N+53))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+55-B_{\bar{N}}(N+52))$
$\displaystyle=B_{\bar{N}}(N+55-(2N+24))+B_{\bar{N}}(N+55-55)+B_{\bar{N}}(N+55-(N+39))$
$\displaystyle=B_{\bar{N}}(-N+31)+B_{\bar{N}}(N)+B_{\bar{N}}(16)=0+N+16=\mathbf{N+16}$
$\displaystyle(N\geq 31)$ $\displaystyle\mathbf{B_{\bar{N}}(N+56)}$
$\displaystyle=B_{\bar{N}}(N+56-B_{\bar{N}}(N+55))+B_{\bar{N}}(N+56-B_{\bar{N}}(N+54))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+56-B_{\bar{N}}(N+53))$
$\displaystyle=B_{\bar{N}}(N+56-(N+16))+B_{\bar{N}}(N+56-(2N+24))+B_{\bar{N}}(N+56-55)$
$\displaystyle=B_{\bar{N}}(40)+B_{\bar{N}}(-N+32)+B_{\bar{N}}(N+1)=40+0+6=\mathbf{46}$
$\displaystyle(N\geq 40)$ $\displaystyle\mathbf{B_{\bar{N}}(N+57)}$
$\displaystyle=B_{\bar{N}}(N+57-B_{\bar{N}}(N+56))+B_{\bar{N}}(N+57-B_{\bar{N}}(N+55))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+57-B_{\bar{N}}(N+54))$
$\displaystyle=B_{\bar{N}}(N+57-46)+B_{\bar{N}}(N+57-(N+16))+B_{\bar{N}}(N+57-(2N+24))$
$\displaystyle=B_{\bar{N}}(N+11)+B_{\bar{N}}(41)+B_{\bar{N}}(-N+33)=(N+8)+41+0=\mathbf{N+49}$
$\displaystyle(N\geq 41)$ $\displaystyle\mathbf{B_{\bar{N}}(N+58)}$
$\displaystyle=B_{\bar{N}}(N+58-B_{\bar{N}}(N+57))+B_{\bar{N}}(N+58-B_{\bar{N}}(N+56))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+58-B_{\bar{N}}(N+55))$
$\displaystyle=B_{\bar{N}}(N+58-(N+49))+B_{\bar{N}}(N+58-46)+B_{\bar{N}}(N+58-(N+16))$
$\displaystyle=B_{\bar{N}}(9)+B_{\bar{N}}(N+12)+B_{\bar{N}}(42)=9+(N+9)+42=\mathbf{N+60}$
$\displaystyle(N\geq 42)$ $\displaystyle\mathbf{B_{\bar{N}}(N+59)}$
$\displaystyle=B_{\bar{N}}(N+59-B_{\bar{N}}(N+58))+B_{\bar{N}}(N+59-B_{\bar{N}}(N+57))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+59-B_{\bar{N}}(N+56))$
$\displaystyle=B_{\bar{N}}(N+59-(N+60))+B_{\bar{N}}(N+59-(N+49))+B_{\bar{N}}(N+59-46)$
$\displaystyle=B_{\bar{N}}(-1)+B_{\bar{N}}(10)+B_{\bar{N}}(N+13)=0+10+15=\mathbf{25}$
$\displaystyle(N\geq 10)$ $\displaystyle\mathbf{B_{\bar{N}}(N+60)}$
$\displaystyle=B_{\bar{N}}(N+60-B_{\bar{N}}(N+59))+B_{\bar{N}}(N+60-B_{\bar{N}}(N+58))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+60-B_{\bar{N}}(N+57))$
$\displaystyle=B_{\bar{N}}(N+60-25)+B_{\bar{N}}(N+60-(N+60))+B_{\bar{N}}(N+60-(N+49))$
$\displaystyle=B_{\bar{N}}(N+35)+B_{\bar{N}}(0)+B_{\bar{N}}(11)=27+0+11=\mathbf{38}$
$\displaystyle(N\geq 11)$ $\displaystyle\mathbf{B_{\bar{N}}(N+61)}$
$\displaystyle=B_{\bar{N}}(N+61-B_{\bar{N}}(N+60))+B_{\bar{N}}(N+61-B_{\bar{N}}(N+59))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+61-B_{\bar{N}}(N+58))$
$\displaystyle=B_{\bar{N}}(N+61-38)+B_{\bar{N}}(N+61-25)+B_{\bar{N}}(N+61-(N+60))$
$\displaystyle=B_{\bar{N}}(N+23)+B_{\bar{N}}(N+36)+B_{\bar{N}}(1)=21+36+1=\mathbf{58}$
$\displaystyle(N\geq 1)$ $\displaystyle\mathbf{B_{\bar{N}}(N+62)}$
$\displaystyle=B_{\bar{N}}(N+62-B_{\bar{N}}(N+61))+B_{\bar{N}}(N+62-B_{\bar{N}}(N+60))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+62-B_{\bar{N}}(N+59))$
$\displaystyle=B_{\bar{N}}(N+62-58)+B_{\bar{N}}(N+62-38)+B_{\bar{N}}(N+62-25)$
$\displaystyle=B_{\bar{N}}(N+4)+B_{\bar{N}}(N+24)+B_{\bar{N}}(N+37)$
$\displaystyle=(N+3)+(2N+11)+(N+37)=\mathbf{4N+51}$ $\displaystyle(N\geq 1)$
$\displaystyle\mathbf{B_{\bar{N}}(N+63)}$
$\displaystyle=B_{\bar{N}}(N+63-B_{\bar{N}}(N+62))+B_{\bar{N}}(N+63-B_{\bar{N}}(N+61))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+63-B_{\bar{N}}(N+60))$
$\displaystyle=B_{\bar{N}}(N+63-(4N+51))+B_{\bar{N}}(N+63-58)+B_{\bar{N}}(N+63-38)$
$\displaystyle=B_{\bar{N}}(-3N+12)+B_{\bar{N}}(N+5)+B_{\bar{N}}(N+25)$
$\displaystyle=0+9+(2N+5)=\mathbf{2N+14}$ $\displaystyle(N\geq 4)$
$\displaystyle\mathbf{B_{\bar{N}}(N+64)}$
$\displaystyle=B_{\bar{N}}(N+64-B_{\bar{N}}(N+63))+B_{\bar{N}}(N+64-B_{\bar{N}}(N+62))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+64-B_{\bar{N}}(N+61))$
$\displaystyle=B_{\bar{N}}(N+64-(2N+14))+B_{\bar{N}}(N+64-(4N+51))+B_{\bar{N}}(N+64-58)$
$\displaystyle=B_{\bar{N}}(-N+50)+B_{\bar{N}}(-3N+13)+B_{\bar{N}}(N+6)$
$\displaystyle=0+0+(N+4)=\mathbf{N+4}$ $\displaystyle(N\geq 50)$
$\displaystyle\mathbf{B_{\bar{N}}(N+65)}$
$\displaystyle=B_{\bar{N}}(N+65-B_{\bar{N}}(N+64))+B_{\bar{N}}(N+65-B_{\bar{N}}(N+63))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+65-B_{\bar{N}}(N+62))$
$\displaystyle=B_{\bar{N}}(N+65-(N+4))+B_{\bar{N}}(N+65-(2N+14))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+65-(4N+51))$
$\displaystyle=B_{\bar{N}}(61)+B_{\bar{N}}(-N+51)+B_{\bar{N}}(-3N+14)=61+0+0=\mathbf{61}$
$\displaystyle(N\geq 61)$ $\displaystyle\mathbf{B_{\bar{N}}(N+66)}$
$\displaystyle=B_{\bar{N}}(N+66-B_{\bar{N}}(N+65))+B_{\bar{N}}(N+66-B_{\bar{N}}(N+64))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+66-B_{\bar{N}}(N+63))$
$\displaystyle=B_{\bar{N}}(N+66-61)+B_{\bar{N}}(N+66-(N+4))+B_{\bar{N}}(N+66-(2N+14))$
$\displaystyle=B_{\bar{N}}(N+5)+B_{\bar{N}}(62)+B_{\bar{N}}(-N+52)=9+62+0=\mathbf{71}$
$\displaystyle(N\geq 62)$ $\displaystyle\mathbf{B_{\bar{N}}(N+67)}$
$\displaystyle=B_{\bar{N}}(N+67-B_{\bar{N}}(N+66))+B_{\bar{N}}(N+67-B_{\bar{N}}(N+65))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+67-B_{\bar{N}}(N+64))$
$\displaystyle=B_{\bar{N}}(N+67-71)+B_{\bar{N}}(N+67-61)+B_{\bar{N}}(N+67-(N+4))$
$\displaystyle=B_{\bar{N}}(N-4)+B_{\bar{N}}(N+6)+B_{\bar{N}}(63)$
$\displaystyle=(N-4)+(N+4)+63=\mathbf{2N+63}$ $\displaystyle(N\geq 63)$
$\displaystyle\mathbf{B_{\bar{N}}(N+68)}$
$\displaystyle=B_{\bar{N}}(N+68-B_{\bar{N}}(N+67))+B_{\bar{N}}(N+68-B_{\bar{N}}(N+66))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+68-B_{\bar{N}}(N+65))$
$\displaystyle=B_{\bar{N}}(N+68-(2N+63))+B_{\bar{N}}(N+68-71)+B_{\bar{N}}(N+68-61)$
$\displaystyle=B_{\bar{N}}(-N+5)+B_{\bar{N}}(N-3)+B_{\bar{N}}(N+7)$
$\displaystyle=0+(N-3)+(N+5)=\mathbf{2N+2}$ $\displaystyle(N\geq 5)$
$\displaystyle\mathbf{B_{\bar{N}}(N+69)}$
$\displaystyle=B_{\bar{N}}(N+69-B_{\bar{N}}(N+68))+B_{\bar{N}}(N+69-B_{\bar{N}}(N+67))$
$\displaystyle\phantom{=}+B_{\bar{N}}(N+69-B_{\bar{N}}(N+66))$
$\displaystyle=B_{\bar{N}}(N+69-(2N+2))+B_{\bar{N}}(N+69-(2N+63))+B_{\bar{N}}(N+69-71)$
$\displaystyle=B_{\bar{N}}(-N+67)+B_{\bar{N}}(-N+6)+B_{\bar{N}}(N-2)=0+0+(N-2)=\mathbf{N-2}$
$\displaystyle(N\geq 67)$

## References