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\begin{document}

\title{Problems in Restricted Partitions}
\author{Dakota Blair}
\begin{abstract}
This note contains discussion on problems concerning partitions whose parts are
powers of a given base $b$.
\end{abstract}
\maketitle

\section{Notation}
\begin{comment}
Let $c$ be an ordinal.
If $c$ is finite, associate it to the corresponding integer $c = |c|$.
Let $s$ be a sequence, that is $s = (s_i)_{i \in I}$ or $s = \seq{s_i}{i \in I}$.
Denote the length of $s$ as $|s| = |I|$.
Further if $s = (s_i)_{i \in I}$ and $i' \in I$ then $s(i') = s_{i'}$.
Let $s^{-1}(x) = \set{i \in I}{s(i) = x}$.
Denote by $c^{<\omega}$ the set of all finite sequences $s = (s(i))_{i \in |s|}$
where $s(i) \in c$, that is,
$c^{<\omega}
= \set{s}{|s| < \omega, s = (s(i))_{i \in |s|} \text{ where } s(i) \in c}$.
Let $s, s' \in c^{<\omega}$.
Denote concatenation by juxtaposition, that is
$s'' = ss' = \{s''(i)\}_{i \in |s|+|s'|}$
where $s''(i) = s(i)$ if $i < |s|$ and $s''(i) = s'(i - |s|)$ otherwise.
If $s \in c^{<\omega}$ is written adjacent to $i < c$, as in $si$,
then $i$ is considered to be a sequence of length $1$ and concatenation
as defined above applies.
The shift left operator,
$\shl$ acts by removing a number of initial items from a sequence.
That is, given $s = (s_i)_{i \in |s|}$
let $s \shl 1 = (s(i+1))_{i \in |s|-1}$.
\end{comment}
Let $p_b(n)$ be the number of $b$-ary partitions of $n$.
Let $B_b(q)$ be the generating function of $p_b(n)$, that is,
\[
B_b(q) = \sum_{n = 0}^\infty p_b(n)q^n = \prod_{i = 0}^\infty \frac{1}{1 - q^{b^i}}.
\]
Abstractify this slightly:
\[
\pg{b}{q}{mn} = \sum_{n \geq 0} p_b(mn)q^n.
\]

Let $\diff$ denote the finite difference operator: $\diff_u = f(u+1) - f(u)$.

\section{Previous Work}
\begin{thm}
The $b$-ary parition function satisfies the following recurrence relation:
\begin{align*}
p_b(n) & = 0 \text{ for } n < 0\\
p_b(n) & = 1 \text{ for } n \in b\\ 
p_b(bn) & = p_b(n) + p_b(bn-1)\\
p_b(bn+i) & = p_b(bn) \text{ for } i \in b\\
\end{align*}
\end{thm}

\begin{thm}[\Ro, Sellers]
Let $r \geq 1$ and suppose that $\sigma_r$ can be expressed as
\[
\sigma_r = \sum_{i = 2}^r \epsv_i b^i
\]
where $\epsv_i \in 2$ for each $i$.
Finally, let $c_r = 2^{\ibra{b \text{ odd}}(r-1)}$.
Then for all $n \geq 1$,
\[
p_b(b^{r+1}n - \sigma_r - b) \in \frac{b^r}{c_r}\integer.
\]
\end{thm}

\begin{cor}
Assume a base $b$, a length $r$ and an integer $m \in 2^r$ are given.
Define $c(r, b) = 2^{\ibra{b \text{ odd}}(r-1)}$,
and let $s = (\floor{\frac{m}{2^i}} \mod b)_{i < \omega}$.
For all $n$
\[
p_b(b^{r+1}n - \nu_b(s) - b) \in \frac{b^r}{c(r, b)}\integer.
\]
\end{cor}
\section{Facts}

\begin{align*}
B_b(q)
 & = B_b(q^{b^m}) \prod_{i \in m} \frac{1}{1 - q^{b^i}}\\
B_b(q^b)
 & = (1-q)B_b(q)\\
\end{align*}

\begin{comment}
Let $\zeta_m$ be a primitive $m$th root of unity.
\[
B_b(q, f(n)) = \sum_{n = 0}^\infty p_b(f(n))q^{f(n)}
\]
\begin{align*}
B_b(q^b, n)
 & = (1-q)B_b(q, n)
   = \sum p_b(n) q^{nb}\\
B_b(q, mn)
 & = \sum p_b(mn) q^{mn}
 = \Re(B_b(\zeta_m q))\\
B_b(q^{1/m}, mn)
 & = \sum p_b(mn) q^{n}
 = \Re(B_b(\zeta_m q^{1/m}))\\
B_b(q, nb^{r+1}+c)
 & = \sum p_b(nb^{r+1} + c) q^{nb^{r+1} + c}\\
q^{-c}B_b(q, n)
 & = \sum_{n=0}^\infty p_b(n) q^{n - c} = \sum_{n = c}^\infty p_b(n-c)q^n
   = q^c B_b(q, n-c) \\
\Re((\zeta_m q^{\frac{1}{m}})^{-c}B_b(q^{\frac{1}{m}}\zeta_m))
 & = \sum p_b(nm + c) q^n\\
\end{align*}
\end{comment}

Here is a table corresponding to identities of the form:
\[
f(q) B_b(q) = (1-q)^m \sum p_b(b^m n) q^n
\]

\[
\begin{array}{llll}
b & m=3 & m=4 & m=5\\
2
 & 1 + 6q + q^2
 & 1 + 31q + 31q^2 + q^3
 & 1 + 196q+ 630q^2 + 196q^2 + q^4\\
3
 & 1 + 19q + 7q^2
 & 1 + 234q + 447q^2 + 47q^3
 & 1 + 5822q + 33504q^2 + 19040q^3 + 682q^4\\
4
 & 1 + 42q + 21q^2
 & 1 + 1081q + 2635q^2 + 379q^3\\
5
 & 1 + 78q + 46q^2
 & 1 + 3072q + 10218q^2 + 1704q^3\\
\end{array}
\]

and in general, these identities are proven below:
\[
\begin{array}{cl}
m & f(q)\\
1 & 1\\
2 & 1 + (b-1)q\\
3 & 1 + \half(b-1)\left((b^2+2b+4)q + (b^2-2)q^2\right)\\
\end{array}
\]

\begin{comment}
\begin{dfn}[Another recurrence]
\[
\R(m,n,k) = p_b(b^mn - kb)
\]
\end{dfn}

\begin{align*}
\R(m,n,k) & = p_b(b^{m-1}n - k) + p_b(b^mn - kb - 1)\\
\R(m,n,k) & = p_b(b^{m-1}n - k) + p_b(b^mn - (k+1)b)\\
\R(m,n,k) & = \R(m-1, n, \floor{\frac{k-1}{b}} + 1) + \R(m, n, k+1)\\
\end{align*}

\begin{align*}
\R(m,n-1,0) & = p_b(b^{m}(n-1))\\
            & = p_b(b^{m}n - b^{m}) = p_b(b^{m}n - b^{m-1}b)\\
            & = \R(m, n, b^{m-1})\\
\end{align*}
\end{comment}

\section{Theorems}
\begin{thm}
\label{thm:base}
\[
B_b(q) = (1-q)\sum p_b(bn) q^{b}\\
\]
\end{thm}

\begin{proof}

\begin{align*}
B_b(q) & = \sum_{n \geq 0} p_b(n) q^n\\
    & = \sum_{n \geq 0} (p_b(bn) - p_b(b(n-1))) q^n\\
    & = 1 + \sum_{n \geq 1} (p_b(bn) - p_b(b(n-1))) q^n\\
    & = 1 + \sum_{n \geq 1} p_b(bn)q^n - \sum_{n \geq 1}p_b(b(n-1)) q^n\\
    & = \sum_{n \geq 0} p_b(bn)q^n - \sum_{n \geq 1}p_b(b(n-1)) q^n\\
    & = \sum_{n \geq 0} p_b(bn)q^n - \sum_{n \geq 0}p_b(bn) q^{n+1}\\
    & = (1 - q)\sum_{n \geq 0} p_b(bn)q^n\\
\end{align*}
\end{proof}

\begin{comment}
\begin{thm}
\[
B_2(q) = \frac{(1-q)^2}{1+q}\sum p_2(4n) q^{n}\\
\]
\end{thm}

\begin{proof}
\begin{align*}
p_2(4n) & = p_2(2n) + p_2(4n - 1)\\
        & = p_2(2n) + p_2(2(2n - 1))\\
        & = p_2(2n) + p_2(2n - 1) + p_2(4n - 3)\\
        & = p_2(2n) + p_2(2(n-1)) + p_2(4(n-1))\\
\end{align*}
\begin{align*}
\sum p_2(4n)q^n
 & = \sum p_2(2n)q^n + q\sum p_2(2n)q^n + q\sum p_2(4(n-1))q^n\\
(1-q) \sum p_2(4n)q^n
 & = (1+q)\sum p_2(2n)q^n\\
(1-q) \sum p_2(4n)q^n
 & = \frac{1+q}{1-q}\sum p_2(n)q^n\\
\end{align*}
\[
\frac{(1-q)^2}{1+q}\sum p_2(4n)q^n
 = \sum p_2(n)q^n = B_2(q)
\]
\end{proof}
\end{comment}

Let 
\[
\pg{b}{q}{mn} = \sum_{n \geq 0} p_b(mn)q^n.
\]

\begin{thm}
\[
(1+(b-1)q)B_b(q) = (1-q)^2\pg{b}{q}{b^2}
\]
\end{thm}

\begin{proof}
\begin{align*}
p_b(b^2n) & = p_b(bn) + p_b(b^2n-1)\\
          & = p_b(bn) + p_b(b(bn-1))\\
          & = p_b(bn) + p_b(bn-1) + p_b(b(bn-2))\\
          & = p_b(bn) + p_b(b(n-1)) + p_b(b(bn-2))\\
          & \text{iterate a total of $b-1$ times}\\
          & = p_b(bn) + (b-1)p_b(b(n-1)) + p_b(b(bn-b))\\
          & = p_b(bn) + (b-1)p_b(b(n-1)) + p_b(b(b(n-1)))\\
          & = p_b(bn) + (b-1)p_b(b(n-1)) + p_b(b^2(n-1))\\
\end{align*}
\begin{align*}
\pgbq{2} & = \pgbq{1} + (b-1)q\pgbq{1} + q\pgbq{2}\\
\pgbq{2} & = \pgbq{1} + (b-1)q\pgbq{1} + q\pgbq{2}\\
(1-q)\pgbq{2} & = (1 + (b-1)q)\pgbq{1}\\
(1-q)\pgbq{2} & = (1 + (b-1)q)(1-q)^{-1}B_b(q)
\end{align*}
\[
(1+(b-1)q)B_b(q) = (1-q)^2\pg{b}{q}{b^2}
\]
\end{proof}

\begin{lem}
\label{lem:evalsum}
\[
%\sum_{k = 1}^{b-1}\R(2, n, k) = (b-1)\R(2, n-1, 0) + {b \choose 2}\R(1, n-1, 0)
\sum_{k=1}^{b-1}p_b(b^2n-kb) = (b-1)p_b(b^2(n-1)) + {b \choose 2}p_b(b(n-1))
\]
\end{lem}

\begin{proof}
Note that when $1 < k < b$
\begin{align*}
p_b(b^2n-kb)
 & = p_b(b^2 n - (k+1)b) + p_b(bn - k)\\
 & = p_b(b^2 n - (k+1)b) + p_b(bn - \floor{\frac{k}{b}}b)\\
 & = p_b(b^2 n - (k+1)b) + p_b(bn - b)\\
 & = p_b(b^2 n - (k+1)b) + p_b(bn - bb^0)\\
 & = p_b(b^2 n - (k+1)b) + p_b(b(n-1))\\
\end{align*}

Therefore
\begin{align*}
\sum_{k = 1}^{b-1}p_b(b^2 n - kb)
 & = \sum_{k = 1}^{b-1}(p_b(b^2 n - (k+1)b) + p_b(b(n-1))) \text{ (descend)}\\
 & = (b-1) p_b(b(n-1)) + \sum_{k = 1}^{b-1}p_b(b^2 n - (k+1)b) \text{ (pop)}\\
 & = (b-1) p_b(b(n-1)) + p_b(b^2 n - b^2) + \sum_{k = 1}^{b-2}p_b(b^2 n - (k+1)b)\\
 & = (b-1) p_b(b(n-1)) + p_b(b^2(n-1)) + \sum_{k = 1}^{b-2}p_b(b^2 n - (k+1)b) \text{ (reindex)}\\
 & = (b-1) p_b(b(n-1)) + p_b(b^2(n-1)) + \sum_{k = 2}^{b-1}p_b(b^2 n - kb)\text{ (descend, pop, reindex)} \\
 & = ((b-1) + (b-2)) p_b(b(n-1)) + 2p_b(b^2(n-1)) + \sum_{k = 3}^{b-1}p_b(b^2 n - kb)\\
 & = \cdots \text{descend, pop, reindex a total of $b-1$ times}\\
 & = {b \choose 2}p_b(b(n-1)) + (b-1)p_b(b^2(n-1))\\
 & = (b-1)p_b(b^2(n-1)) + {b \choose 2} p_b(b(n-1))\\
\end{align*}

\begin{comment}
\begin{align*}
\R(2, n, k)
 & = \R(2, n, k+1) + \R(1, n, \floor{\frac{k-1}{b}} + 1)\\
 & = \R(2, n, k+1) + \R(1, n, 1)\\
 & = \R(2, n, k+1) + \R(1, n, b^0)\\
\R(2, n, k)
 & = \R(2, n, k+1) + \R(1, n-1, 0)
\end{align*}
\begin{align*}
\sum_{k = 1}^{b-1}\R(2, n, k)
 & = \sum_{k = 1}^{b-1}(\R(2, n, k+1) + \R(1, n-1, 0))\\
 & = (b-1) \R(1, n-1, 0) + \sum_{k = 1}^{b-1}\R(2, n, k+1)\\
 & = (b-1) \R(1, n-1, 0) + \R(2, n, b) + \sum_{k = 1}^{b-2}\R(2, n, k+1)\\
 & = (b-1) \R(1, n-1, 0) + \R(2, n-1, 0) + \sum_{k = 1}^{b-2}\R(2, n, k)\\
 & = (b-1) \R(1, n-1, 0) + \R(2, n-1, 0) + \sum_{k = 2}^{b-1}\R(2, n, k)\\
 & = ((b-1) + (b-2)) \R(1, n-1, 0) + 2\R(2, n-1, 0) + \sum_{k = 3}^{b-1}\R(2, n, k)\\
 & = \cdots \\
 & = (b-1)\R(2, n-1, 0) + {b \choose 2} \R(1, n-1, 0)\\
\end{align*}
\end{comment}
\end{proof}

\begin{dfn}[The recurrence applied $l$ times]
\[
F^l(p_b(b^m n)) = p_b(b^m n - bl) + \sum_{k = 0}^{l -1}p_b(b^{m-1}n - k)
\]
\end{dfn}

\begin{thm}
\[
f(q)B_b(q) = (1-q)^3\pg{b}{q}{b^3}
\]
where 
\[
f(q) = 1 + \half(b-1)\left((b^2+2b+4)q + (b^2-2)q^2\right)
\]
\end{thm}

\begin{proof}

\begin{comment}
Let's prove this first:
\[
 b\left(\sum_{k = 1}^{b-1}p_b(b^2 n - kb)\right) + (b-1)p_b(b^2n - bb)
 = (b+1)(b-1)p_b(b^2(n-1)) + {b \choose 2}(b-1) p_b(b(n-1))
\]

\[
 b\left(\sum_{k = 1}^{b-1}\R(2, n, k)\right) + (b-1)R(2, n, b)
 = (b+1)(b-1)\R(2, n-1, 0) + {b \choose 2}(b-1) \R(1, n-1, 0)
\]

\begin{align*}
\text{LHS}
    & = b\left(\sum_{k = 1}^{b-1}\R(2, n, k)\right) + (b-1)R(2, n, b)\\
    & = b\left(\sum_{k = 1}^{b-1}\R(2, n, k)\right) + (b-1)R(2, n-1, 0)\\
    & = b(b-1) \R(2, n-1, 0) + b{b \choose 2} \R(1, n-1, 0) + (b-1)R(2, n-1, 0)\\
    & = (b+1)(b-1)\R(2, n-1, 0) + {b \choose 2}(b-1) \R(1, n-1, 0)\\
    & = \text{RHS}
\end{align*}

Then:
\begin{align*}
\R(3,n,0) & = \R(2, n, 0) + \R(3, n, 1)\\
          & = \R(2, n, 0) + \R(3, n, b^2) + b\left(\sum_{k = 1}^{b-1}\R(2, n, k)\right) + (b-1)R(2, n, b)\\
          & = \R(2, n, 0) + \R(3, n-1, 0) + b\left(\sum_{k = 1}^{b-1}\R(2, n, k)\right) + (b-1)R(2, n, b)\\
          & = \R(3, n-1, 0) + \R(2, n, 0) + (b+1)(b-1)\R(2, n-1, 0) + {b \choose 2}(b-1) \R(1, n-1, 0)\\
\end{align*}
\end{comment}

\begin{align*}
p_b(b^3 n)
 & = p_b(b^3 n - b) + p_b(b^2 n)= F(p_b(b^3 n))\\
 & = p_b(b^3 n - b^3) + \sum_{k = 0}^{b^2-1}p_b(b^{2}n - k) = F^{b^2}(p_b(b^3 n))\\
 & = p_b(b^3(n - 1)) + \sum_{k = 0}^{b^2-1}p_b(b^{2}n - k)\\
 & = p_b(b^3(n - 1)) + \sum_{k = 0}^{b^2-1}p_b(b^{2}n - \ceil{\frac{k}{b}}b)\\
 & = p_b(b^3(n - 1)) + p_b(b^2n) + \sum_{k = 1}^{b^2-1}p_b(b^{2}n - \ceil{\frac{k}{b}}b)\\
 & = p_b(b^3(n - 1)) + p_b(b^2n) + \sum_{k = 1}^{(b-1)(b+1)}p_b(b^{2}n - \ceil{\frac{k}{b}}b)\\
 & = p_b(b^3(n - 1)) + p_b(b^2n) + (b-1)p_b(b^2n - b^2) + \sum_{k = 1}^{(b-1)(b)}p_b(b^{2}n - \ceil{\frac{k}{b}}b)\\
 & = p_b(b^3(n - 1)) + p_b(b^2n) + (b-1)p_b(b^2(n - 1)) + \sum_{k = 1}^{(b-1)(b)}p_b(b^{2}n - \ceil{\frac{k}{b}}b)\\
 & = p_b(b^3(n - 1)) + p_b(b^2n) + (b-1)p_b(b^2(n - 1)) + b\sum_{k = 1}^{b-1}p_b(b^{2}n - kb)\\
 & \tab \text{ (so by Lemma \ref{lem:evalsum})}\\
 & = p_b(b^3(n - 1)) + p_b(b^2n) + (b-1)p_b(b^2(n - 1)) + b\left((b-1)p_b(b^2(n-1)) + {b \choose 2}p_b(b(n-1))\right)\\
p_b(b^3 n)
 & = p_b(b^3(n - 1)) + p_b(b^2n) + (b+1)(b-1)p_b(b^2(n - 1)) + b{b \choose 2}p_b(b(n-1))\\
\end{align*}

Finally, upon passing to generating functions:
\begin{align*}
\pgbq{3} & = q\pgbq{3} + \pgbq{2} + (b+1)(b-1)q\pgbq{2} + b{b \choose 2}q\pgbq{1}\\
(1-q)\pgbq{3} & = (1 + (b+1)(b-1)q)\pgbq{2} + b{b \choose 2}q\pgbq{1}\\
(1-q)\pgbq{3} & = (1 + (b+1)(b-1)q)\frac{1+(b-1)q}{(1-q)^2}B_b(q) + b{b \choose 2}q(1-q)^{-1}B_b(q)\\
(1-q)^3\pgbq{3} & = (1 + (b+1)(b-1)q)(1+(b-1))qB_b(q) + b{b \choose 2}q(1-q)B_b(q)\\
(1-q)^3\pgbq{3} & = \left(1 + \left((b^2-1) + (b-1) + b{b \choose 2}\right)q + \left((b^2-1)(b-1) - b{b \choose 2}\right)q^2\right)B_b(q)\\
(1-q)^3\pgbq{3} & = \left(1
    + \left(\half(b-1)(b^2+2b+4)\right)q
    + \left(\half(b-1)(b^2-2)\right)q^2
\right)B_b(q)\\
(1-q)^3\pgbq{3}
& = \left(1 + (\half(b-1)\left(
    + (b^2+2b+4)q
    + (b^2-2)q^2
    \right)\right)B_b(q)
\end{align*}
\end{proof}

\begin{comment}
Also notice that this expansion shows
\begin{thm}
\[
f(q)B_b(q) = (1-q)^m \pgbq{m}
\]
where $f(q)$ is a degree $m-1$ polynomial in $q$.
\end{thm}

\begin{proof}
The base case is established by \ref{thm:base}.
Now observe that applying the recurrence property $\ceil{\frac{j}{b}}$ times to
$p_b(b^m n + j)$
will yield
\[
p_b(b^m n + j) = \sum_{i = 0}^{m} a_i p_b(b^i).
\]
Passing to generating functions
\begin{align*}
 \sum p_b(b^m n) q^n
 & = \sum p_b(b^m n - b^m) q^n + \sum \sum_{k = 0}^{b^m-1}p_b(b^{m-1}n - k)q^n\\
(1-q)\pgbq{m}
 & = \sum_{k=0}^{b^m-1}\sum_{i = 0}^{m-1} a_i p_b(b^i)q^n
\end{align*}
By the induction hypothesis the right hand side is a sum
of rational functions with denominator $(1-q)^{m'}$
where $m'$ is at most $m-1$.
Thus
\[
(1-q)^{m-1}\pgbq{m} = f(q)B_b(q).
\]
BY INDUCTION.
\end{proof}

\end{comment}

The goal now is to evaluate the expression
\begin{align*}
\lhs
 & = p_b(b^m n)\\
 & = F^{b^{m-1}}(p_b(b^m n))\\
 & = p_b(b^m n - b^m) + \sum_{k=0}^{b^{m-1}-1}p_b(b^{m-1}n - k)\\
 & = p_b(b^m(n{-}1)) + \sum_{k=0}^{b^{m-1}-1}p_b(b^{m-1}n - k)\\
 & = p_b(b^m(n{-}1)) + \sum_{k=1}^{b^{m-1}}p_b(b^{m-1}(n-1) + k)\\
 & = p_b(b^m(n{-}1)) + p_b(b^{m-1}n) + \sum_{k=1}^{b-1}p_b(b^{m-1}(n-1)+k) + \ibra{m\geq2}\sum_{k=b}^{b^{m-1}}p_b(b^{m-1}(n-1)+k)\\
 & \text{note the last term is 0 if $m > 2$ thus the case $m=2$ is finished here.}\\
 & = p_b(b^m(n{-}1)) + p_b(b^{m-1}n) + \sum_{k=1}^{b-1}p_b(b^{m-1}(n-1)+k) + \sum_{k=b}^{b^{m-1}}p_b(b^{m-1}(n-1)+k)\\
 & = p_b(b^m(n{-}1)) + p_b(b^{m-1}n) + (b-1)p_b(b^{m-1}(n{-}1)) + \sum_{k=b}^{b^{m-1}}p_b(b^{m-1}(n-1)+k)\\
 & = p_b(b^m(n{-}1)) + p_b(b^{m-1}n) + (b-1)p_b(b^{m-1}(n{-}1)) + \sum_{u=1}^{b^{m-2}}\sum_{v=0}^{b-1}p_b(b^{m-1}(n-1)+ub+v)\\
 & = p_b(b^m(n{-}1)) + p_b(b^{m-1}n) + (b-1)p_b(b^{m-1}(n{-}1)) + \sum_{u=1}^{b^{m-2}}\sum_{v=0}^{b-1}p_b(b^{m-1}(n-1)+ub)\\
 & = p_b(b^m(n{-}1)) + p_b(b^{m-1}n) + (b-1)p_b(b^{m-1}(n{-}1)) + b\sum_{u=1}^{b^{m-2}}p_b(b^{m-1}(n-1)+ub)\\
 & = p_b(b^m(n{-}1)) + p_b(b^{m-1}n) + (b-1)p_b(b^{m-1}(n{-}1)) + bp_b(b^{m-1}n) + b\sum_{u=1}^{b^{m-2}-1}p_b(b^{m-1}(n-1)+ub)\\
 & = p_b(b^m(n{-}1)) + (b+1)p_b(b^{m-1}n) + (b-1)p_b(b^{m-1}(n-1)) + b\sum_{u=1}^{b^{m-2}-1}p_b(b^{m-1}(n-1)+ub)\\
\end{align*}

The expression of interest is now:
\begin{align*}
p_b(b^m n)
 & = p_b(b^m(n{-}1)) + (b+1)p_b(b^{m-1}n) + (b-1)p_b(b^{m-1}(n-1)) + b\sum_{u=1}^{b^{m-2}-1}p_b(b^{m-1}(n-1)+ub)\\
 & \tab \\
\end{align*}

Look at the last term here separately:
\begin{align*}
\lhs
 & = \sum_{k=1}^{b^{m-2}-1}p_b(b^{m-1}(n-1)+kb)\\
 & = \sum_{k=1}^{b^{m-2}-1}\left(p_b(b^{m-1}(n-1)+(k-1)b) + p_b(b^{m-2}(n-1)+k)\right)\\
 & = \sum_{k=1}^{b^{m-2}-1}\left(\ibra{k\geq2}p_b(b^{m-1}(n-1)+(k-2)b) + p_b(b^{m-2}(n-1)+k-1) + p_b(b^{m-2}(n-1)+k)\right)\\
 & = \sum_{k=1}^{b^{m-2}-1}\left(p_b(b^{m-1}(n-1)+(k-2)) + \sum_{j=0}^1\ibra{k\geq j}p_b(b^{m-2}(n-1)+k-j)\right)\\
 & \ldots\\
 & = \sum_{k=1}^{b^{m-2}-1}\left(p_b(b^{m-1}(n-1)) + \sum_{j=0}^{k-1}\ibra{k\geq j}p_b(b^{m-2}(n-1)+k-j)\right)\\
 & = (b^{m-2}-1)p_b(b^{m-1}(n-1)) + \sum_{k=1}^{b^{m-2}}\sum_{j=0}^{k-1}\ibra{k\geq j}p_b(b^{m-2}(n-1)+k-j)\\
\end{align*}

The expression of interest is now:
\begin{align*}
p_b(b^m n)
 & = p_b(b^m(n{-}1)) + (b+1)p_b(b^{m-1}n) + (b-1)p_b(b^{m-1}(n-1))\\
 & \tab + b(b^{m-2}-1)p_b(b^{m-1}(n-1)) + b\sum_{k=1}^{b^{m-2}}\sum_{j=0}^{k-1}\ibra{k\geq j}p_b(b^{m-2}(n-1)+k-j)\\
\end{align*}

Again, the action is in the last term:
\begin{align*}
\lhs
 & = \sum_{k=1}^{b^{m-2}}\sum_{j=0}^{k-1}\ibra{k\geq j}p_b(b^{m-2}(n-1)+k-j)\Let u=k-j\\
 & = \sum_{k=1}^{b^{m-2}-1}\sum_{u=1}^{k}\ibra{u\geq0}p_b(b^{m-2}(n-1)+u)\Let j = u\\
 & = \sum_{k=1}^{b^{m-2}-1}\sum_{j=1}^{k}\ibra{j\geq1}p_b(b^{m-2}(n-1)+j)\\
 & = \sum_{k=1}^{b^{m-2}-1}\sum_{j=1}^{k}\ibra{1\leq j\leq k}p_b(b^{m-2}(n-1)+j)\\
 & = \sum_{k=1}^{b^{m-2}-1}\sum_{j=1}^{b^{m-2}-1}\ibra{1\leq j\leq k}p_b(b^{m-2}(n-1)+j)\\
 & = \sum_{j=1}^{b^{m-2}-1}\sum_{k=1}^{b^{m-2}-1}\ibra{1\leq j\leq k}p_b(b^{m-2}(n-1)+j)\\
 & = \sum_{j=1}^{b^{m-2}-1}p_b(b^{m-2}(n-1)+j)\sum_{k=1}^{b^{m-2}-1}\ibra{1\leq j\leq k}\\
 & = \sum_{j=1}^{b^{m-2}-1}p_b(b^{m-2}(n-1)+j)\sum_{k=j}^{b^{m-2}-1}1\\
 & = \sum_{j=1}^{b^{m-2}-1}p_b(b^{m-2}(n-1)+j)(b^{m-2}-1-j+1)\\%\star why?
 & = \sum_{j=1}^{b^{m-2}-1}(b^{m-2}-j)p_b(b^{m-2}(n-1)+j) \Let k=b^{m-2}-j\\
 & = \sum_{k=1}^{b^{m-2}-1}kp_b(b^{m-2}(n-1)+b^{m-2}-k)\\
 & = \sum_{k=1}^{b^{m-2}-1}kp_b(b^{m-2}n-k)\\
\end{align*}

\begin{align*}
\lhs
 & = \sum_{k=1}^{b^{m-2}-1}kp_b(b^{m-2}n-k)\\
 & = \sum_{k=1}^{b^{m-2}-1}kp_b(b^{m-2}(n-1)+b^{m-2}-k) \Let j=b^{m-2}-k\\
 & = \sum_{j=1}^{b^{m-2}-1}(b^{m-2}-j)p_b(b^{m-2}(n-1)+j)\\
 & = \sum_{j=1}^{b-1}(b^{m-2}-j)p_b(b^{m-2}(n-1)+j) + \ibra{m\geq3}\sum_{j=b}^{b^{m-2}-1}(b^{m-2}-j)p_b(b^{m-2}(n-1)+j)\\
 & = \sum_{j=1}^{b-1}(b^{m-2}-j)p_b(b^{m-2}(n-1)) + \ibra{m\geq3}\sum_{j=b}^{b^{m-2}-1}(b^{m-2}-j)p_b(b^{m-2}(n-1)+j)\\
 & = p_b(b^{m-2}(n-1))\sum_{j=1}^{b-1}(b^{m-2}-j) + \ibra{m\geq3}\sum_{j=b}^{b^{m-2}-1}(b^{m-2}-j)p_b(b^{m-2}(n-1)+j)\\
 & = \left({b^{m-2} \choose 2} - {b^{m-2}-b \choose 2}\right)p_b(b^{m-2}(n-1)) + \ibra{m\geq3}\sum_{j=b}^{b^{m-2}-1}(b^{m-2}-j)p_b(b^{m-2}(n-1)+j)\\
 %& = p_b(b^{m-2}(n-1))\sum_{j=1}^{b-1}(b^{m-2}-b+b-j) + \sum_{j=b}^{b^{m-2}-1}(b^{m-2}-j)p_b(b^{m-2}(n-1)+j)\\
 %& = p_b(b^{m-2}(n-1))(b-1)(b^{m-2}-b)\sum_{j=1}^{b-1}(b-j) + \sum_{j=b}^{b^{m-2}-1}(b^{m-2}-j)p_b(b^{m-2}(n-1)+j)\\
 %& = p_b(b^{m-2}(n-1))(b-1)(b^{m-2}-b)\sum_{j=1}^{b-1}j + \sum_{j=b}^{b^{m-2}-1}(b^{m-2}-j)p_b(b^{m-2}(n-1)+j)\\
 %& = (b-1)(b^{m-2}-b){b-1 \choose 2}p_b(b^{m-2}(n-1)) + \ibra{m\geq3}\sum_{j=b}^{b^{m-2}-1}(b^{m-2}-j)p_b(b^{m-2}(n-1)+j)\\
\end{align*}

The expression of interest is now:
\begin{align*}
p_b(b^m n)
 & = p_b(b^m(n{-}1)) + (b+1)p_b(b^{m-1}n) + (b-1)p_b(b^{m-1}(n-1)) + b(b^{m-2}-1)p_b(b^{m-1}(n-1))\\
 & \tab + b(b-1)(b^{m-2}-b){b-1 \choose 2}p_b(b^{m-2}(n-1)) + b\sum_{j=b}^{b^{m-2}-1}(b^{m-2}-j)p_b(b^{m-2}(n-1)+j)\\
\end{align*}

Again, the action is in the last term:
\begin{align*}
\lhs
 & = \sum_{j=b}^{b^{m-2}-1}(b^{m-2}-j)p_b(b^{m-2}(n-1)+j)\\
 & = \sum_{u=1}^{b^{m-3}-1}\sum_{v=0}^{b-1}(b^{m-2}-ub-v)p_b(b^{m-2}(n-1)+ub+v)\\
 & = \sum_{u=1}^{b^{m-3}-1}\sum_{v=0}^{b-1}(b^{m-2}-ub-v)p_b(b^{m-2}(n-1)+ub)\\
 & = \sum_{u=1}^{b^{m-3}-1}p_b(b^{m-2}(n-1)+ub)\sum_{v=0}^{b-1}(b^{m-2}-ub-v)\\
 %& = \sum_{u=1}^{b^{m-3}-1}p_b(b^{m-2}(n-1)+ub)\sum_{v=0}^{b-1}(b^{m-2}-(u-1)b+b-v)\\
 %& = \sum_{u=1}^{b^{m-3}-1}p_b(b^{m-2}(n-1)+ub)\left((b^{m-2}-(u-1)b)b + \sum_{v=0}^{b-1}(b-v)\right)\\
 %& = \sum_{u=1}^{b^{m-3}-1}p_b(b^{m-2}(n-1)+ub)\left((b^{m-2}-(u-1)b)b + \sum_{v=1}^{b}v\right)\\
 & = \sum_{u=1}^{b^{m-3}-1}p_b(b^{m-2}(n-1)+ub)\left((b^{m-2}-(u-1)b)b + {b+1 \choose 2}\right)\\
\end{align*}

Let $u_0 = ub + v$:
\begin{align*}
p_b(b^{m-1}(n-1)+u_0)
& = p_b(b^{m-1}(n-1)+ub+v)\\
& = p_b(b^{m-1}(n-1)+\floor{\frac{ub+v}{b}}b)\\
& = p_b(b^{m-1}(n-1)+ub)\\
& = p_b(b^{m-2}(n-1)+u) + p_b(b^{m-1}(n-1)+(u-1)b) \\
& = p_b(b^{m-2}(n-1)+u) + p_b(b^{m-2}(n-1)+(u-1)) + p_b(b^{m-1}(n-1)+(u-2)b) \\
& = \sum_{k=0}^{u}p_b(b^{m-2}(n-1)+(u-k)) + p_b(b^{m-1}(n-1)) \\
& = \sum_{k=0}^{u}p_b(b^{m-2}(n-1)+(u-k)) + p_b(b^{m-2}(n-2)) + p_b(b^{m-1}(n-1)-b) \\
& = \sum_{k=0}^{u}p_b(b^{m-2}(n-1)+(u-k)) + p_b(b^{m-2}(n-2)) + p_b(b^{m-1}(n-2)+(b^{m-1}-b)) \\
\end{align*}

\begin{align*}
p_b(b^m n)
 & = p_b(b^m(n - 1)) + \sum_{k = 0}^{b^m-1}p_b(b^{m-1}n - k)\\
 & = p_b(b^m(n - 1)) + \sum_{k = 0}^{b^m-1}p_b(b^{m-1}(n-1) + k)\\
\end{align*}

Note
\[
k = \sum_{i=0}^{l-1}\eps_i b^i = \sum_{i=0}^{l-1}\floor{\frac{k - \floor{\frac{k}{b^{i+1}}}b^{i+1}}{b^{i}}}b^i
\]
Let $k = \sum_{i=0}^{l-1}\eps_i b^i$ where $\eps_i \in b$. 
Then
\begin{align*}
p_b(b^{m}n + k)
 & = p_b(b^{m}n + \sum_{i=0}^{l-1}\eps_i b^i)\\
 & = p_b(b^{m}n + \eps_0 + b\sum_{i=1}^{l-1}\eps_i b^{i-1})\\
 & = p_b(b^{m}n + b\sum_{i=1}^{l-1}\eps_i b^{i-1})\\
 & = p_b(b^{m}n + b(\eps_1 - 1) + b^2\sum_{i=2}^{l-1}\eps_i b^{i-2}) + p_b(b^{m-1}n + \sum_{i=1}^{l-1}\eps_i b^{i-1})\\
 & = p_b(b^{m}n + b(\eps_1 - 1) + b^2\floor{\frac{k}{b^2}}) + p_b(b^{m-1}n + \floor{\frac{k}{b}})\\
 & = p_b(b^{m}n + b(\eps_1 - 2) + b^2\floor{\frac{k}{b^2}}) + p_b(b^{m-1}n + \floor{\frac{k}{b}} - 1) + p_b(b^{m-1}n + \floor{\frac{k}{b}})\\
 & = p_b(b^{m}n + b^2\floor{\frac{k}{b^2}}) + \eps_1 p_b(b^{m-1}n + b\floor{\frac{k}{b^2}})\\
 & = p_b(b^{m}n + b(b-1) + b^2(\eps_2-1) + b^3\floor{\frac{k}{b^3}})
   + p_b(b^{m-1}n + b\floor{\frac{k}{b^2}}) + \eps_1 p_b(b^{m-1}n + b\floor{\frac{k}{b^2}})\\
\end{align*}

If $\eps_1 > 1$
\begin{align*}
p_b(b^{m}n + k)
 & = p_b(b^{m}n + \sum_{i=0}^{l-1}\eps_i b^i) \text{ case }(m, 1)\\
 & = p_b(b^{m}n + b^2\floor{\frac{k}{b^2}}) + \eps_1 p_b(b^{m-1}n + b\floor{\frac{k}{b^2}})\\
\end{align*}

Then $p_b(b^{m}n + b^2\floor{\frac{k}{b^2}})$ is like the case $(m, 2)$.
$p_b(b^mn + k) = p_b(b^mn) + \text{ $k$ terms}$
\[
 p_b(b^mn + \eps_{m-1}b^{m-1}) = p_b(b^mn) + \text{ $\eps_{m-1}b^{m-1}$ terms}
\]

\begin{align*}
\text{LHS}
 & = p_b(b^{m}n + \eps_{m-1}b^{m-1})\\
 & = p_b(b^{m}n+\eps_{m-1}b^{m-1}-1) + p_b(b^{m-1}n+\eps_{m-1}b^{m-2})\\
 & = p_b(b^{m}n+\eps_{m-1}b^{m-1}-b) + p_b(b^{m-1}n+\eps_{m-1}b^{m-2})\\
 & = p_b(b^{m}n+(\eps_{m-1}-1)b^{m-1}+b^{m-1}-b) + p_b(b^{m-1}n+\eps_{m-1}b^{m-2})\\
 & = p_b(b^{m}n+(\eps_{m-1}-1)b^{m-1}+b(b^{m-2}-1)) + p_b(b^{m-1}n+\eps_{m-1}b^{m-2}) \tab \star\\
 & = p_b(b^{m}n+(\eps_{m-1}-1)b^{m-1}+b(b^{m-2}-2))\\
 & \tab + p_b(b^{m-1}n+(\eps_{m-1}-1)b^{m-2}+b^{m-2}-1) + p_b(b^{m-1}n+\eps_{m-1}b^{m-2}) \\
 & = p_b(b^{m}n+(\eps_{m-1}-1)b^{m-1}+b(b^{m-2}-3))\\
 & \tab + \sum_{k=0}^{2} p_b(b^{m-1}n+(\eps_{m-1}-1)b^{m-2}+b^{m-2}-k)\\
 & = p_b(b^{m}n+(\eps_{m-1}-1)b^{m-1})\\
 & \tab + \sum_{k=0}^{b^{m-2}-1} p_b(b^{m-1}n+(\eps_{m-1}-1)b^{m-2}+b^{m-2}-k)\\
 & = p_b(b^{m}n+(\eps_{m-1}-1)b^{m-1}) + \sum_{k=1}^{b^{m-2}} p_b(b^{m-1}n+(\eps_{m-1}-1)b^{m-2}+k) \tab \star\star \\
 & = p_b(b^{m}n+(\eps_{m-1}-2)b^{m-1})\\
 & \tab + \sum_{j=1}^{2}\sum_{k=1}^{b^{m-2}} p_b(b^{m-1}n+(\eps_{m-1}-j)b^{m-2}+k)\\
 & = p_b(b^{m}n) + \sum_{j=1}^{\eps_{m-1}}\sum_{k=1}^{b^{m-2}} p_b(b^{m-1}n+(\eps_{m-1}-j)b^{m-2}+k)\\
 & = p_b(b^{m}n) + \sum_{j=0}^{\eps_{m-1}-1}\sum_{k=1}^{b^{m-2}} p_b(b^{m-1}n+jb^{m-2}+k)\\
 & = p_b(b^{m}n) + \sum_{j=0}^{\eps_{m-1}-1}\sum_{k=1}^{b^{m-2}} p_b(b^{m-1}n+jb^{m-2}+b\floor{\frac{k}{b}})\\
\end{align*}

\begin{lem}

\end{lem}

%\begin{proof}
\begin{align*}
\lhs(n, m, j, k)
 & = p_b(b^mn+b^{m-j}\floor{\frac{k}{b^{m-j}}})\\
 & = p_b(b^mn+b(b^{m-j-1}\floor{\frac{k}{b^{m-j}}}))\\
 & = p_b(b^mn+b(b^{m-j-1}\left(b\floor{\frac{k}{b^{m-j+1}}}+\eps_{m-j}\right)))\\
 & = p_b(b^mn+b\left(b^{m-j}\floor{\frac{k}{b^{m-j+1}}}+b^{m-j-1}\eps_{m-j}\right))\\
 & = p_b(b^mn+b\left(b^{m-j}\floor{\frac{k}{b^{m-j+1}}}+b^{m-j-1}\eps_{m-j}-1\right))\\
 & \tab + p_b(b^{m-1}n+b^{m-j}\floor{\frac{k}{b^{m-j+1}}}+b^{m-j-1}\eps_{m-j})\\
 & = p_b(b^mn+b\left(b^{m-j}\floor{\frac{k}{b^{m-j+1}}}+b^{m-j-1}\eps_{m-j}-2\right))\\
 & \tab + \sum_{t=0}^{1}p_b(b^{m-1}n+b^{m-j}\floor{\frac{k}{b^{m-j+1}}}+b^{m-j-1}\eps_{m-j}-t)\\
 & \ldots\\
 & = p_b(b^mn+b^{m-j+1}\floor{\frac{k}{b^{m-j+1}}})\\
 & \tab + \sum_{t=0}^{b^{m-j-1}\eps_{m-j}-1}p_b(b^{m-1}n+b^{m-j}\floor{\frac{k}{b^{m-j+1}}}+b^{m-j-1}\eps_{m-j}-t)\\
 & = p_b(b^mn+b^{m-j+1}\floor{\frac{k}{b^{m-j+1}}})
 + \sum_{t=1}^{b^{m-j-1}\eps_{m-j}}p_b(b^{m-1}n+b^{m-j}\floor{\frac{k}{b^{m-j+1}}}+t)\\
 & = \lhs(n, m, j-1, k)
 + \sum_{t=1}^{b^{m-j-1}\eps_{m-j}}p_b(b^{m-1}n+b^{m-j}\floor{\frac{k}{b^{m-j+1}}}+t)\\
 & = \lhs(n, m, j-1, k)
 + \sum_{t=b}^{b^{m-j-1}\eps_{m-j}-1}p_b(b^{m-1}n+b^{m-j}\floor{\frac{k}{b^{m-j+1}}}+t)\\
 & \tab + (b-1) p_b(b^{m-1}n+b^{m-j}\floor{\frac{k}{b^{m-j+1}}})
 + p_b(b^{m-1}n+b^{m-j}\floor{\frac{k}{b^{m-j+1}}}+b^{m-j-1}\eps_{m-j})\\
 & = \lhs(n, m, j-1, k)
 + \sum_{t=b}^{b^{m-j-1}\eps_{m-j}-1}p_b(b^{m-1}n+b^{m-j}\floor{\frac{k}{b^{m-j+1}}}+t)\\
 & \tab + (b-1) p_b(b^{m-1}n+b^{m-j}\floor{\frac{k}{b^{m-j+1}}})
 + p_b(b^{m-1}n+b^{m-j-1}\floor{\frac{k}{b^{m-j}}})\\
\end{align*}
\begin{align*}
\lhs(n, m, j, k)
 & = \lhs(n, m, j-1, k)
 + \sum_{t=b}^{b^{m-j-1}\eps_{m-j}-1}p_b(b^{m-1}n+b^{m-j}\floor{\frac{k}{b^{m-j+1}}}+t)\\
 & \tab + (b-1)\lhs(n, m-1, j-1, \frac{k}{b})
 + p_b(b^{m-1}n+b^{m-j-1}\floor{\frac{k}{b^{m-j}}})\\
 & = \lhs(n, m, j-1, k) + (b-1)\lhs(n,m-1,j-1,\frac{k}{b}) + \lhs(n,m-1,j,\frac{k}{b})\\
 & \tab + \sum_{t=b}^{b^{m-j-1}\eps_{m-j}-1}p_b(b^{m-1}n+b^{m-j}\floor{\frac{k}{b^{m-j+1}}}+t)\\
 & = \lhs(n, m, j-1, k) + (b-1)\lhs(n,m-1,j-1,\frac{k}{b}) + \lhs(n,m-1,j,\frac{k}{b})\\
 & \tab + \sum_{u=1}^{b^{m-j-2}\eps_{m-j}}\sum_{v=0}^{b-1}p_b(b^{m-1}n+b^{m-j}\floor{\frac{k}{b^{m-j+1}}}+ub+v)\\
 & = \lhs(n, m, j-1, k) + (b-1)\lhs(n,m-1,j-1,\frac{k}{b}) + \lhs(n,m-1,j,\frac{k}{b})\\
 & \tab + b\sum_{u=1}^{b^{m-j-2}\eps_{m-j}}p_b(b^{m-1}n+b^{m-j}\floor{\frac{k}{b^{m-j+1}}}+ub)\\
\lhs(n, m, j, k)
 & = \lhs(n, m, j-1, k) + (b-1)\lhs(n,m-1,j-1,\frac{k}{b}) + \lhs(n,m-1,j,\frac{k}{b})\\
 & \tab + b\sum_{u=1}^{b^{m-j-2}\eps_{m-j}}\lhs(n,m-1,m-1,b^{m-j}\floor{\frac{k}{b^{m-j+1}}}+ub)\\
\end{align*}
%\end{proof}

These quantities appear often, so name them:
\begin{align*}
       P(m) & = p_b(b^m(n-1))\\
  T_b(m, n) & = \sum_{k=1}^{b^{m-1}-1}p_b(b^{m-1}n-k)\\
U_b(k_0, m) & = \sum_{k=k_0}^{b^{m-2}-1} p_b(b^{m-1}n - kb)\\
V_b(k_0, m) & = \sum_{k=k_0}^{b^{m-2}-1} p_b(b^{m-2}n - k)\\
S_b(k_0, m) & = \sum_{k=k_0}^{b^{m-2}-1} p_b(b^{m-1}n - (k+1)b)\\
\end{align*}

The expression of interest is now:
\begin{align*}
p_b(b^m n)
 & = p_b(b^m(n - 1)) + p_b(b^{m-1} n) + \sum_{k = 1}^{b^m-1}p_b(b^{m-1}n - k)\\
% & = P(m) + p_b(b^{m-1} n) + T_b(m, n)\\
 & = P(m) + p_b(b^{m-1} n) + V_b(1, m+1)\\
\end{align*}

So it begins:
\begin{align*}
V_b(1, m+1)
 & = \sum_{k = 1}^{b^{m-1}-1}p_b(b^{m-1}n-k)\\
 & = \sum_{k = 1}^{b^{m-1}-1}p_b(b^{m-1}n-\ceil{\frac{k}{b}}b) \text{ (pop)}\\
 & = (b-1)p_b(b^{m-1}(n-1)) + \sum_{k = 1}^{b^{m-1}-b}p_b(b^{m-1}n-\ceil{\frac{k}{b}}b)\\
 & = (b-1)P(m-1) + \sum_{k = 1}^{b(b^{m-2}-1)}p_b(b^{m-1}n-\ceil{\frac{k}{b}}b)\\
 & = (b-1)P(m-1) + bU_b(1, m)\\
\end{align*}

It makes sense to perform this argument in more generality:
\begin{lem}
\label{lem:sumrecV}
Let $1 \leq k_0 = qb + r < b^{m-1}-b$.
Then
\[
V_b(k_0, m+1) = (b-1)p_b(b^{m-1}(n-1)) + (b-r+1)p_b(b^{m-1}n - (q+1)b) + b U_b(q+2, m).
\]
\end{lem}
\begin{proof}
\begin{align*}
V_b(k_0, m+1)
 & = \sum_{k=k_0}^{b^{m-1}-1} p_b(b^{m-1}n - k)\\
 & = \sum_{k=k_0}^{b^{m-1}-1} p_b(b^{m-1}n - \ceil{\frac{k}{b}}b)\\
 & = \sum_{k=k_0}^{(q+1)b} p_b(b^{m-1}n - (q+1)b) + \sum_{k=(q+1)b + 1}^{b^{m-1}-1} p_b(b^{m-1}n - \ceil{\frac{k}{b}}b)\\
 & = (qb+b-(qb+r)+1)p_b(b^{m-1}n - (q+1)b) + \sum_{k=(q+1)b + 1}^{b^{m-1}-1} p_b(b^{m-1}n - \ceil{\frac{k}{b}}b)\\
 & = (b-r+1)p_b(b^{m-1}n - (q+1)b) + \sum_{k=(q+1)b + 1}^{b^{m-1}-1} p_b(b^{m-1}n - \ceil{\frac{k}{b}}b)\\
\end{align*}
Note
\begin{align*}
(b^{m-1}-1) - (b^{m-1}-b+1) + 1
 %& = (b^{m-1}-1) - (b^{m-1}-b+1) + 1\\
 & = b^{m-1} - 1 - b^{m-1} + b - 1 + 1\\
 & = b - 1\\
\end{align*}
Therefore
\begin{align*}
\sum_{k=(q+1)b + 1}^{b^{m-1}-1} p_b(b^{m-1}n - \ceil{\frac{k}{b}}b)
&= \sum_{k=b^{m-1}-b+1}^{b^{m-1}-1} p_b(b^{m-1}n - \ceil{\frac{k}{b}}b) + \sum_{k=(q+1)b + 1}^{b^{m-1}-b} p_b(b^{m-1}n - \ceil{\frac{k}{b}}b)\\
&= \sum_{k=b^{m-1}-b+1}^{b^{m-1}-1} p_b(b^{m-1}n - b^{m-1}) + \sum_{k=(q+1)b + 1}^{b^{m-1}-b} p_b(b^{m-1}n - \ceil{\frac{k}{b}}b)\\
&= \sum_{k=b^{m-1}-b+1}^{b^{m-1}-1} p_b(b^{m-1}(n - 1)) + \sum_{k=(q+1)b + 1}^{b^{m-1}-b} p_b(b^{m-1}n - \ceil{\frac{k}{b}}b)\\
&= (b-1) p_b(b^{m-1}(n-1)) + \sum_{k=(q+1)b + 1}^{b^{m-1}-b} p_b(b^{m-1}n - \ceil{\frac{k}{b}}b)\\
\end{align*}
Note
\begin{align*}
b^{m-1} - b - ((q+1)b + 1) + 1
 & = b(b^{m-2} - 1) - (q+1)b - 1 + 1\\
 & = b(b^{m-2} - 1 - (q+1))\\
 & = b(b^{m-2} - 2 - (q+1) + 1)\\
\end{align*}
Then
\begin{align*}
\sum_{k=(q+1)b + 1}^{b^{m-1}-b} p_b(b^{m-1}n - \ceil{\frac{k}{b}}b)
 & = \sum_{k=q+1}^{b^{m-2}-2} \sum_{j=1}^b p_b(b^{m-1}n - \ceil{\frac{kb + j}{b}}b)\\
 & = \sum_{k=q+1}^{b^{m-2}-2} \sum_{j=1}^b p_b(b^{m-1}n - \ceil{\frac{(k+1)b}{b}}b)\\
 & = \sum_{k=q+1}^{b^{m-2}-2} \sum_{j=1}^b p_b(b^{m-1}n - (k+1)b)\\
 & = b \sum_{k=q+1}^{b^{m-2}-2} p_b(b^{m-1}n - (k+1)b)\\
 & = b \sum_{k=q+2}^{b^{m-2}-1} p_b(b^{m-1}n - kb)\\
 & = b U_b(q+2, m)
\end{align*}
Consequently
\[
V_b(k_0, m+1) = (b-1)p_b(b^{m-1}(n-1)) + (b-r+1)p_b(b^{m-1}n - (q+1)b) + b U_b(q+2, m)
\]
\end{proof}

\begin{cor}
\label{cor:sumrecV}
\[
V_b(1, m+1) = (b-1)p_b(b^{m-1}(n-1)) + b U_b(1, m)
\]
\end{cor}

\begin{proof}
Setting $k_0 = 1$ implies $q = 0$ and $r = 1$.
Therefore
\begin{align*}
V_b(1, m+1)
 & = (b-1)p_b(b^{m-1}(n-1)) + bp_b(b^{m-1}n - b) + b U_b(2, m)\\
 & = (b-1)p_b(b^{m-1}(n-1)) + bp_b(b^{m-1}n - b) + b \sum_{k=2}^{b^{m-2}-1} p_b(b^{m-1}n - kb)\\
 & = (b-1)p_b(b^{m-1}(n-1)) + b \sum_{k=1}^{b^{m-2}-1} p_b(b^{m-1}n - kb)\\
 & = (b-1)p_b(b^{m-1}(n-1)) + b U_b(1, m)\\
\end{align*}
\end{proof}

The expression of interest is now:
\begin{align*}
p_b(b^m n)
 & = P(m) + p_b(b^{m-1} n) + V_b(1, m+1)\\
 & = P(m) + p_b(b^{m-1} n) + (b-1)P(m-1) + bU_b(1, m)\\
\end{align*}

Here is a lemma to deal with $U$:
\begin{lem}[Pop, reindex, descend]
If $k_0 \leq b^{m-2}-1$ Then
\label{lem:sumrecU}
\[
U_b(k_0, m) = P(m-1) + V_b(k_0, m) + U_b(k_0+1, m)
\]
\end{lem}

\begin{proof}
Applying the recurrence to $U$:
\begin{align*}
U_b(k_0, m)
 & = \sum_{k=k_0}^{b^{m-2}-1} p_b(b^{m-1}n - kb)\\
 & = \sum_{k=k_0}^{b^{m-2}-1} (p_b(b^{m-2}n - k) + p_b(b^{m-1}n - (k+1)b))\text{ (descend)}\\
 & = \sum_{k=k_0}^{b^{m-2}-1} p_b(b^{m-1}n - (k+1)b) + \sum_{k=k_0}^{b^{m-2}-1} p_b(b^{m-2}n - k)\\
 & = \sum_{k=k_0}^{b^{m-2}-1} p_b(b^{m-1}n - (k+1)b) + V_b(k_0, m)\\
 & = \sum_{k=k_0}^{b^{m-2}-2} p_b(b^{m-1}n - (k+1)b) + p_b(b^{m-1}(n-1)) + V_b(k_0, m) \text{ (pop)}\\
 & = \sum_{k=k_0+1}^{b^{m-2}-1} p_b(b^{m-1}n - kb) + p_b(b^{m-1}(n-1)) + V_b(k_0, m) \text{ (reindex)}\\
 & = U_b(k_0+1, m) + p_b(b^{m-1}(n-1)) + V_b(k_0, m)\\
\end{align*}

Consequently,
\[
U_b(k_0, m) = U_b(k_0+1, m) + p_b(b^{m-1}(n-1)) + V_b(k_0, m)
\]

\end{proof}

Therefore:
\begin{lem}
\label{lem:u1}
\[
U_b(1, m) = (b^{m-2}-1)P(m-1) + \sum_{k=1}^{b^{m-2}-1} V_b(k, m)\\
\]
\end{lem}
\begin{proof}
\begin{align*}
U_b(1, m)
 & = \sum_{k=1}^{b^{m-2}-1}p_b(b^{m-1}n-kb)\\
 & = P(m-1) + V_b(1, m) + U(2, m)\\
 & = 2P(m-1) + V_b(1, m) + V_b(2, m) + U(3, m)\\
 & = \cdots\\
 & = (b^{m-2}-2)P(m-1) + \left(\sum_{k=1}^{b^{m-2}-2} V_b(k, m)\right) + U(b^{m-2}-1, m)\\
%& = (b^{m-2}-2)P(m-1) + \left(\sum_{k=1}^{b^{m-2}-2} V_b(k, m)\right) + \sum_{k=b^{m-2}-1}^{b^{m-2}-1} p_b(b^{m-1}n - kb)\\
%& = (b^{m-2}-2)P(m-1) + \left(\sum_{k=1}^{b^{m-2}-2} V_b(k, m)\right) + p_b(b^{m-1}n - (b^{m-2}-1)b)\\
%& = (b^{m-2}-2)P(m-1) + \left(\sum_{k=1}^{b^{m-2}-2} V_b(k, m)\right) + p_b(b^{m-1}n - b^{m-1}) + p_b(b^{m-2}n-(b^{m-2}-1))\\
%& = (b^{m-2}-1)P(m-1) + \left(\sum_{k=1}^{b^{m-2}-2} V_b(k, m)\right) + p_b(b^{m-2}n-(b^{m-2}-1))\\
%& = (b^{m-2}-1)P(m-1) + \left(\sum_{k=1}^{b^{m-2}-2} V_b(k, m)\right) + V_b(b^{m-2}-1, m)\\
 & = (b^{m-2}-1)P(m-1) + \sum_{k=1}^{b^{m-2}-1} V_b(k, m)\\
\end{align*}
\end{proof}

By combining Lemmas \ref{cor:sumrecV} and \ref{lem:u1}
\begin{lem}
\label{lem:V1rec}
\[
 V_b(1, m+1) = (b^{m-1}-1)p_b(b^{m-1}(n-1)) + b \sum_{k=1}^{b^{m-2}-1} V_b(k, m)\\
\]
\end{lem}
\begin{proof}

\begin{align*}
V_b(1, m+1)
 & = (b-1)p_b(b^{m-1}(n-1)) + b U_b(1, m)\\
 & = (b-1)p_b(b^{m-1}(n-1)) + b \left((b^{m-2}-1)P(m-1) + \sum_{k=1}^{b^{m-2}-1} V_b(k, m)\right)\\
 & = (b^{m-1}-1)p_b(b^{m-1}(n-1)) + b \sum_{k=1}^{b^{m-2}-1} V_b(k, m)\\
\end{align*}
\end{proof}

This nearly provides a recursion for $V$,
but currently the expression for $V_b(k, m)$ for $k>1$
still contains terms of $U$.
These can be eliminated by applying Lemma \ref{lem:sumrecU} to the $U$ term in Lemma \ref{lem:sumrecV}.

\begin{lem}
\label{lem:Uelimination}
Let $0 \leq q < b^{m-2}-1$.
\[
U_b(q+2, m)
 = \sum_{k=q+2}^{b^{m-2}-1} V_b(k, m) + (b^{m-2}-q-2)p_b(b^{m-1}(n-1))
\]
\end{lem}

\begin{proof}
\begin{align*}
U_b(q+2, m)
 & = \sum_{k = q+2}^{b^{m-2}-1}p_b(b^{m-1}n-kb)\\
 & = U_b(q+3, m) + p_b(b^{m-1}(n-1)) + V_b(q+2, m)\\
 & = U_b(q+4, m) + 2p_b(b^{m-1}(n-1)) + V_b(q+2, m) + V_b(q+3, m)\\
 & = U_b(q+4, m) + 2p_b(b^{m-1}(n-1)) + \sum_{k=q+2}^{q+3}V_b(q+2+k, m)\\
 & = \ldots \text{ then after $b^{m-2}-(q+2)$ steps:}\\
 & = (b^{m-2}-q-2)p_b(b^{m-1}(n-1)) + \sum_{k=q+2}^{b^{m-2}-1} V_b(k, m)\\
\end{align*}
\end{proof}

INCEPTION
\begin{align*}
V_b(1, m+1)
 & = (b^{m-1}-1)p_b(b^{m-1}(n-1)) + b \sum_{k=1}^{b^{m-2}-1} V_b(k, m)\\
 & = (b^{m-1}-1)p_b(b^{m-1}(n-1))\\
 & \tab + b \left((b^{m-2}-1)p_b(b^{m-2}(n-1)) + b \sum_{k_1=1}^{b^{m-3}-1} V_b(k_1, m-1)\right) + \sum_{k=2}^{b^{m-2}-1}V_b(k, m)\\
%& \tab + b \sum_{k=1}^{b^{m-2}-1}\left((b^{m-2}-1)p_b(b^{m-2}(n-1)) + b \sum_{k_1=1}^{b^{m-3}-1} V_b(k_1, m-1)\right)\\
%& = (b^{m-1}-1)p_b(b^{m-1}(n-1)) + b(b^{m-2}-1)^2p_b(b^{m-2}(n-1))\\
%& \tab + b^2 \sum_{k=1}^{b^{m-2}-1} \sum_{k_1=1}^{b^{m-3}-1} V_b(k_1, m-1)\\
%& = (b^{m-1}-1)p_b(b^{m-1}(n-1)) + b(b^{m-2}-1)^2p_b(b^{m-2}(n-1))\\
%& \tab + b^2 \sum_{k=1}^{b^{m-2}-1} \sum_{k_1=1}^{b^{m-3}-1}\left((b^{m-3}-1)p_b(b^{m-3}(n-1)) + b \sum_{k_2=1}^{b^{m-4}-1} V_b(k_2, m-2)\right)\\
%& = (b^{m-1}-1)p_b(b^{m-1}(n-1)) + b(b^{m-2}-1)^2p_b(b^{m-2}(n-1))\\
%& \tab + b^2 \sum_{k=1}^{b^{m-2}-1} \sum_{k_1=1}^{b^{m-3}-1}\left((b^{m-3}-1)p_b(b^{m-3}(n-1)) + b \sum_{k_2=1}^{b^{m-4}-1} V_b(k_2, m-2)\right)\\
%& = (b^{m-1}-1)p_b(b^{m-1}(n-1)) + b(b^{m-2}-1)^2p_b(b^{m-2}(n-1))\\
%& \tab + b^2(b^{m-2}-1)(b^{m-3}-1)^2p_b(b^{m-3}(n-1)) + b^3 \sum_{k=1}^{b^{m-2}-1} \sum_{k_1=1}^{b^{m-3}-1}\sum_{k_2=1}^{b^{m-4}-1} V_b(k_2, m-2)\\
\end{align*}

The expression of interest is now:
\begin{align*}
p_b(b^m n)
 & = P(m) + p_b(b^{m-1} n) + V_b(1, m+1)\\
 & = P(m) + p_b(b^{m-1} n) + (b-1)P(m-1) + bU_b(1, m)\\
 & = P(m) + p_b(b^{m-1} n) + (b-1)P(m-1) + b\left((b^{m-2}-1)P(m-1) + \sum_{k=1}^{b^{m-2}-1} V_b(k, m)\right)\\
 & = P(m) + p_b(b^{m-1} n) + (b^{m-1}-b+b-1)P(m-1) + b\sum_{k=1}^{b^{m-2}-1} V_b(k, m)\\
 & = P(m) + p_b(b^{m-1} n) + (b^{m-1}-1)P(m-1) + b\sum_{k=1}^{b^{m-2}-1} V_b(k, m)\\
\end{align*}

Evaluating this sum:
\begin{align*}
\sum_{k=1}^{b^{m-2}-1} V_b(k, m)
 & = \sum_{j=1}^{b^{m-2}-1} \sum_{k=j}^{b^{m-2}-1} p_b(b^{m-2}n - k)\\
 & = \sum_{j=1}^{b^{m-2}-1} \sum_{k=1}^{b^{m-2}-1} \ibra{k\geq j} p_b(b^{m-2}n - k)\\
 & = \sum_{k=1}^{b^{m-2}-1} p_b(b^{m-2}n - k) \sum_{j=1}^{b^{m-2}-1} \ibra{k\geq j}\\
 & = \sum_{k=1}^{b^{m-2}-1} p_b(b^{m-2}n - k) \sum_{j=1}^{k} 1\\
 & = \sum_{k=1}^{b^{m-2}-1} k p_b(b^{m-2}n - k)\\
\end{align*}

The expression of interest is now:
\begin{align*}
p_b(b^m n)
 & = P(m) + p_b(b^{m-1} n) + (b^{m-1}-1)P(m-1) + b\sum_{k=1}^{b^{m-2}-1} V_b(k, m)\\
 & = P(m) + p_b(b^{m-1} n) + (b^{m-1}-1)P(m-1) + b\sum_{k=1}^{b^{m-2}-1} k p_b(b^{m-2}n - k)\\
\end{align*}

Let
\begin{align*}
      P(m) & = p_b(b^m(n-1))\\
 X_b(m, n) & = \sum_{k = 1}^{b^{m-2}-1} k p_b(b^{m-2}n - k)\\
%\Trp{a}{b} & = {a \choose 2} - {b \choose 2}\\
\end{align*}

Then $U_b$ takes the following shape:

\begin{align*}
U_b(1, m)
 & = \sum_{k=1}^{b^{m-2}-1} p_b(b^{m-1}n - kb)\\
 & = (b^{m-2}-1)P(m-1) + \sum_{k=1}^{b^{m-2}-1} V_b(k, m)\\
 %& = (b^{m-2}-1)P(m-1) + \sum_{k=1}^{b^{m-2}-1} k p_b(b^{m-2}n - k)\\
 & = (b^{m-2}-1)P(m-1) + X_b(m, n)\\
\end{align*}

The expression of interest is now:
\begin{align*}
p_b(b^m n)
 & = P(m) + p_b(b^{m-1} n) + (b^{m-1}-1)P(m-1) + b\sum_{k=1}^{b^{m-2}-1} k p_b(b^{m-2}n - k)\\
 & = P(m) + p_b(b^{m-1} n) + (b^{m-1}-1)P(m-1) + bX_b(m, n)\\
\end{align*}

Examine the highlights of how the expression of interest has evolved:

\begin{align*}
p_b(b^m n)
 & = p_b(b^m(n - 1)) + p_b(b^{m-1} n) + \sum_{k = 1}^{b^m-1}p_b(b^{m-1}n - k)\\
 & = P(m) + p_b(b^{m-1} n) + V_b(1, m+1)\\
 & = P(m) + p_b(b^{m-1} n) + (b-1)P(m-1) + bU_b(1, m)\\
 & = P(m) + p_b(b^{m-1} n) + (b-1)P(m-1) + b\left((b^{m-2}-1)P(m-1) + \sum_{k=1}^{b^{m-2}-1} V_b(k, m)\right)\\
 & = P(m) + p_b(b^{m-1} n) + (b^{m-1}-1)P(m-1) + b\sum_{k=1}^{b^{m-2}-1} V_b(k, m)\\
 & = P(m) + p_b(b^{m-1} n) + (b^{m-1}-1)P(m-1) + b\sum_{k=1}^{b^{m-2}-1} k p_b(b^{m-2}n - k)\\
 & = P(m) + p_b(b^{m-1} n) + (b^{m-1}-1)P(m-1) + bX_b(m, n)\\
 %& = P(m) + p_b(b^{m-1} n) + (b^{m-1}-1)P(m-1) + b\left(b^2{b+1 \choose 2}S_b(0, m-1) - (b^{m-2})P(m-2)\right)\\
 %& = P(m) + p_b(b^{m-1} n) + (b^{m-1}-1)P(m-1) + b^3{b+1 \choose 2}S_b(0, m-1) - (b^{m-1})P(m-2)\\
 %& = P(m) + p_b(b^{m-1} n) + (b^{m-1}-1)P(m-1) - b^{m-1}P(m-2) + b^3{b+1 \choose 2}S_b(0, m-1)\\
 %& = P(m) + p_b(b^{m-1} n) + (b^{m-1}-1)P(m-1) - b^{m-1}P(m-2)\\
 %& \tab + b^3{b+1 \choose 2}(P(m-2) + U_b(1, m-1))\\
 %& = P(m) + p_b(b^{m-1} n) + (b^{m-1}-1)P(m-1) - b^{m-1}P(m-2) + b^3{b+1 \choose 2}P(m-2)\\
 %& \tab + b^3 {b+1 \choose 2} U_b(1, m-1)\\
\end{align*}

Let's see if doing stuff leads to a general expression.
\begin{align*}
X_b(m, n) {+} (b^{m{-}2})P(m{-}2)
 & = \sum_{k=1}^{b^{m-2}} k p_b(b^{m-2}n - k)\\
 & = \sum_{k=1}^{b^{m-2}} k p_b(b^{m-2}n - \ceil{\frac{k}{b}}b) \text{ case $m=3$ done here}\\
 & = \sum_{k=1}^{b(b^{m-3})} k p_b(b^{m-2}n - \ceil{\frac{k}{b}}b)\\
 & = \sum_{k=0}^{b^{m-3}-1}\sum_{j=1}^{b} (kb + j) p_b(b^{m-2}n - \ceil{\frac{kb+j}{b}}b)\\
 & = \sum_{k=0}^{b^{m-3}-1}\sum_{j=1}^{b} (kb + j) p_b(b^{m-2}n - (k+1)b)\\
 & = \sum_{k=0}^{b^{m-3}-1}p_b(b^{m-2}n - (k+1)b)\sum_{j=1}^{b} (kb + j) \\
 & = \sum_{k=0}^{b^{m-3}-1}p_b(b^{m-2}n - (k+1)b)\left(kb^2 + \sum_{j=1}^{b}j\right)\\
 & = \sum_{k=0}^{b^{m-3}-1}p_b(b^{m-2}n - (k+1)b)\left(kb^2 + {b+1 \choose 2}\right)\\
 & = {b+1 \choose 2}\sum_{k=0}^{b^{m-3}-1}p_b(b^{m-2}n - (k+1)b) + b^2\sum_{k=0}^{b^{m-3}-1}kp_b(b^{m-2}n - (k+1)b)\\
 & = {b+1 \choose 2}\sum_{k=1}^{b^{m-3}}p_b(b^{m-2}n - kb) + b^2\sum_{k=0}^{b^{m-3}-1}kp_b(b^{m-2}n - (k+1)b)\\
 & = {b+1 \choose 2}(p_b(b^{m-2}(n-1)){+}U_b(1, m{-}1)) + b^2\sum_{k=0}^{b^{m-3}-1}kp_b(b^{m-2}n - (k+1)b)\\
\end{align*}

This last term seems promising:
\begin{align*}
W
 & = \sum_{k=0}^{b^{m-3}-1}kp_b(b^{m-2}n - (k+1)b)\\
 & = \sum_{k=1}^{b^{m-3}}(k-1)p_b(b^{m-2}n - kb)\\
 & = \sum_{k=1}^{b^{m-3}}kp_b(b^{m-2}n - kb) - p_b(b^{m-2}n - kb)\\
 & = \sum_{k=1}^{b^{m-3}}kp_b(b^{m-2}n - kb) - \sum_{k=1}^{b^{m-3}}p_b(b^{m-2}n - kb)\\
 & = \sum_{k=1}^{b^{m-3}}kp_b(b^{m-2}n - kb) - \left(p_b(b^{m-2}(n-1)){+}U_b(1, m{-}1)\right)\\
 & = \sum_{k=1}^{b^{m-3}-1}kp_b(b^{m-2}n - kb) + b^{m-3}p_b(b^{m-2}(n-1)) - p_b(b^{m-2}(n-1)){-}U_b(1, m{-}1)\\
 & = \sum_{k=1}^{b^{m-3}-1}kp_b(b^{m-2}n - kb) + \left(b^{m-3}{-}1\right)p_b(b^{m-2}(n-1)) - U_b(1, m{-}1)\\
\end{align*}

Then note:
\begin{align*}
A
 & = \sum_{k=1}^{b^{m-3}-1}kp_b(b^{m-2}n - kb)\\
 & = \sum_{k=1}^{b^{m-3}-1}k(p_b(b^{m-2}n - (k+1)b) + p_b(b^{m-3}-k))\\
 & = \sum_{k=1}^{b^{m-3}-1}kp_b(b^{m-2}n - (k+1)b) + \sum_{k=1}^{b^{m-3}-1}kp_b(b^{m-3}-k)\\
 & = \sum_{k=1}^{b^{m-3}-1}kp_b(b^{m-2}n - (k+1)b) + X_b(m-1, n)\\
 & = \sum_{k=2}^{b^{m-3}}(k-1)p_b(b^{m-2}n - kb) + X_b(m-1, n)\\
 & = \sum_{k=2}^{b^{m-3}-1}(k-1)p_b(b^{m-2}n - kb) + (b^{m-3}-1)p_b(b^{m-2}(n-1)) + X_b(m-1, n)\\
\end{align*}

Then $U_b$ takes the following shape:

\begin{align*}
U_b(1, m)
 & = \sum_{k=1}^{b^{m-2}-1} p_b(b^{m-1}n - kb)\\
 & = (b^{m-2}-1)P(m-1) + \sum_{k=1}^{b^{m-2}-1} V_b(k, m)\\
 %& = (b^{m-2}-1)P(m-1) + \sum_{k=1}^{b^{m-2}-1} k p_b(b^{m-2}n - k)\\
 & = (b^{m-2}-1)P(m-1) + X_b(m, n)\\
 & = (b^{m-2}-1)P(m-1) - p_b(b^{m-2}(n-1)) + {b+1 \choose 2}(p_b(b^{m-2}(n-1)){+}U_b(1, m{-}1))\\
 & \tab + b^2\sum_{k=0}^{b^{m-3}-1}kp_b(b^{m-2}n - (k+1)b)\\
 & = (b^{m-2}-1)P(m-1) - p_b(b^{m-2}(n-1)) + {b+1 \choose 2}(p_b(b^{m-2}(n-1)){+}U_b(1, m{-}1)) + b^2W\\
 & = (b^{m-2}{-}1)P(m{-}1) + \left({b+1 \choose 2}-1\right)(p_b(b^{m-2}(n-1)){+}U_b(1, m{-}1)) + b^2W\\
 & = (b^{m-2}{-}1)P(m{-}1) + \left({b+1 \choose 2}-1\right)(p_b(b^{m-2}(n-1)){+}U_b(1, m{-}1))\\
 & \tab b^2\left(A + (b^{m-3}-1)p_b(b^{m-2}(n-1)) - U_b(1, m-1)\right)\\
 & = (b^{m-2}{-}1)P(m{-}1) + \left({b+1 \choose 2}-1\right)(p_b(b^{m-2}(n-1)){+}U_b(1, m{-}1))\\
 & \tab (b^{m-1}-1)p_b(b^{m-2}(n-1)) - b^2U_b(1, m-1) + b^2A\\
\end{align*}

\begin{comment}
The expression of interest is now:
\begin{align*}
p_b(b^m n)
 & = P(m) + p_b(b^{m-1} n) + (b^{m-1}-1)P(m-1) + bX_b(m, n)\\
 & = P(m) + p_b(b^{m-1} n) + (b^{m-1}-1)P(m-1) + b\left(b^2{b+1 \choose 2}S_b(0, m-1) - (b^{m-2})P(m-2)\right)\\
 & = P(m) + p_b(b^{m-1} n) + (b^{m-1}-1)P(m-1) + b^3{b+1 \choose 2}S_b(0, m-1) - (b^{m-1})P(m-2)\\
 & = P(m) + p_b(b^{m-1} n) + (b^{m-1}-1)P(m-1) - b^{m-1}P(m-2) + b^3{b+1 \choose 2}S_b(0, m-1)\\
 %& = P(m) + p_b(b^{m-1} n) + (b^{m-1}-1)P(m-1) - b^{m-1}P(m-2) + b^3{b+1 \choose 2}(P(m-2) + U_b(1, m-1))
 %& = P(m) + p_b(b^{m-1} n) + (b^{m-1}-1)P(m-1) - b^{m-1}P(m-2) + b^3{b+1 \choose 2}P(m-2) + b^3 {b+1 \choose 2} U_b(1, m-1)
 & = P(m) + p_b(b^{m-1} n) + (b^{m-1}-1)P(m-1) - b^{m-1}P(m-2)\\
 & \tab + b^3{b+1 \choose 2}(P(m-2) + U_b(1, m-1))\\
 & = P(m) + p_b(b^{m-1} n) + (b^{m-1}-1)P(m-1) - b^{m-1}P(m-2) + b^3{b+1 \choose 2}P(m-2)\\
 & \tab + b^3 {b+1 \choose 2} U_b(1, m-1)\\
\end{align*}

In particular
\begin{align*}
p_b(b^m n)
 & = P(m) + p_b(b^{m-1} n) + (b-1)P(m-1) + bU_b(1, m)\\
 & = P(m) + p_b(b^{m-1} n) + (b^{m-1}-1)P(m-1) - b^{m-1}P(m-2) + b^3{b+1 \choose 2}P(m-2)\\
 & \tab + b^3 {b+1 \choose 2} U_b(1, m-1)\\
% & = P(m) + p_b(b^{m-1} n) + (b^{m-1}-1)P(m-1) - b^{m-1}P(m-2) + b^3{b+1 \choose 2}P(m-2)\\
% & \tab + b^3 {b+1 \choose 2} U_b(1, m-1)\\
\end{align*}

Therefore
\[
U_b(1, m)
   = (b^{m-2} - 1)p_b(b^{m-1}(n-1))
     + \left(b^2 {b+1 \choose 2} - b^{m-2}\right)p_b(b^{m-2}(n-1))
     + b^2 {b+1 \choose 2} U_b(1, m-1)
\]


Thus for $m=4$
\[
p_b(b^4n) = p_b(b^4(n-1)
+ p_b(b^3n)
+ (b-1)p_b(b^3(n-1))
+ b((b^2 - 1)p_b(b^3(n-1)) + \left(b^2 {b+1 \choose 2} - b^2\right)p_b(b^2(n-1))
+ b^2 {b+1 \choose 2} (
   (b - 1)p_b(b^2(n-1)) + \left(b^2 {b+1 \choose 2} - b\right)p_b(b(n-1)) + b^2 {b+1 \choose 2} p_b(b(n-1))
))
\]
And passing to generating functions:
\begin{align*}
(1-q)B(4)
 & = B(3) + (b-1)qB(3) \\
 & \tab + b(\\
 & \tab \tab (b^2 - 1)qB(3)\\
 & \tab \tab + \left(b^2 {b+1 \choose 2} - b^2\right)qB(2)
    + b^2 {b+1 \choose 2} (
      (b - 1)qB(2)\\
 & \tab \tab
      + \left(b^2 {b+1 \choose 2} - b\right)qB(1)
      + b^2 {b+1 \choose 2} qB(1)
    )\\
 & \tab )\\
%
 & = B(3) + (b-1)qB(3) + (b^3-b)qB(3)\\
 & \tab + b(\\
 & \tab \tab + \left(b^2 {b+1 \choose 2} - b^2\right)qB(2)
    + b^2 {b+1 \choose 2} (
      (b - 1)qB(2)\\
 & \tab \tab
      + \left(b^2 {b+1 \choose 2} - b\right)qB(1)
      + b^2 {b+1 \choose 2} qB(1)
    )\\
 & \tab )\\
%
 & = (1 + (b^3 - 1)q)B(3)\\ 
 & \tab + (\\
 & \tab \tab + b\left(b^2 {b+1 \choose 2} - b^2\right)qB(2)
      + b^3 {b+1 \choose 2} (b - 1)qB(2)\\
 & \tab \tab
      + b^3 {b+1 \choose 2} \left(b^2 {b+1 \choose 2} - b\right)qB(1)
      + b^3 {b+1 \choose 2} b^2 {b+1 \choose 2} qB(1)
    \\
 & \tab )\\
%
 & = (1 + (b^3 - 1)q)B(3)\\ 
 & \tab + (\\
 & \tab \tab + b^3\left({b+1 \choose 2} - 1\right)qB(2)
      + b^3 {b+1 \choose 2} (b - 1)qB(2)\\
 & \tab \tab
      + b^3 {b+1 \choose 2} \left(b^2 {b+1 \choose 2} - b\right)qB(1)
      + b^3 {b+1 \choose 2} b^2 {b+1 \choose 2} qB(1)
    \\
 & \tab )\\
%
 & = (1 + (b^3 - 1)q)B(3)\\ 
 & \tab + (\\
 & \tab \tab + b^3\left(b{b+1 \choose 2} - 1\right)qB(2)\\
 & \tab \tab
      + b^3 {b+1 \choose 2} \left(b^2 {b+1 \choose 2} - b\right)qB(1)
      + b^3 {b+1 \choose 2} b^2 {b+1 \choose 2} qB(1)
    \\
 & \tab )\\
\end{align*}
Continued...
\begin{align*}
(1-q)B(4)
 & = (1 + (b^3 - 1)q)B(3)\\ 
 & \tab + b^3\left(b{b+1 \choose 2} - 1\right)qB(2)\\
 & \tab + b^3 {b+1 \choose 2} \left(2b^2 {b+1 \choose 2} - b\right)qB(1)\\
\end{align*}
So
\begin{align*}
(1-q)^4B(4)
 & = (1 + (b^3 - 1)q)(1 + \half(b-1)\left((b^2+2b+4)q + (b^2-2)q^2\right)B(0)\\
 & \tab + b^3\left(b{b+1 \choose 2} - 1\right)q(1-(b-1)q)B(0)\\
 & \tab + b^3 {b+1 \choose 2} \left(2b^2 {b+1 \choose 2} - b\right)q(1-q)B(0)\\
%
 & = [(1 + (b^3 - 1)q)(1 + \half(b-1)\left((b^2+2b+4)q + (b^2-2)q^2\right)\\
 & \tab + b^3\left(b{b+1 \choose 2} - 1\right)q(1-(b-1)q)\\
 & \tab + b^3 {b+1 \choose 2} \left(2b^2 {b+1 \choose 2} - b\right)q(1-q)]B(0)\\
\end{align*}
"""(b^{two} - 1)p_b(b^{one}(n-1))
+ \left(b^2 {b+1 \choose 2} - b^{two}\right)p_b(b^{two}(n-1))
+ b^2 {b+1 \choose 2} U_b(1, one)
""".format(one=(m-1), two=(m-2))

 & = \sum_{k=0}^{b^{m-3}-1}p_b(b^{m-2}n - (k+1)b)\Trp{(k+1)b}{kb+1}\text{ (pop)}\\
 & = \Trp{b^{m-2}}{b^{m-2}-b+1}P(m-2)\\
 & \tab + \sum_{k=0}^{b^{m-3}-1}\Trp{(k+1)b}{kb+1}p_b(b^{m-2}n - (k+1)b)\text{ (reindex)}\\
 & = \Trp{b^{m-2}}{b^{m-2}-b+1}P(m-2)\\
 & \tab + \sum_{k=1}^{b^{m-3}}\Trp{kb}{kb-b+1}p_b(b^{m-2}n - kb) \text{ (descend)}\\
 & = \Trp{b^{m-2}}{b^{m-2}-b+1}P(m-2)\\
 & \tab + \sum_{k=1}^{b^{m-3}}\Trp{kb}{kb-b+1}(p_b(b^{m-2}n - (k+1)b) + p_b(b^{m-3}n-k))\\
 & = \Trp{b^{m-2}}{b^{m-2}-b+1}P(m-2)\\
 & \tab + \sum_{k=1}^{b^{m-3}}\Trp{kb}{kb-b+1}p_b(b^{m-2}n - (k+1)b)\\
 & \tab + \sum_{k=1}^{b^{m-3}}\Trp{kb}{kb-b+1}p_b(b^{m-3}n-k))\\

Let's try another attack.
\begin{align*}
F(p_b(b^m-kb)) &= p_b(b^m n-(k+1)b) + p_b(b^{m-1}n-k)\\
F^{b^{m-1}-k}(p_b(b^m-kb)) &= p_b(b^m(n-1)) + \sum_{j=0}^{b^{m-1}-k-1}p_b(b^{m-1}n-(j+k))\\
\end{align*}

\begin{align*}
V_b(1, m+1)
 & = \sum_{k = 1}^{b^{m-1}-1}p_b(b^{m-1}n-k)\\
 & = \sum_{k = 1}^{b^{m-1}-1}p_b(b^{m-1}n-\ceil{\frac{k}{b}}b) \text{ (pop)}\\
 & = (b-1)p_b(b^{m-1}(n-1)) + \sum_{k = 1}^{b^{m-1}-b}p_b(b^{m-1}n-\ceil{\frac{k}{b}}b)\\
 & = (b-1)P(m-1) + \sum_{k=1}^{b(b^{m-2}-1)}p_b(b^{m-1}n-\ceil{\frac{k}{b}}b)\\
 & = (b-1)P(m-1) + \sum_{k=0}^{(b^{m-2}-1)}\sum_{j=1}^b p_b(b^{m-1}n-\ceil{\frac{kb+j}{b}}b)\\
 & = (b-1)P(m-1) + \sum_{k=0}^{(b^{m-2}-1)}\sum_{j=1}^b p_b(b^{m-1}n-\ceil{\frac{(k+1)b}{b}}b)\\
 & = (b-1)P(m-1) + b \sum_{k=0}^{(b^{m-2}-1)}p_b(b^{m-1}n-\ceil{\frac{(k+1)b}{b}}b)\\
 & = (b-1)P(m-1) + b \sum_{k=0}^{(b^{m-2}-1)}p_b(b^{m-1}n-(k+1)b)\\
 & = (b-1)P(m-1) + b S(0, m)\\
 & = (b-1)P(m-1) + b \sum_{k=1}^{b^{m-2}}p_b(b^{m-1}n-kb)\\
 & = (b-1)P(m-1) + b p_b(b^{m-1}(n-1))  + b \sum_{k=1}^{b^{m-2}-1}p_b(b^{m-1}n-kb)\\
 & = (2b-1)P(m-1) + b U_b(1, m)\\
\cdots\\
\end{align*}

Let $p(n, k) (= {n \choose k})?$ be the number of partitions of $n$ in to exactly $k$ parts.
\begin{align*}
%S_b(0, m)
%& = P(m-1) + U_b(1, m)\\
%& = \sum_{k=0}^{b^{m-2}-1}p_b(b^{m-1}n-(k+1)b)\\
%& = \sum_{k=0}^{b^{m-2}-1}F^{b^{m-1}-(k+1)}(p_b(b^m-(k+1)b))\\
%& = \sum_{k=0}^{b^{m-2}-1}p_b(b^m(n-1)) + \sum_{j=0}^{b^{m-1}-(k+1)-1}p_b(b^{m-1}n-(j+k+1))\\
%\cdots\\
U_b(1, m)
 & = \sum_{k=1}^{b^{m-2}-1} p_b(b^{m-1}n - kb)\\
 & = \sum_{k=1}^{b^{m-2}-1} F^{b^{m-2}-k}(p_b(b^{m-1}n - kb))\\
 & = \sum_{k=1}^{b^{m-2}-1} p_b(b^{m-1}(n-1)) + \sum_{j=0}^{b^{m-2}-k-1}p_b(b^{m-2}n-(j+k))\\
 & = (b^{m-2}-1)p_b(b^{m-1}(n-1)) + \sum_{k=1}^{(b^{m-2}-1)}\sum_{j=0}^{(b^{m-2}-k-1)}p_b(b^{m-2}n-(j+k))\\
 & = (b^{m-2}-1)P(m-1)) + \sum_{k=1}^{(b^{m-2}-1)}\sum_{j=0}^{(b^{m-2}-1)}\ibra{j < b^{m-2}-k}p_b(b^{m-2}n-(j+k))\\
 & = (b^{m-2}-1)P(m-1)) + \sum_{k=1}^{(b^{m-2}-1)}p_b(b^{m-2}n-k) + \sum_{k=1}^{(b^{m-2}-1)}\sum_{j=1}^{(b^{m-2}-1)}\ibra{j < b^{m-2}-k}p_b(b^{m-2}n-(j+k))\\
 & = (b^{m-2}-1)P(m-1)) + V_b(1, m) + \sum_{k=1}^{(b^{m-2}-1)}\sum_{j=1}^{(b^{m-2}-1)}\ibra{j<b^{m-2}-k}p_b(b^{m-2}n-(j+k))\\
 & = (b^{m-2}-1)P(m-1)) + V_b(1, m) + \sum_{k=1}^{(b^{m-2}-1)}\sum_{j=1}^{(b^{m-2}-1)}\ibra{j+k<b^{m-2}}p_b(b^{m-2}n-(j+k))\\
 & = (b^{m-2}-1)P(m-1)) + V_b(1, m) + \sum_{k=1}^{b^{m-2}-1}p(k,2)p_b(b^{m-2}n-k)\\
\end{align*}

Note that
\begin{align*}
V_b(1, m)
 & = \sum_{k = 1}^{b^{m-2}-1}p_b(b^{m-2}n-k)\\
 & = \sum_{k = 1}^{b^{m-2}-1}p_b(b^{m-1}n-kb) -\sum_{k=1}^{b^{m-2}-1} p_b(b^{m-1}n - (k+1)b)\\
 & = \sum_{k = 1}^{b^{m-2}-1}(p_b(b^{m-1}n-kb) - p_b(b^{m-1}n - (k+1)b))\\
 & = -\sum_{k = 1}^{b^{m-2}-1}\diff_k(p_b(b^{m-1}n-kb))
\end{align*}

Alternatively:
\begin{align*}
\diff_k(p_b(b^{m-1}n-kb))
& = p_b(b^{m-1}n-(k+1)b) - p_b(b^{m-1}n-kb)\\
& = p_b(b^{m-2}n-(k+1)) + p_b(b^{m-1}n-(k+2)b)\\
& \tab - (p_b(b^{m-2}n-k) + p_b(b^{m-1}n-(k+1)b))\\
& = p_b(b^{m-2}n-(k+1)) + p_b(b^{m-1}n-(k+2)b)\\
& \tab - (p_b(b^{m-2}n-k) + p_b(b^{m-2}n-(k+1)) + p_b(b^{m-1}n-(k+2)b))\\
& = - p_b(b^{m-2}n-k)\\
\end{align*}

Then let $u(k) = k, \diff_k u(k) = 1, \diff_k v(k) = p_b(b^{m-2}n - k), v(k) = - p_b(b^{m-2}n-k)$.
Then by summation by parts
\begin{align*}
%\sum u(k)\diff_k v(k) & = uv - \sum v(k+1)\diff_k u(k)\\
X_b(m, n)
 & = \sum_{k = 1}^{b^{m-2}-1} k p_b(b^{m-2}n - k)\\
 & = -k p_b(b^{m-1}n - kb) - \sum_{k=1}^{b^m-2}p_b(b^{m-2}n - (k+1)b)\\
 & = -(b^{m-2}p_b(b^{m-1}(n-1))) + p_b(b^{m-1}n - b) - p_b(b^{m-1}(n-1)) - \sum_{k=2}^{b^{m-2}-1}p_b(b^{m-1}n-kb)\\
\end{align*}

Then, once more, with feeling:
\begin{align*}
%T_b(m, n)
 & = (b^{m-1}-1)p_b(b^{m-1}(n-1)) + b\sum_{k = 1}^{b^{m-2}-1} k p_b(b^{m-2}n - k)\\
 & = (b^{m-1}-1)p_b(b^{m-1}(n-1))\\
 & \tab -b(b^{m-2}p_b(b^{m-1}(n-1))) + bp_b(b^{m-1}n - b) - bp_b(b^{m-1}(n-1)) - b\sum_{k=2}^{b^{m-2}-1}p_b(b^{m-1}n-kb))\\
 & = (b^{m-1}-1)p_b(b^{m-1}(n-1)) - b(b^{m-2}-1)p_b(b^{m-1}(n-1)) + bp_b(b^{m-1}n - b)\\
 & \tab - b\sum_{k=2}^{b^{m-2}-1}p_b(b^{m-1}n-kb))\\
 & = (b-1)p_b(b^{m-1}(n-1)) + bp_b(b^{m-1}n - b) + b\sum_{k=2}^{b^{m-2}-1}p_b(b^{m-1}n-kb))\\
 & = (b-1)p_b(b^{m-1}(n-1)) + b\sum_{k=1}^{b^{m-2}-1}p_b(b^{m-1}n-kb))\\
\end{align*}
On the bright side, the arithmetic is correct.
This result suggest the present summation by parts argument involves a circular dependency.

And the rest!
\[
%\diff_k S_b(k, m) = F^{k+1}(p_b(b^{m-1}n)) + T_b(m-1, n) - p_b(b^{m-1}(n-1)) - p_b(b^{m-1}n - (k+1)b) - p_b(b^{m-2}n)
\]
\begin{align*}
F^{k_0+1}(p_b(b^{m-1}n))
 & = p_b(b^{m-1}n - (k_0+1)b) + p_b(b^{m-2}n) + \sum_{k=1}^{k_0} p_b(b^{m-2}n - k)\\
\sum_{k=1}^{k_0} p_b(b^{m-2}n - k)
 & = p_b(b^{m-1}n - (k_0+1)b) + p_b(b^{m-2}n) - F^{k_0+1}(p_b(b^{m-1}n))\\
\end{align*}

\begin{align*}
%T_b(m-1, n)
& = \sum_{k=1}^{b^{m-2}-1} p_b(b^{m-2}n - k)\\
& = \sum_{k=1}^{k_0} p_b(b^{m-2}n - k) + \sum_{k=k_0 + 1}^{b^{m-2}-1} p_b(b^{m-2}n - k)\\
& = p_b(b^{m-1}n - (k_0+1)b) + p_b(b^{m-2}n) - F^{k_0+1}(p_b(b^{m-1}n)) + \sum_{k=k_0 + 1}^{b^{m-2}-1} p_b(b^{m-2}n - k)\\
\end{align*}

So
\begin{align*}
%T_b(m-1, n) & = p_b(b^{m-1}n - (k_0+1)b) + p_b(b^{m-2}n) - F^{k_0+1}(p_b(b^{m-1}n)) + \sum_{k=k_0 + 1}^{b^{m-2}-1} p_b(b^{m-2}n - k)\\
\sum_{k=k_0 + 1}^{b^{m-2}-1} p_b(b^{m-2}n - k)
%& = p_b(b^{m-1}n - (k_0+1)b) + p_b(b^{m-2}n) - F^{k_0+1}(p_b(b^{m-1}n)) - T_b(m-1, n)\\
\end{align*}

Note
\[
F(p_b(b^{m-2}n)) = p_b(b^{m-3}n) + p_b(b^{m-2}n-b)
\]
\end{comment}
\pagebreak
\begin{sideways}
\begin{minipage}{\textheight}
\[
\begin{array}{llllll}
m & b=2 & b=3 & b=4 & b=5\\
3
 & 1 + 6q + q^2
 & 1 + 19q + 7q^2
 & 1 + 42q + 21q^2
 & 1 + 78q + 46q^2\\
4
 & 1 + 31q + 31q^2 + q^3
 & 1 + 234q + 447q^2 + 47q^3
 & 1 + 1081q + 2635q^2 + 379q^3
 & 1 + 3072q + 10218q^2 + 1704q^3\\
5
 & 1 + 196q+ 630q^2 + 196q^2 + q^4
 & 1 + 5822q + 33504q^2 + 19040q^3 + 682q^4\\
\end{array}
\]
\end{minipage}
\end{sideways}
\pagebreak

\end{document}
