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In lemma 2.1 of this paper by McKay, the author mentions that the number of closed walks of length $2s$ starting at a fixed vertex in an infinite $v$ regular tree is given by

$$ \sum_{k=1}^s \binom{2s-k}{s}\frac{k}{2s-k}v^k(v-1)^{s-k} $$ This can be proved by counting the number of sequences of $+1$ and $-1$'s of length $2s$ where the partial sums are always nonnegative and exactly $k$ of the partial sums are $0$, and I have a proof for this expression. The author claims that this expression by algebraic manipulation can be shown to be equal to the following two expressions: \begin{gather} v\sum_{k=0}^{s-1} \binom{2s}{k}\frac{s-k}{s}(v-1)^k \quad\text{ and }\quad \sum_{k=1}^s \binom{2s}{k}\frac{2s-2k+1}{2s-k+1} (v-1)^k. \end{gather}

I am stuck with showing that these three expressions are equal to one another. There is another MathSE question showing that the first expression is equal to the second, but I am unable to show that the third one is equal to either of the first or the second expression. Any hints/full solutions will be appreciated!

Blue
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Mutasim Mim
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1 Answers1

4

Starting the computation we get for the third one

$$P(v) = \sum_{k=0}^n {2n\choose k} \frac{2n-2k+1}{2n-k+1} (v-1)^k \\ = \sum_{k=0}^n {2n\choose k} (v-1)^k - \sum_{k=1}^n {2n\choose k} \frac{k}{2n-k+1} (v-1)^k \\ = \sum_{k=0}^n {2n\choose k} (v-1)^k - \sum_{k=1}^n {2n\choose k-1} (v-1)^k.$$

We now examine the coefficient on $[v^q]$ where $0\le q\le n$ by construction. We get for the first piece

$$[v^q] [z^n] \frac{1}{1-z} (1+z(v-1))^{2n} = [v^q] [z^n] \frac{1}{1-z} (1-z+vz)^{2n} \\ = [z^n] {2n\choose q} z^q (1-z)^{2n-1-q} = {2n\choose q} (-1)^{n-q} {2n-1-q\choose n-q}.$$

We have for the second piece

$$[v^q] (v-1) [z^{n-1}] \frac{1}{1-z} (1+z(v-1))^{2n} = [v^q] (v-1) [z^{n-1}] \frac{1}{1-z} (1-z+vz)^{2n} \\ = [z^{n-1}] {2n\choose q-1} z^{q-1} (1-z)^{2n-(q-1)-1} - [z^{n-1}] {2n\choose q} z^q (1-z)^{2n-q-1} \\ = {2n\choose q-1} (-1)^{n-q} {2n-q\choose n-q} - {2n\choose q} (-1)^{n-1-q} {2n-q-1\choose n-1-q}.$$

We have shown that

$$\bbox[5px,border:2px solid #00A000]{ \begin{align} [v^q] P(v) = (-1)^{n-q} \left[ {2n\choose q} {2n-1-q\choose n-q} - {2n\choose q-1} {2n-q\choose n-q} \\ - {2n\choose q} {2n-1-q\choose n-1-q} \right]. \end{align}}$$

We continue with

$$Q(v) = v \sum_{k=0}^{n-1} {2n\choose k} \frac{n-k}{n} (v-1)^k.$$

and do another coefficient extraction on two pieces, the first is (here we may take $q\ge 1$ by inspection as $[v^0] P(v) = 0$ same as $Q(v)$)

$$[v^{q-1}] [z^{n-1}] \frac{1}{1-z} (1+z(v-1))^{2n} = [v^{q-1}] [z^{n-1}] \frac{1}{1-z} (1-z+vz)^{2n} \\ = [z^{n-1}] {2n\choose q-1} z^{q-1} (1-z)^{2n-(q-1)-1} = {2n\choose q-1} (-1)^{n-q} {2n-q\choose n-q}.$$

The second is

$$[v^q] 2v \sum_{k=1}^{n-1} {2n-1\choose k-1} (v-1)^k = 2[v^{q-1}] (v-1) \sum_{k=0}^{n-2} {2n-1\choose k} (v-1)^k \\ = 2[v^{q-1}] (v-1) [z^{n-2}] \frac{1}{1-z} (1+z(v-1))^{2n-1} \\ = 2[v^{q-1}] (v-1) [z^{n-2}] \frac{1}{1-z} (1-z+vz)^{2n-1} \\ = 2 {2n-1\choose q-2} [z^{n-2}] z^{q-2} (1-z)^{2n-1-(q-2)-1} - 2 {2n-1\choose q-1} [z^{n-2}] z^{q-1} (1-z)^{2n-1-(q-1)-1} \\ = 2 {2n-1\choose q-2} (-1)^{n-q} {2n-q\choose n-q} - 2 {2n-1\choose q-1} (-1)^{n-q-1} {2n-1-q\choose n-q-1}.$$

We have shown that

$$\bbox[5px,border:2px solid #00A000]{ \begin{align} [v^q] Q(v) = (-1)^{n-q} \left[ {2n\choose q-1} {2n-q\choose n-q} - 2 {2n-1\choose q-2} {2n-q\choose n-q} \\ - 2 {2n-1\choose q-1} {2n-1-q\choose n-1-q} \right] \end{align}.}$$

Now we just need to verify that these are the same. Dividing by $(-1)^{n-q} {2n-q\choose n-q}$ we get from $P(v)$

$${2n\choose q} \frac{n}{2n-q} - {2n\choose q-1} - {2n\choose q} \frac{n-q}{2n-q} $$

and from $Q(v)$

$${2n\choose q-1} - 2{2n-1\choose q-2} - 2 {2n-1\choose q-1} \frac{n-q}{2n-q}.$$

Next divide by ${2n\choose q}$ for the new pair

$$\frac{n}{2n-q} - \frac{q}{2n+1-q} - \frac{n-q}{2n-q}$$

and

$$\frac{q}{2n+1-q} - 2 \frac{q(q-1)}{2n(2n+1-q)} - 2 \frac{q}{2n} \frac{n-q}{2n-q}.$$

The rest is certainly feasible by hand or may be entrusted to a computer algebra system and we may conclude. With the above computation it is hoped that something simpler might appear. A single lemma might encapsulate the parts that repeat.

Marko Riedel
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