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‎404.html‎

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‎authors/admin/index.html‎

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‎content/post/three-var-recursive.md‎

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@@ -17,7 +17,8 @@ This post extends the generating-function technique from the [two-variable recur
1717

1818
We want to solve the recurrence
1919

20-
$$a(m,n,k) = 2a(m-1,n-1,k-1) + a(m-1,n-1,k) + a(m-1,n,k-1) + a(m,n-1,k-1)$$
20+
$a(m,n,k) = 2a(m-1,n-1,k-1) + a(m-1,n-1,k)$
21+
$ + a(m-1,n,k-1) + a(m,n-1,k-1)$
2122

2223
where $m$, $n$, $k$ are nonnegative integers, with boundary conditions:
2324

@@ -33,34 +34,34 @@ Define
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3435
$$\Phi(x,y,z) = \sum_{m,n,k \geq 0} a(m,n,k) \cdot x^m y^n z^k$$
3536

36-
Using the initial values above, we can write the recurrence including boundary terms as:
37+
Using the initial values above, we can write the recurrence including boundary terms as follow:
3738

38-
$$a(m, n, k) = 2a(m-1, n-1, k-1) + a(m-1, n-1, k) + a(m-1, n, k-1) + a(m, n-1, k-1) + [m=n=k=0] + [m=n=0 \wedge k=1] + [m=k=0 \wedge n=1] + [n=k=0 \wedge m=1]$$
39+
$a(m, n, k) = 2a(m-1, n-1, k-1) + a(m-1, n-1, k)$
40+
$ + a(m-1, n, k-1) + a(m, n-1, k-1)$
41+
$ + [m=n=k=0] + [m=n=0 \wedge k=1]$
42+
$ + [m=k=0 \wedge n=1] + [n=k=0 \wedge m=1]$
3943

40-
I believe there are still some initial conditions missing, since for example $a(0,1,1)$ is not well defined. Computing its value will result in negative arguments:
44+
<!-- I believe there are still some initial conditions missing, since for example $a(0,1,1)$ is not well defined. Computing its value will result in negative arguments:
4145
42-
$a(0,1,1) = 2a(-1, 0, 0) + a(-1, 0, 1) + a(-1, 1, 0) + a(0, 0, 0) = 2a(-1, 0, 0) + 2a(-1, 0, 1) + 1$.
46+
$a(0,1,1) = 2a(-1, 0, 0) + a(-1, 0, 1) + a(-1, 1, 0) + a(0, 0, 0)$
47+
$ = 2a(-1, 0, 0) + 2a(-1, 0, 1) + 1$
4348
44-
Adding the extra condition that $a(m,n,k)=0$ for any negative argument(s) solves the issue.
49+
Adding the extra condition that $a(m,n,k)=0$ for any negative argument(s) solves the issue. -->
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4651
Substituting the recurrence into the generating function and collecting terms:
4752

53+
$\Phi(x,y,z) = \sum_{m,n,k} a(m,n,k) \cdot x^m y^n z^k $
54+
$ = 2\sum_{m,n,k} a(m,n,k) \cdot x^{m+1} y^{n+1} z^{k+1} $
55+
$ + \sum_{m,n,k} a(m,n,k) \cdot x^{m+1} y^{n+1} z^k $
56+
$ + \sum_{m,n,k} a(m,n,k) \cdot x^{m+1} y^n z^{k+1} $
57+
$ + \sum_{m,n,k} a(m,n,k) \cdot x^m y^{n+1} z^{k+1} $
58+
$ + 1 + x + y + z $
59+
$ = 2 \Phi \cdot x y z + \Phi \cdot x y + \Phi \cdot x z + \Phi\ cdot y z + 1 + x + y + z$
60+
$ = \Phi \cdot ( 2 x y z + x y + x z + y z ) + 1 + x + y + z$
4861

49-
$$ \Phi(x,y,z) =\sum_{m,n,k}a(m,n,k) \cdot x^m y^n z^k $$
50-
$$= 2\sum_{m,n,k}a(m,n,k) \cdot x^{m+1} y^{n+1} z^{k+1} + \sum_{m,n,k}a(m,n,
51-
k) \cdot x^{m+1} y^{n+1} z^k + \sum_{m,n,k}a(m,n,k) \cdot x^{m+1} y^n z^{k+1}
52-
+\sum_{m,n,k}a(m,n,k) \cdot x^m y^{n+1} z^{k+1} + 1 + x + y + z $$
53-
$$ 2\Phi(x,y,z) \cdot x y z + \Phi(x,y,z)\cdot x y + \Phi(x,y,z) \cdot x z +
54-
\Phi(x,y,z) \cdot y z + 1 + x + y + z $$
55-
$$ \Phi(x,y,z)\left(2 x y z + x y + x z + y z\right) + 1 + x + y + z$$
62+
where the boundary terms $1 + x + y + z$ come from $a(0,0,0)$, $a(1,0,0)$, $a(0,1,0)$, $a(0,0,1)$.
5663

57-
58-
59-
$\Phi(x,y,z) = \sum_{m,n,k} a(m,n,k)\, x^m y^n z^k$
60-
61-
$= 2\Phi\, xyz + \Phi\, xy + \Phi\, xz + \Phi\, yz + 1 + x + y + z$
62-
63-
where the boundary terms $1 + x + y + z$ come from $a(0,0,0)$, $a(1,0,0)$, $a(0,1,0)$, $a(0,0,1)$. Solving for $\Phi$:
64+
Solving for $\Phi$:
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6566
$$\Phi(x,y,z) = \frac{1 + x + y + z}{1 - 2xyz - xy - xz - yz}$$
6667

@@ -70,39 +71,44 @@ Using $\frac{1}{1-\rho} = \sum_{i \geq 0} \rho^i$ and the multinomial expansion
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7172
$$(x_1+x_2+x_3+x_4)^N = \sum_{k_1+k_2+k_3+k_4=N} \binom{N}{k_1,k_2,k_3,k_4} x_1^{k_1} x_2^{k_2} x_3^{k_3} x_4^{k_4}$$
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73-
with $\binom{N}{k_1,k_2,k_3,k_4} = \frac{N!}{k_1!\cdot k_2!\cdot k_3!\cdot k_4!}$, we expand the denominator. Let $\rho = 2xyz + xy + xz + yz$. Then
74+
with
7475

75-
$$\Phi = (1+x+y+z) \sum_{N \geq 0} \rho^N$$
76+
$$\binom{N}{k_1,k_2,k_3,k_4} = \frac{N!}{k_1!\cdot k_2!\cdot k_3!\cdot k_4!}$$
7677

77-
Expanding $\rho^N$ with the multinomial theorem (and writing $k_4 = N - k_1 - k_2 - k_3$):
78+
we expand the denominator. Let $\rho = 2xyz + xy + xz + yz$. Then
7879

79-
$\Phi = \frac{1 + x + y + z}{1-2xyz - xy - xz - yz}$
80+
$$\Phi = \frac{1+x+y+z}{1-\rho} = (1+x+y+z) \sum_{N \geq 0} \rho^N$$
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81-
$= (1 + x + y + z) \sum_N (2xyz + xy + xz + yz)^N$
82+
Expanding $\rho^N$ with the multinomial theorem (and writing $k_4 = N - k_1 - k_2 - k_3$):
8283

83-
$= (1 + x + y + z) \sum_{k_1+k_2+k_3+k_4=N} \binom{N}{k_1,k_2,k_3,k_4} (2xyz)^{k_1}(xy)^{k_2}(xz)^{k_3}(yz)^{k_4}$
84+
$\sum_{N \geq 0} \rho^N = \sum_{N}(2 x y z + x y + x z + y z)^N $
85+
$ = \sum_{k_1+k_2+k_3+k_4=N} \binom{N} {k_1,k_2,k_3,k_4} (2 x y z)^{k_1} \cdot (x y)^{k_2} \cdot (x z)^{k_3} \cdot (y z)^{k_4}$
86+
$ = \sum_{k_1+k_2+k_3+k_4=N} \binom{N} {k_1,k_2,k_3,k_4} 2^{k_1} x^{k_1+k_2+k_3} y^{k_1+k_2+k_4} z^{k_1+k_3+k_4}$
87+
$ = \sum_{k_1+k_2+k_3 \leq N} \binom{N} {k_1,k_2,k_3, N-k_1-k_2-k_3} 2^{k_1} x^{k_1+k_2+k_3} y^{N-k_3} z^{N-k_2}$
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85-
$= (1 + x + y + z) \sum_{k_1+k_2+k_3 \leq N} \binom{N}{k_1,k_2,k_3,N-k_1-k_2-k_3} 2^{k_1} x^{k_1+k_2+k_3} y^{N-k_3} z^{N-k_2}$
89+
So we have
8690

91+
$$ \Phi = (1 + x + y + z) \sum_{k_1+k_2+k_3 \leq N} \binom{N} {k_1,k_2,k_3, N-k_1-k_2-k_3} 2^{k_1} x^{k_1+k_2+k_3} y^{N-k_3} z^{N-k_2}$$
8792

88-
$$= \frac{1 + x + y + z}{1-2 x y z - x y - x z - y z} $$
89-
$$= (1 + x + y + z) \sum_{N}(2 x y z + x y + x z + y z)^N $$
90-
$$= (1 + x + y + z) \sum_{k_1+k_2+k_3+k_4=N} \binom{N} {k_1,k_2,k_3,k_4} (2 x
91-
y z)^{k_1} \cdot (x y)^{k_2} \cdot (x z)^{k_3} \cdot (y z)^{k_4}$$
92-
$$= (1 + x + y + z) \sum_{k_1+k_2+k_3+k_4=N} \binom{N} {k_1,k_2,k_3,k_4} 2^
93-
{k_1} x^{k_1+k_2+k_3} y^{k_1+k_2+k_4} z^{k_1+k_3+k_4}$$
94-
$$= (1 + x + y + z) \sum_{k_1+k_2+k_3\leq N} \binom{N} {k_1,k_2,k_3,
95-
N-k_1-k_2-k_3} 2^{k_1} x^{k_1+k_2+k_3} y^{N-k_3} z^{N-k_2}$$
9693

97-
Extracting the coefficient of $x^m y^n z^k$ gives the closed form. The full expression has four sums (from the numerator $1+x+y+z$):
94+
<!-- $\Phi = \frac{1 + x + y + z}{1-2xyz - xy - xz - yz}$
95+
$ = (1 + x + y + z) \sum_N (2xyz + xy + xz + yz)^N$
96+
$ = (1 + x + y + z) \sum_{k_1+k_2+k_3+k_4=N} \binom{N}{k_1,k_2,k_3,k_4} (2xyz)^{k_1}(xy)^{k_2}(xz)^{k_3}(yz)^{k_4}$
97+
$ = (1 + x + y + z) \sum_{k_1+k_2+k_3 \leq N} \binom{N}{k_1,k_2,k_3,N-k_1-k_2-k_3} 2^{k_1} x^{k_1+k_2+k_3} y^{N-k_3} z^{N-k_2}$
9898
99-
$$a(m,n,k) = \sum_{N=\max(m,n,k)}^{\lfloor (m+n+k)/2 \rfloor} \binom{N}{m+n+k-2N,\, N-m,\, N-n,\, N-k} 2^{m+n+k-2N}$$
10099
101-
$$+ \sum_{N=\max(m-1,n,k)}^{\lfloor (m+n+k-1)/2 \rfloor} \binom{N}{m+n+k-2N-1,\, N-m+1,\, N-n,\, N-k} 2^{m+n+k-2N-1}$$
100+
$ = \frac{1 + x + y + z}{1-2 x y z - x y - x z - y z} $
101+
$ = (1 + x + y + z) \sum_{N}(2 x y z + x y + x z + y z)^N $
102+
$ = (1 + x + y + z) \sum_{k_1+k_2+k_3+k_4=N} \binom{N} {k_1,k_2,k_3,k_4} (2 x y z)^{k_1} \cdot (x y)^{k_2} \cdot (x z)^{k_3} \cdot (y z)^{k_4}$
103+
$ = (1 + x + y + z) \sum_{k_1+k_2+k_3+k_4=N} \binom{N} {k_1,k_2,k_3,k_4} 2^{k_1} x^{k_1+k_2+k_3} y^{k_1+k_2+k_4} z^{k_1+k_3+k_4}$
104+
$ = (1 + x + y + z) \sum_{k_1+k_2+k_3\leq N} \binom{N} {k_1,k_2,k_3, N-k_1-k_2-k_3} 2^{k_1} x^{k_1+k_2+k_3} y^{N-k_3} z^{N-k_2}$ -->
102105

103-
$$+ \sum_{N=\max(m,n-1,k)}^{\lfloor (m+n+k-1)/2 \rfloor} \binom{N}{m+n+k-2N-1,\, N-m,\, N-n+1,\, N-k} 2^{m+n+k-2N-1}$$
106+
Extracting the coefficient of $x^m y^n z^k$ gives the closed form. The full expression has four sums (from the numerator $1+x+y+z$):
104107

105-
$$+ \sum_{N=\max(m,n,k-1)}^{\lfloor (m+n+k-1)/2 \rfloor} \binom{N}{m+n+k-2N-1,\, N-m,\, N-n,\, N-k+1} 2^{m+n+k-2N-1}$$
108+
$$a(m,n,k) = \sum_{N=\max(m,n,k)}^{ (m+n+k)/2 } \binom{N}{m+n+k-2N, N-m, N-n, N-k} 2^{m+n+k-2N}$$
109+
$$ + \sum_{N=\max(m-1,n,k)}^{ (m+n+k-1)/2 } \binom{N}{m+n+k-2N-1, N-m+1, N-n, N-k} 2^{m+n+k-2N-1}$$
110+
$$ + \sum_{N=\max(m,n-1,k)}^{ (m+n+k-1)/2 } \binom{N}{m+n+k-2N-1, N-m, N-n+1, N-k} 2^{m+n+k-2N-1}$$
111+
$$ + \sum_{N=\max(m,n,k-1)}^{ (m+n+k-1)/2 } \binom{N}{m+n+k-2N-1, N-m, N-n, N-k+1} 2^{m+n+k-2N-1}$$
106112

107113
There may be room to simplify this further; the symmetry in $m,n,k$ could help.
108114

‎index.html‎

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<!DOCTYPE html>
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<!-- This site was created with Hugo Blox. https://hugoblox.com -->
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<head>
@@ -984,7 +984,8 @@ <h1 class="mb-0">Recent Posts</h1>
984984
<p>This post extends the generating-function technique from the <a href="/post/two-var-recursive-func/">two-variable recursion</a> to a three-variable case. I originally wrote this as an answer to a <a href="https://math.stackexchange.com/questions/1093271/how-to-solve-this-multivariable-recursion/2730331#2730331" target="_blank" rel="noopener">Math Stack Exchange question</a>; here it is adapted for the blog with clearer exposition and code.</p>
985985
<h2 id="the-problem">The Problem</h2>
986986
<p>We want to solve the recurrence</p>
987-
<p>$$a(m,n,k) = 2a(m-1,n-1,k-1) + a(m-1,n-1,k) + a(m-1,n,k-1) + a(m,n-1,k-1)$$</p>
987+
<p>$a(m,n,k) = 2a(m-1,n-1,k-1) + a(m-1,n-1,k)$
988+
$ + a(m-1,n,k-1) + a(m,n-1,k-1)$</p>
988989
<p>where $m$, $n$, $k$ are nonnegative integers, with boundary conditions:</p>
989990
<ul>
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<li>$a(0,0,0) = a(1,0,0) = a(0,1,0) = a(0,0,1) = 1$</li>
@@ -1024,7 +1025,7 @@ <h2 id="the-problem">The Problem</h2>
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<span class="middot-divider"></span>
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<span class="article-reading-time">
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6 min read
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5 min read
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</span>
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‎index.xml‎

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2020
<description>&lt;p&gt;This post extends the generating-function technique from the &lt;a href=&#34;/post/two-var-recursive-func/&#34;&gt;two-variable recursion&lt;/a&gt; to a three-variable case. I originally wrote this as an answer to a &lt;a href=&#34;https://math.stackexchange.com/questions/1093271/how-to-solve-this-multivariable-recursion/2730331#2730331&#34; target=&#34;_blank&#34; rel=&#34;noopener&#34;&gt;Math Stack Exchange question&lt;/a&gt;; here it is adapted for the blog with clearer exposition and code.&lt;/p&gt;
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&lt;h2 id=&#34;the-problem&#34;&gt;The Problem&lt;/h2&gt;
2222
&lt;p&gt;We want to solve the recurrence&lt;/p&gt;
23-
&lt;p&gt;$$a(m,n,k) = 2a(m-1,n-1,k-1) + a(m-1,n-1,k) + a(m-1,n,k-1) + a(m,n-1,k-1)$$&lt;/p&gt;
23+
&lt;p&gt;$a(m,n,k) = 2a(m-1,n-1,k-1) + a(m-1,n-1,k)$
24+
$ + a(m-1,n,k-1) + a(m,n-1,k-1)$&lt;/p&gt;
2425
&lt;p&gt;where $m$, $n$, $k$ are nonnegative integers, with boundary conditions:&lt;/p&gt;
2526
&lt;ul&gt;
2627
&lt;li&gt;$a(0,0,0) = a(1,0,0) = a(0,1,0) = a(0,0,1) = 1$&lt;/li&gt;
@@ -31,46 +32,42 @@
3132
&lt;h2 id=&#34;the-generating-function&#34;&gt;The Generating Function&lt;/h2&gt;
3233
&lt;p&gt;Define&lt;/p&gt;
3334
&lt;p&gt;$$\Phi(x,y,z) = \sum_{m,n,k \geq 0} a(m,n,k) \cdot x^m y^n z^k$$&lt;/p&gt;
34-
&lt;p&gt;Using the initial values above, we can write the recurrence including boundary terms as:&lt;/p&gt;
35-
&lt;p&gt;$$a(m, n, k) = 2a(m-1, n-1, k-1) + a(m-1, n-1, k) + a(m-1, n, k-1) + a(m, n-1, k-1) + [m=n=k=0] + [m=n=0 \wedge k=1] + [m=k=0 \wedge n=1] + [n=k=0 \wedge m=1]$$&lt;/p&gt;
36-
&lt;p&gt;I believe there are still some initial conditions missing, since for example $a(0,1,1)$ is not well defined. Computing its value will result in negative arguments:&lt;/p&gt;
37-
&lt;p&gt;$a(0,1,1) = 2a(-1, 0, 0) + a(-1, 0, 1) + a(-1, 1, 0) + a(0, 0, 0) = 2a(-1, 0, 0) + 2a(-1, 0, 1) + 1$.&lt;/p&gt;
38-
&lt;p&gt;Adding the extra condition that $a(m,n,k)=0$ for any negative argument(s) solves the issue.&lt;/p&gt;
35+
&lt;p&gt;Using the initial values above, we can write the recurrence including boundary terms as follow:&lt;/p&gt;
36+
&lt;p&gt;$a(m, n, k) = 2a(m-1, n-1, k-1) + a(m-1, n-1, k)$
37+
$ + a(m-1, n, k-1) + a(m, n-1, k-1)$
38+
$ + [m=n=k=0] + [m=n=0 \wedge k=1]$
39+
$ + [m=k=0 \wedge n=1] + [n=k=0 \wedge m=1]$&lt;/p&gt;
3940
&lt;p&gt;Substituting the recurrence into the generating function and collecting terms:&lt;/p&gt;
40-
&lt;p&gt;$$ \Phi(x,y,z) =\sum_{m,n,k}a(m,n,k) \cdot x^m y^n z^k $$
41-
$$= 2\sum_{m,n,k}a(m,n,k) \cdot x^{m+1} y^{n+1} z^{k+1} + \sum_{m,n,k}a(m,n,
42-
k) \cdot x^{m+1} y^{n+1} z^k + \sum_{m,n,k}a(m,n,k) \cdot x^{m+1} y^n z^{k+1}
43-
+\sum_{m,n,k}a(m,n,k) \cdot x^m y^{n+1} z^{k+1} + 1 + x + y + z $$
44-
$$ 2\Phi(x,y,z) \cdot x y z + \Phi(x,y,z)\cdot x y + \Phi(x,y,z) \cdot x z +
45-
\Phi(x,y,z) \cdot y z + 1 + x + y + z $$
46-
$$ \Phi(x,y,z)\left(2 x y z + x y + x z + y z\right) + 1 + x + y + z$$&lt;/p&gt;
47-
&lt;p&gt;$\Phi(x,y,z) = \sum_{m,n,k} a(m,n,k), x^m y^n z^k$&lt;/p&gt;
48-
&lt;p&gt;$= 2\Phi, xyz + \Phi, xy + \Phi, xz + \Phi, yz + 1 + x + y + z$&lt;/p&gt;
49-
&lt;p&gt;where the boundary terms $1 + x + y + z$ come from $a(0,0,0)$, $a(1,0,0)$, $a(0,1,0)$, $a(0,0,1)$. Solving for $\Phi$:&lt;/p&gt;
41+
&lt;p&gt;$\Phi(x,y,z) = \sum_{m,n,k} a(m,n,k) \cdot x^m y^n z^k $
42+
$ = 2\sum_{m,n,k} a(m,n,k) \cdot x^{m+1} y^{n+1} z^{k+1} $
43+
$ + \sum_{m,n,k} a(m,n,k) \cdot x^{m+1} y^{n+1} z^k $
44+
$ + \sum_{m,n,k} a(m,n,k) \cdot x^{m+1} y^n z^{k+1} $
45+
$ + \sum_{m,n,k} a(m,n,k) \cdot x^m y^{n+1} z^{k+1} $
46+
$ + 1 + x + y + z $
47+
$ = 2 \Phi \cdot x y z + \Phi \cdot x y + \Phi \cdot x z + \Phi\ cdot y z + 1 + x + y + z$
48+
$ = \Phi \cdot ( 2 x y z + x y + x z + y z ) + 1 + x + y + z$&lt;/p&gt;
49+
&lt;p&gt;where the boundary terms $1 + x + y + z$ come from $a(0,0,0)$, $a(1,0,0)$, $a(0,1,0)$, $a(0,0,1)$.&lt;/p&gt;
50+
&lt;p&gt;Solving for $\Phi$:&lt;/p&gt;
5051
&lt;p&gt;$$\Phi(x,y,z) = \frac{1 + x + y + z}{1 - 2xyz - xy - xz - yz}$$&lt;/p&gt;
5152
&lt;h2 id=&#34;from-generating-function-to-closed-form&#34;&gt;From Generating Function to Closed Form&lt;/h2&gt;
5253
&lt;p&gt;Using $\frac{1}{1-\rho} = \sum_{i \geq 0} \rho^i$ and the multinomial expansion&lt;/p&gt;
5354
&lt;p&gt;$$(x_1+x_2+x_3+x_4)^N = \sum_{k_1+k_2+k_3+k_4=N} \binom{N}{k_1,k_2,k_3,k_4} x_1^{k_1} x_2^{k_2} x_3^{k_3} x_4^{k_4}$$&lt;/p&gt;
54-
&lt;p&gt;with $\binom{N}{k_1,k_2,k_3,k_4} = \frac{N!}{k_1!\cdot k_2!\cdot k_3!\cdot k_4!}$, we expand the denominator. Let $\rho = 2xyz + xy + xz + yz$. Then&lt;/p&gt;
55-
&lt;p&gt;$$\Phi = (1+x+y+z) \sum_{N \geq 0} \rho^N$$&lt;/p&gt;
55+
&lt;p&gt;with&lt;/p&gt;
56+
&lt;p&gt;$$\binom{N}{k_1,k_2,k_3,k_4} = \frac{N!}{k_1!\cdot k_2!\cdot k_3!\cdot k_4!}$$&lt;/p&gt;
57+
&lt;p&gt;we expand the denominator. Let $\rho = 2xyz + xy + xz + yz$. Then&lt;/p&gt;
58+
&lt;p&gt;$$\Phi = \frac{1+x+y+z}{1-\rho} = (1+x+y+z) \sum_{N \geq 0} \rho^N$$&lt;/p&gt;
5659
&lt;p&gt;Expanding $\rho^N$ with the multinomial theorem (and writing $k_4 = N - k_1 - k_2 - k_3$):&lt;/p&gt;
57-
&lt;p&gt;$\Phi = \frac{1 + x + y + z}{1-2xyz - xy - xz - yz}$&lt;/p&gt;
58-
&lt;p&gt;$= (1 + x + y + z) \sum_N (2xyz + xy + xz + yz)^N$&lt;/p&gt;
59-
&lt;p&gt;$= (1 + x + y + z) \sum_{k_1+k_2+k_3+k_4=N} \binom{N}{k_1,k_2,k_3,k_4} (2xyz)^{k_1}(xy)^{k_2}(xz)^{k_3}(yz)^{k_4}$&lt;/p&gt;
60-
&lt;p&gt;$= (1 + x + y + z) \sum_{k_1+k_2+k_3 \leq N} \binom{N}{k_1,k_2,k_3,N-k_1-k_2-k_3} 2^{k_1} x^{k_1+k_2+k_3} y^{N-k_3} z^{N-k_2}$&lt;/p&gt;
61-
&lt;p&gt;$$= \frac{1 + x + y + z}{1-2 x y z - x y - x z - y z} $$
62-
$$= (1 + x + y + z) \sum_{N}(2 x y z + x y + x z + y z)^N $$
63-
$$= (1 + x + y + z) \sum_{k_1+k_2+k_3+k_4=N} \binom{N} {k_1,k_2,k_3,k_4} (2 x
64-
y z)^{k_1} \cdot (x y)^{k_2} \cdot (x z)^{k_3} \cdot (y z)^{k_4}$$
65-
$$= (1 + x + y + z) \sum_{k_1+k_2+k_3+k_4=N} \binom{N} {k_1,k_2,k_3,k_4} 2^
66-
{k_1} x^{k_1+k_2+k_3} y^{k_1+k_2+k_4} z^{k_1+k_3+k_4}$$
67-
$$= (1 + x + y + z) \sum_{k_1+k_2+k_3\leq N} \binom{N} {k_1,k_2,k_3,
68-
N-k_1-k_2-k_3} 2^{k_1} x^{k_1+k_2+k_3} y^{N-k_3} z^{N-k_2}$$&lt;/p&gt;
60+
&lt;p&gt;$\sum_{N \geq 0} \rho^N = \sum_{N}(2 x y z + x y + x z + y z)^N $
61+
$ = \sum_{k_1+k_2+k_3+k_4=N} \binom{N} {k_1,k_2,k_3,k_4} (2 x y z)^{k_1} \cdot (x y)^{k_2} \cdot (x z)^{k_3} \cdot (y z)^{k_4}$
62+
$ = \sum_{k_1+k_2+k_3+k_4=N} \binom{N} {k_1,k_2,k_3,k_4} 2^{k_1} x^{k_1+k_2+k_3} y^{k_1+k_2+k_4} z^{k_1+k_3+k_4}$
63+
$ = \sum_{k_1+k_2+k_3 \leq N} \binom{N} {k_1,k_2,k_3, N-k_1-k_2-k_3} 2^{k_1} x^{k_1+k_2+k_3} y^{N-k_3} z^{N-k_2}$&lt;/p&gt;
64+
&lt;p&gt;So we have&lt;/p&gt;
65+
&lt;p&gt;$$ \Phi = (1 + x + y + z) \sum_{k_1+k_2+k_3 \leq N} \binom{N} {k_1,k_2,k_3, N-k_1-k_2-k_3} 2^{k_1} x^{k_1+k_2+k_3} y^{N-k_3} z^{N-k_2}$$&lt;/p&gt;
6966
&lt;p&gt;Extracting the coefficient of $x^m y^n z^k$ gives the closed form. The full expression has four sums (from the numerator $1+x+y+z$):&lt;/p&gt;
70-
&lt;p&gt;$$a(m,n,k) = \sum_{N=\max(m,n,k)}^{\lfloor (m+n+k)/2 \rfloor} \binom{N}{m+n+k-2N,, N-m,, N-n,, N-k} 2^{m+n+k-2N}$$&lt;/p&gt;
71-
&lt;p&gt;$$+ \sum_{N=\max(m-1,n,k)}^{\lfloor (m+n+k-1)/2 \rfloor} \binom{N}{m+n+k-2N-1,, N-m+1,, N-n,, N-k} 2^{m+n+k-2N-1}$$&lt;/p&gt;
72-
&lt;p&gt;$$+ \sum_{N=\max(m,n-1,k)}^{\lfloor (m+n+k-1)/2 \rfloor} \binom{N}{m+n+k-2N-1,, N-m,, N-n+1,, N-k} 2^{m+n+k-2N-1}$$&lt;/p&gt;
73-
&lt;p&gt;$$+ \sum_{N=\max(m,n,k-1)}^{\lfloor (m+n+k-1)/2 \rfloor} \binom{N}{m+n+k-2N-1,, N-m,, N-n,, N-k+1} 2^{m+n+k-2N-1}$$&lt;/p&gt;
67+
&lt;p&gt;$$a(m,n,k) = \sum_{N=\max(m,n,k)}^{ (m+n+k)/2 } \binom{N}{m+n+k-2N, N-m, N-n, N-k} 2^{m+n+k-2N}$$
68+
$$ + \sum_{N=\max(m-1,n,k)}^{ (m+n+k-1)/2 } \binom{N}{m+n+k-2N-1, N-m+1, N-n, N-k} 2^{m+n+k-2N-1}$$
69+
$$ + \sum_{N=\max(m,n-1,k)}^{ (m+n+k-1)/2 } \binom{N}{m+n+k-2N-1, N-m, N-n+1, N-k} 2^{m+n+k-2N-1}$$
70+
$$ + \sum_{N=\max(m,n,k-1)}^{ (m+n+k-1)/2 } \binom{N}{m+n+k-2N-1, N-m, N-n, N-k+1} 2^{m+n+k-2N-1}$$&lt;/p&gt;
7471
&lt;p&gt;There may be room to simplify this further; the symmetry in $m,n,k$ could help.&lt;/p&gt;
7572
&lt;h2 id=&#34;complexity&#34;&gt;Complexity&lt;/h2&gt;
7673
&lt;ul&gt;

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