partitions of positive integer $n$ with respect to a multiset
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Recently, I think on a new problem related to partitions.
Let $n$ be a non-negative integer and $mathbb{A}={a_1,ldots,a_k}$ be a multiset with $k$
not necessarily distinct positive integers. We denote by $D(nmidmathbb{A})$, the number
of ways to partition $n$ in the form $a_1x_1+cdots+a_kx_k$, where $x_i$'s are positive integers
and $x_ileqslant x_{i+1}$ whenever $a_i=a_{i+1}$.
I would like to calculate generation function of $D(n|A)$.
combinatorics generating-functions integer-partitions
$endgroup$
add a comment |
$begingroup$
Recently, I think on a new problem related to partitions.
Let $n$ be a non-negative integer and $mathbb{A}={a_1,ldots,a_k}$ be a multiset with $k$
not necessarily distinct positive integers. We denote by $D(nmidmathbb{A})$, the number
of ways to partition $n$ in the form $a_1x_1+cdots+a_kx_k$, where $x_i$'s are positive integers
and $x_ileqslant x_{i+1}$ whenever $a_i=a_{i+1}$.
I would like to calculate generation function of $D(n|A)$.
combinatorics generating-functions integer-partitions
$endgroup$
$begingroup$
Have you tried ignoring the semimonotonicity condition while putting together your product of elementwise generating functions, then dividing out the overcounts at the end?
$endgroup$
– Eric Towers
Aug 27 '17 at 20:29
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Why wouldn't you divide out the coefficient you extract by $n_i!$ where the $n_i$ are the repetition counts in the multiset $A$, rather than trying to do this with the gf?
$endgroup$
– Eric Towers
Aug 27 '17 at 20:50
add a comment |
$begingroup$
Recently, I think on a new problem related to partitions.
Let $n$ be a non-negative integer and $mathbb{A}={a_1,ldots,a_k}$ be a multiset with $k$
not necessarily distinct positive integers. We denote by $D(nmidmathbb{A})$, the number
of ways to partition $n$ in the form $a_1x_1+cdots+a_kx_k$, where $x_i$'s are positive integers
and $x_ileqslant x_{i+1}$ whenever $a_i=a_{i+1}$.
I would like to calculate generation function of $D(n|A)$.
combinatorics generating-functions integer-partitions
$endgroup$
Recently, I think on a new problem related to partitions.
Let $n$ be a non-negative integer and $mathbb{A}={a_1,ldots,a_k}$ be a multiset with $k$
not necessarily distinct positive integers. We denote by $D(nmidmathbb{A})$, the number
of ways to partition $n$ in the form $a_1x_1+cdots+a_kx_k$, where $x_i$'s are positive integers
and $x_ileqslant x_{i+1}$ whenever $a_i=a_{i+1}$.
I would like to calculate generation function of $D(n|A)$.
combinatorics generating-functions integer-partitions
combinatorics generating-functions integer-partitions
edited Apr 20 '18 at 11:21
d.y
asked Aug 27 '17 at 20:14
d.yd.y
328111
328111
$begingroup$
Have you tried ignoring the semimonotonicity condition while putting together your product of elementwise generating functions, then dividing out the overcounts at the end?
$endgroup$
– Eric Towers
Aug 27 '17 at 20:29
$begingroup$
Why wouldn't you divide out the coefficient you extract by $n_i!$ where the $n_i$ are the repetition counts in the multiset $A$, rather than trying to do this with the gf?
$endgroup$
– Eric Towers
Aug 27 '17 at 20:50
add a comment |
$begingroup$
Have you tried ignoring the semimonotonicity condition while putting together your product of elementwise generating functions, then dividing out the overcounts at the end?
$endgroup$
– Eric Towers
Aug 27 '17 at 20:29
$begingroup$
Why wouldn't you divide out the coefficient you extract by $n_i!$ where the $n_i$ are the repetition counts in the multiset $A$, rather than trying to do this with the gf?
$endgroup$
– Eric Towers
Aug 27 '17 at 20:50
$begingroup$
Have you tried ignoring the semimonotonicity condition while putting together your product of elementwise generating functions, then dividing out the overcounts at the end?
$endgroup$
– Eric Towers
Aug 27 '17 at 20:29
$begingroup$
Have you tried ignoring the semimonotonicity condition while putting together your product of elementwise generating functions, then dividing out the overcounts at the end?
$endgroup$
– Eric Towers
Aug 27 '17 at 20:29
$begingroup$
Why wouldn't you divide out the coefficient you extract by $n_i!$ where the $n_i$ are the repetition counts in the multiset $A$, rather than trying to do this with the gf?
$endgroup$
– Eric Towers
Aug 27 '17 at 20:50
$begingroup$
Why wouldn't you divide out the coefficient you extract by $n_i!$ where the $n_i$ are the repetition counts in the multiset $A$, rather than trying to do this with the gf?
$endgroup$
– Eric Towers
Aug 27 '17 at 20:50
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let's try again. (The original text is gone. If one would like to waste one's time, it is available through the "edited" link just below the examples.)
Instead of starting with a structureless multiset, let's go with: $I$ is the cardinality of the set of distinct elements in $A$, and $B = {(b_i, m_i), i in I}$ where the $b_i$ are the distinct elements of $A$ and $m_i$ is the multiplicity of $b_i$ in $A$. So, for example, $sum_{i in I} m_i = |A|$.
For each $i$ in $I$, we want a monotonically nondecreasing sequence $n_{i,1} leq n_{i,2} leq cdots leq n_{i,m_i}$. We make the change of variables begin{align*}
d_{i,1} &= n_{i,1} text{,} \
d_{i,2} &= n_{i,2} - n_{i,1} text{,} \
&vdots \
d_{i,j} &= n_{i,j} - n_{i,j-1} & (2 leq j leq m_i) text{,} \
&vdots \
d_{i,m_i} &= n_{i,m_i} - n_{i,m_i-1} text{.}
end{align*}
Then the monotonically nondecreasing condition on the $(n_{i,j})_j$ becomes $d_{i,1} geq 1$ and $d_{i,j} geq 0$ for $1 leq j leq m_i$. Observe that
begin{align*}
sum_{j=1}^{m_i} n_{i,j} &= (d_{i,1}) + (d_{i,1} + d_{i,2}) + cdots + (d_{i,1} + d_{i,2} + cdots + d_{i,m_i}) \
&= sum_{j=1}^{m_i} (m_i - j +1) d_{i,j}
end{align*}
Then $D(n|A)$ is the number of ways of choosing all these $d_{i,j}$ such that begin{align*}
n &= sum_{i in I} b_i sum_{j =1}^{m_i} n_{i,j} \
&= sum_{i in I} b_i sum_{j=1}^{m_i} (m_i - j +1) d_{i,j} \
&= sum_{i in I} left( b_i m_i d_{i,1} + sum_{j=2}^{m_i} b_i (m_i - j +1) d_{i,j} right) text{,} \
d_{i,1} &geq 1 qquad (i in I) & text{, and } \
d_{i,j} &geq 0 qquad (i in I, 2 leq j leq m_i) text{.}
end{align*}
The generating function for $D$ is then
$$ prod_{i in I} left( frac{x^{b_i m_i}}{1-x^{b_i m_i}} prod_{j = 2}^{m_i} frac{1}{1-x^{b_i(m_i - j +1)}} right) text{.} $$
Examples:
begin{align*}
mathrm{gf}(D(n|{1,2})) &= x^3 + x^4 + 2 x^5 + 2 x^6 + 3 x^7 + 3 x^8 + 4 x^9 + 4 x^{10} + cdots text{,} \
mathrm{gf}(D(n|{1,2,2})) &= x^5 + x^6 + 2 x^7 + 2 x^8 + 4 x^9 + 4 x^{10} + 6 x^{11} + 6 x^{12} + 9 x^{13} \
&qquad + 9 x^{14} + 12 x^{15} + 12 x^{16} + 16 x^{17} + 16 x^{18} + 20 x^{19} + 20 x^{20} + cdots text{, and} \
mathrm{gf}(D(n|{1,1,1})) &= x^3 + x^4 + 2 x^5 + 3 x^6 + 4 x^7 + 5 x^8 + 7 x^9 + 8 x^{10} + cdots text{.}
end{align*}
Continuing:
We can simplify the gf a bit. begin{align*}
prod_{i in I} left( frac{x^{b_i m_i}}{1-x^{b_i m_i}} prod_{j = 2}^{m_i} frac{1}{1-x^{b_i(m_i - j +1)}} right)
&= prod_{i in I} frac{x^{b_i m_i}}{1-x^{b_i m_i}} frac{1}{prod_{j = 2}^{m_i}1-x^{b_i(m_i - j +1)}} \
&= prod_{i in I} frac{x^{b_i m_i}}{prod_{j = 1}^{m_i}1-x^{b_i(m_i - j +1)}} \
&= prod_{i in I} prod_{j = 1}^{m_i} frac{x^{b_i}}{1-x^{b_i(m_i - j +1)}} text{.}
end{align*}
$endgroup$
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@d.y : I'd rather say $B = {(1,ell)}$ so that $B$ is a set of ordered pairs $(b_i, m_i)$ (with distinct $b_i$s). The gf for this $B$ is exactly the one given and is the same as the one you write. The given gf is a character-by-character translation of the last sum in the display above it.
$endgroup$
– Eric Towers
Aug 28 '17 at 14:41
$begingroup$
@d.y : I can't right now. Have to get back to work. I'll check in in several hours.
$endgroup$
– Eric Towers
Aug 28 '17 at 14:45
add a comment |
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1 Answer
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$begingroup$
Let's try again. (The original text is gone. If one would like to waste one's time, it is available through the "edited" link just below the examples.)
Instead of starting with a structureless multiset, let's go with: $I$ is the cardinality of the set of distinct elements in $A$, and $B = {(b_i, m_i), i in I}$ where the $b_i$ are the distinct elements of $A$ and $m_i$ is the multiplicity of $b_i$ in $A$. So, for example, $sum_{i in I} m_i = |A|$.
For each $i$ in $I$, we want a monotonically nondecreasing sequence $n_{i,1} leq n_{i,2} leq cdots leq n_{i,m_i}$. We make the change of variables begin{align*}
d_{i,1} &= n_{i,1} text{,} \
d_{i,2} &= n_{i,2} - n_{i,1} text{,} \
&vdots \
d_{i,j} &= n_{i,j} - n_{i,j-1} & (2 leq j leq m_i) text{,} \
&vdots \
d_{i,m_i} &= n_{i,m_i} - n_{i,m_i-1} text{.}
end{align*}
Then the monotonically nondecreasing condition on the $(n_{i,j})_j$ becomes $d_{i,1} geq 1$ and $d_{i,j} geq 0$ for $1 leq j leq m_i$. Observe that
begin{align*}
sum_{j=1}^{m_i} n_{i,j} &= (d_{i,1}) + (d_{i,1} + d_{i,2}) + cdots + (d_{i,1} + d_{i,2} + cdots + d_{i,m_i}) \
&= sum_{j=1}^{m_i} (m_i - j +1) d_{i,j}
end{align*}
Then $D(n|A)$ is the number of ways of choosing all these $d_{i,j}$ such that begin{align*}
n &= sum_{i in I} b_i sum_{j =1}^{m_i} n_{i,j} \
&= sum_{i in I} b_i sum_{j=1}^{m_i} (m_i - j +1) d_{i,j} \
&= sum_{i in I} left( b_i m_i d_{i,1} + sum_{j=2}^{m_i} b_i (m_i - j +1) d_{i,j} right) text{,} \
d_{i,1} &geq 1 qquad (i in I) & text{, and } \
d_{i,j} &geq 0 qquad (i in I, 2 leq j leq m_i) text{.}
end{align*}
The generating function for $D$ is then
$$ prod_{i in I} left( frac{x^{b_i m_i}}{1-x^{b_i m_i}} prod_{j = 2}^{m_i} frac{1}{1-x^{b_i(m_i - j +1)}} right) text{.} $$
Examples:
begin{align*}
mathrm{gf}(D(n|{1,2})) &= x^3 + x^4 + 2 x^5 + 2 x^6 + 3 x^7 + 3 x^8 + 4 x^9 + 4 x^{10} + cdots text{,} \
mathrm{gf}(D(n|{1,2,2})) &= x^5 + x^6 + 2 x^7 + 2 x^8 + 4 x^9 + 4 x^{10} + 6 x^{11} + 6 x^{12} + 9 x^{13} \
&qquad + 9 x^{14} + 12 x^{15} + 12 x^{16} + 16 x^{17} + 16 x^{18} + 20 x^{19} + 20 x^{20} + cdots text{, and} \
mathrm{gf}(D(n|{1,1,1})) &= x^3 + x^4 + 2 x^5 + 3 x^6 + 4 x^7 + 5 x^8 + 7 x^9 + 8 x^{10} + cdots text{.}
end{align*}
Continuing:
We can simplify the gf a bit. begin{align*}
prod_{i in I} left( frac{x^{b_i m_i}}{1-x^{b_i m_i}} prod_{j = 2}^{m_i} frac{1}{1-x^{b_i(m_i - j +1)}} right)
&= prod_{i in I} frac{x^{b_i m_i}}{1-x^{b_i m_i}} frac{1}{prod_{j = 2}^{m_i}1-x^{b_i(m_i - j +1)}} \
&= prod_{i in I} frac{x^{b_i m_i}}{prod_{j = 1}^{m_i}1-x^{b_i(m_i - j +1)}} \
&= prod_{i in I} prod_{j = 1}^{m_i} frac{x^{b_i}}{1-x^{b_i(m_i - j +1)}} text{.}
end{align*}
$endgroup$
$begingroup$
@d.y : I'd rather say $B = {(1,ell)}$ so that $B$ is a set of ordered pairs $(b_i, m_i)$ (with distinct $b_i$s). The gf for this $B$ is exactly the one given and is the same as the one you write. The given gf is a character-by-character translation of the last sum in the display above it.
$endgroup$
– Eric Towers
Aug 28 '17 at 14:41
$begingroup$
@d.y : I can't right now. Have to get back to work. I'll check in in several hours.
$endgroup$
– Eric Towers
Aug 28 '17 at 14:45
add a comment |
$begingroup$
Let's try again. (The original text is gone. If one would like to waste one's time, it is available through the "edited" link just below the examples.)
Instead of starting with a structureless multiset, let's go with: $I$ is the cardinality of the set of distinct elements in $A$, and $B = {(b_i, m_i), i in I}$ where the $b_i$ are the distinct elements of $A$ and $m_i$ is the multiplicity of $b_i$ in $A$. So, for example, $sum_{i in I} m_i = |A|$.
For each $i$ in $I$, we want a monotonically nondecreasing sequence $n_{i,1} leq n_{i,2} leq cdots leq n_{i,m_i}$. We make the change of variables begin{align*}
d_{i,1} &= n_{i,1} text{,} \
d_{i,2} &= n_{i,2} - n_{i,1} text{,} \
&vdots \
d_{i,j} &= n_{i,j} - n_{i,j-1} & (2 leq j leq m_i) text{,} \
&vdots \
d_{i,m_i} &= n_{i,m_i} - n_{i,m_i-1} text{.}
end{align*}
Then the monotonically nondecreasing condition on the $(n_{i,j})_j$ becomes $d_{i,1} geq 1$ and $d_{i,j} geq 0$ for $1 leq j leq m_i$. Observe that
begin{align*}
sum_{j=1}^{m_i} n_{i,j} &= (d_{i,1}) + (d_{i,1} + d_{i,2}) + cdots + (d_{i,1} + d_{i,2} + cdots + d_{i,m_i}) \
&= sum_{j=1}^{m_i} (m_i - j +1) d_{i,j}
end{align*}
Then $D(n|A)$ is the number of ways of choosing all these $d_{i,j}$ such that begin{align*}
n &= sum_{i in I} b_i sum_{j =1}^{m_i} n_{i,j} \
&= sum_{i in I} b_i sum_{j=1}^{m_i} (m_i - j +1) d_{i,j} \
&= sum_{i in I} left( b_i m_i d_{i,1} + sum_{j=2}^{m_i} b_i (m_i - j +1) d_{i,j} right) text{,} \
d_{i,1} &geq 1 qquad (i in I) & text{, and } \
d_{i,j} &geq 0 qquad (i in I, 2 leq j leq m_i) text{.}
end{align*}
The generating function for $D$ is then
$$ prod_{i in I} left( frac{x^{b_i m_i}}{1-x^{b_i m_i}} prod_{j = 2}^{m_i} frac{1}{1-x^{b_i(m_i - j +1)}} right) text{.} $$
Examples:
begin{align*}
mathrm{gf}(D(n|{1,2})) &= x^3 + x^4 + 2 x^5 + 2 x^6 + 3 x^7 + 3 x^8 + 4 x^9 + 4 x^{10} + cdots text{,} \
mathrm{gf}(D(n|{1,2,2})) &= x^5 + x^6 + 2 x^7 + 2 x^8 + 4 x^9 + 4 x^{10} + 6 x^{11} + 6 x^{12} + 9 x^{13} \
&qquad + 9 x^{14} + 12 x^{15} + 12 x^{16} + 16 x^{17} + 16 x^{18} + 20 x^{19} + 20 x^{20} + cdots text{, and} \
mathrm{gf}(D(n|{1,1,1})) &= x^3 + x^4 + 2 x^5 + 3 x^6 + 4 x^7 + 5 x^8 + 7 x^9 + 8 x^{10} + cdots text{.}
end{align*}
Continuing:
We can simplify the gf a bit. begin{align*}
prod_{i in I} left( frac{x^{b_i m_i}}{1-x^{b_i m_i}} prod_{j = 2}^{m_i} frac{1}{1-x^{b_i(m_i - j +1)}} right)
&= prod_{i in I} frac{x^{b_i m_i}}{1-x^{b_i m_i}} frac{1}{prod_{j = 2}^{m_i}1-x^{b_i(m_i - j +1)}} \
&= prod_{i in I} frac{x^{b_i m_i}}{prod_{j = 1}^{m_i}1-x^{b_i(m_i - j +1)}} \
&= prod_{i in I} prod_{j = 1}^{m_i} frac{x^{b_i}}{1-x^{b_i(m_i - j +1)}} text{.}
end{align*}
$endgroup$
$begingroup$
@d.y : I'd rather say $B = {(1,ell)}$ so that $B$ is a set of ordered pairs $(b_i, m_i)$ (with distinct $b_i$s). The gf for this $B$ is exactly the one given and is the same as the one you write. The given gf is a character-by-character translation of the last sum in the display above it.
$endgroup$
– Eric Towers
Aug 28 '17 at 14:41
$begingroup$
@d.y : I can't right now. Have to get back to work. I'll check in in several hours.
$endgroup$
– Eric Towers
Aug 28 '17 at 14:45
add a comment |
$begingroup$
Let's try again. (The original text is gone. If one would like to waste one's time, it is available through the "edited" link just below the examples.)
Instead of starting with a structureless multiset, let's go with: $I$ is the cardinality of the set of distinct elements in $A$, and $B = {(b_i, m_i), i in I}$ where the $b_i$ are the distinct elements of $A$ and $m_i$ is the multiplicity of $b_i$ in $A$. So, for example, $sum_{i in I} m_i = |A|$.
For each $i$ in $I$, we want a monotonically nondecreasing sequence $n_{i,1} leq n_{i,2} leq cdots leq n_{i,m_i}$. We make the change of variables begin{align*}
d_{i,1} &= n_{i,1} text{,} \
d_{i,2} &= n_{i,2} - n_{i,1} text{,} \
&vdots \
d_{i,j} &= n_{i,j} - n_{i,j-1} & (2 leq j leq m_i) text{,} \
&vdots \
d_{i,m_i} &= n_{i,m_i} - n_{i,m_i-1} text{.}
end{align*}
Then the monotonically nondecreasing condition on the $(n_{i,j})_j$ becomes $d_{i,1} geq 1$ and $d_{i,j} geq 0$ for $1 leq j leq m_i$. Observe that
begin{align*}
sum_{j=1}^{m_i} n_{i,j} &= (d_{i,1}) + (d_{i,1} + d_{i,2}) + cdots + (d_{i,1} + d_{i,2} + cdots + d_{i,m_i}) \
&= sum_{j=1}^{m_i} (m_i - j +1) d_{i,j}
end{align*}
Then $D(n|A)$ is the number of ways of choosing all these $d_{i,j}$ such that begin{align*}
n &= sum_{i in I} b_i sum_{j =1}^{m_i} n_{i,j} \
&= sum_{i in I} b_i sum_{j=1}^{m_i} (m_i - j +1) d_{i,j} \
&= sum_{i in I} left( b_i m_i d_{i,1} + sum_{j=2}^{m_i} b_i (m_i - j +1) d_{i,j} right) text{,} \
d_{i,1} &geq 1 qquad (i in I) & text{, and } \
d_{i,j} &geq 0 qquad (i in I, 2 leq j leq m_i) text{.}
end{align*}
The generating function for $D$ is then
$$ prod_{i in I} left( frac{x^{b_i m_i}}{1-x^{b_i m_i}} prod_{j = 2}^{m_i} frac{1}{1-x^{b_i(m_i - j +1)}} right) text{.} $$
Examples:
begin{align*}
mathrm{gf}(D(n|{1,2})) &= x^3 + x^4 + 2 x^5 + 2 x^6 + 3 x^7 + 3 x^8 + 4 x^9 + 4 x^{10} + cdots text{,} \
mathrm{gf}(D(n|{1,2,2})) &= x^5 + x^6 + 2 x^7 + 2 x^8 + 4 x^9 + 4 x^{10} + 6 x^{11} + 6 x^{12} + 9 x^{13} \
&qquad + 9 x^{14} + 12 x^{15} + 12 x^{16} + 16 x^{17} + 16 x^{18} + 20 x^{19} + 20 x^{20} + cdots text{, and} \
mathrm{gf}(D(n|{1,1,1})) &= x^3 + x^4 + 2 x^5 + 3 x^6 + 4 x^7 + 5 x^8 + 7 x^9 + 8 x^{10} + cdots text{.}
end{align*}
Continuing:
We can simplify the gf a bit. begin{align*}
prod_{i in I} left( frac{x^{b_i m_i}}{1-x^{b_i m_i}} prod_{j = 2}^{m_i} frac{1}{1-x^{b_i(m_i - j +1)}} right)
&= prod_{i in I} frac{x^{b_i m_i}}{1-x^{b_i m_i}} frac{1}{prod_{j = 2}^{m_i}1-x^{b_i(m_i - j +1)}} \
&= prod_{i in I} frac{x^{b_i m_i}}{prod_{j = 1}^{m_i}1-x^{b_i(m_i - j +1)}} \
&= prod_{i in I} prod_{j = 1}^{m_i} frac{x^{b_i}}{1-x^{b_i(m_i - j +1)}} text{.}
end{align*}
$endgroup$
Let's try again. (The original text is gone. If one would like to waste one's time, it is available through the "edited" link just below the examples.)
Instead of starting with a structureless multiset, let's go with: $I$ is the cardinality of the set of distinct elements in $A$, and $B = {(b_i, m_i), i in I}$ where the $b_i$ are the distinct elements of $A$ and $m_i$ is the multiplicity of $b_i$ in $A$. So, for example, $sum_{i in I} m_i = |A|$.
For each $i$ in $I$, we want a monotonically nondecreasing sequence $n_{i,1} leq n_{i,2} leq cdots leq n_{i,m_i}$. We make the change of variables begin{align*}
d_{i,1} &= n_{i,1} text{,} \
d_{i,2} &= n_{i,2} - n_{i,1} text{,} \
&vdots \
d_{i,j} &= n_{i,j} - n_{i,j-1} & (2 leq j leq m_i) text{,} \
&vdots \
d_{i,m_i} &= n_{i,m_i} - n_{i,m_i-1} text{.}
end{align*}
Then the monotonically nondecreasing condition on the $(n_{i,j})_j$ becomes $d_{i,1} geq 1$ and $d_{i,j} geq 0$ for $1 leq j leq m_i$. Observe that
begin{align*}
sum_{j=1}^{m_i} n_{i,j} &= (d_{i,1}) + (d_{i,1} + d_{i,2}) + cdots + (d_{i,1} + d_{i,2} + cdots + d_{i,m_i}) \
&= sum_{j=1}^{m_i} (m_i - j +1) d_{i,j}
end{align*}
Then $D(n|A)$ is the number of ways of choosing all these $d_{i,j}$ such that begin{align*}
n &= sum_{i in I} b_i sum_{j =1}^{m_i} n_{i,j} \
&= sum_{i in I} b_i sum_{j=1}^{m_i} (m_i - j +1) d_{i,j} \
&= sum_{i in I} left( b_i m_i d_{i,1} + sum_{j=2}^{m_i} b_i (m_i - j +1) d_{i,j} right) text{,} \
d_{i,1} &geq 1 qquad (i in I) & text{, and } \
d_{i,j} &geq 0 qquad (i in I, 2 leq j leq m_i) text{.}
end{align*}
The generating function for $D$ is then
$$ prod_{i in I} left( frac{x^{b_i m_i}}{1-x^{b_i m_i}} prod_{j = 2}^{m_i} frac{1}{1-x^{b_i(m_i - j +1)}} right) text{.} $$
Examples:
begin{align*}
mathrm{gf}(D(n|{1,2})) &= x^3 + x^4 + 2 x^5 + 2 x^6 + 3 x^7 + 3 x^8 + 4 x^9 + 4 x^{10} + cdots text{,} \
mathrm{gf}(D(n|{1,2,2})) &= x^5 + x^6 + 2 x^7 + 2 x^8 + 4 x^9 + 4 x^{10} + 6 x^{11} + 6 x^{12} + 9 x^{13} \
&qquad + 9 x^{14} + 12 x^{15} + 12 x^{16} + 16 x^{17} + 16 x^{18} + 20 x^{19} + 20 x^{20} + cdots text{, and} \
mathrm{gf}(D(n|{1,1,1})) &= x^3 + x^4 + 2 x^5 + 3 x^6 + 4 x^7 + 5 x^8 + 7 x^9 + 8 x^{10} + cdots text{.}
end{align*}
Continuing:
We can simplify the gf a bit. begin{align*}
prod_{i in I} left( frac{x^{b_i m_i}}{1-x^{b_i m_i}} prod_{j = 2}^{m_i} frac{1}{1-x^{b_i(m_i - j +1)}} right)
&= prod_{i in I} frac{x^{b_i m_i}}{1-x^{b_i m_i}} frac{1}{prod_{j = 2}^{m_i}1-x^{b_i(m_i - j +1)}} \
&= prod_{i in I} frac{x^{b_i m_i}}{prod_{j = 1}^{m_i}1-x^{b_i(m_i - j +1)}} \
&= prod_{i in I} prod_{j = 1}^{m_i} frac{x^{b_i}}{1-x^{b_i(m_i - j +1)}} text{.}
end{align*}
edited Aug 29 '17 at 4:05
answered Aug 27 '17 at 22:19
Eric TowersEric Towers
33.3k22370
33.3k22370
$begingroup$
@d.y : I'd rather say $B = {(1,ell)}$ so that $B$ is a set of ordered pairs $(b_i, m_i)$ (with distinct $b_i$s). The gf for this $B$ is exactly the one given and is the same as the one you write. The given gf is a character-by-character translation of the last sum in the display above it.
$endgroup$
– Eric Towers
Aug 28 '17 at 14:41
$begingroup$
@d.y : I can't right now. Have to get back to work. I'll check in in several hours.
$endgroup$
– Eric Towers
Aug 28 '17 at 14:45
add a comment |
$begingroup$
@d.y : I'd rather say $B = {(1,ell)}$ so that $B$ is a set of ordered pairs $(b_i, m_i)$ (with distinct $b_i$s). The gf for this $B$ is exactly the one given and is the same as the one you write. The given gf is a character-by-character translation of the last sum in the display above it.
$endgroup$
– Eric Towers
Aug 28 '17 at 14:41
$begingroup$
@d.y : I can't right now. Have to get back to work. I'll check in in several hours.
$endgroup$
– Eric Towers
Aug 28 '17 at 14:45
$begingroup$
@d.y : I'd rather say $B = {(1,ell)}$ so that $B$ is a set of ordered pairs $(b_i, m_i)$ (with distinct $b_i$s). The gf for this $B$ is exactly the one given and is the same as the one you write. The given gf is a character-by-character translation of the last sum in the display above it.
$endgroup$
– Eric Towers
Aug 28 '17 at 14:41
$begingroup$
@d.y : I'd rather say $B = {(1,ell)}$ so that $B$ is a set of ordered pairs $(b_i, m_i)$ (with distinct $b_i$s). The gf for this $B$ is exactly the one given and is the same as the one you write. The given gf is a character-by-character translation of the last sum in the display above it.
$endgroup$
– Eric Towers
Aug 28 '17 at 14:41
$begingroup$
@d.y : I can't right now. Have to get back to work. I'll check in in several hours.
$endgroup$
– Eric Towers
Aug 28 '17 at 14:45
$begingroup$
@d.y : I can't right now. Have to get back to work. I'll check in in several hours.
$endgroup$
– Eric Towers
Aug 28 '17 at 14:45
add a comment |
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$begingroup$
Have you tried ignoring the semimonotonicity condition while putting together your product of elementwise generating functions, then dividing out the overcounts at the end?
$endgroup$
– Eric Towers
Aug 27 '17 at 20:29
$begingroup$
Why wouldn't you divide out the coefficient you extract by $n_i!$ where the $n_i$ are the repetition counts in the multiset $A$, rather than trying to do this with the gf?
$endgroup$
– Eric Towers
Aug 27 '17 at 20:50