All Questions
Tagged with integer-partitions generating-functions
211
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Generating function and currency
We assume that we have a country's currency that contains three coins worth 1, 3, and 4. How many ways can we get an amount of $n$ using these three pieces?
In others words what is the number of ...
- 37
-1
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48
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proving $2$ generating functions are equal [duplicate]
I have been doing a problem on generating functions and I need to prove that these are equal:
$\displaystyle\prod_{i=1}^\infty\frac{1+x^{2i-1}}{1-x^{2i}}$
and
$\displaystyle\prod_{i=1}^\infty\frac{1-x^...
- 27
1
vote
1
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46
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Generating function of partitions of $n$ in $k$ prime parts.
I have been looking for the function that generates the partitions of $n$ into $k$ parts of prime numbers (let's call it $Pi_k(n)$). For example: $Pi_3(9)=2$, since $9=5+2+2$ and $9=3+3+3$.
I know ...
2
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1
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154
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About the product $\prod_{k=1}^n (1-x^k)$
In this question asked by S. Huntsman, he asks about an expression for the product:
$$\prod_{k=1}^n (1-x^k)$$
Where the first answer made by Mariano Suárez-Álvarez states that given the Pentagonal ...
2
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1
answer
37
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Prove $\sum_{j=0}^{n} q^{j^{2}}\binom{n}{j}_{q^{2}}$ generates the self-conjugate partitions with part at most $n$.
Prove $\sum_{j=0}^{n} q^{j^{2}}\binom{n}{j}_{q^{2}}$ generates the self-conjugate partitions with part at most $n$, and that it equals $(1+q)(1+q^{3})\cdot\cdot\cdot(1+q^{2n-1})$.
For the first part, ...
- 795
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Need help with part of a proof that $p(5n+4)\equiv 0$ mod $5$
Some definitions:
$p(n)$ denotes the number of partitions of $n$.
Let $f(q)$ and $h(q)$ be polynomials in $q$, so $f(q)=\sum_0^\infty a_n q^n$ and $h(q)=\sum_0^\infty b_n q^n$. Then, we say that $f(q)\...
- 350
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33
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Proving an Identity on Partitions with Durfee Squares Using $q$-Binomial Coefficients and Generating Functions
Using the Durfee square, prove that
$$
\sum_{j=0}^n\left[\begin{array}{l}
n \\
j
\end{array}\right] \frac{t^j q^{j^2}}{(1-t q) \cdots\left(1-t q^j\right)}=\prod_{i=1}^n \frac{1}{1-t q^i} .
$$
My ...
- 195
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43
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Find generating function for the number of partitions which are not divisible by $3$. [duplicate]
I'm trying to find the generating function for the number of partitions into parts, which are not divisible by $3,$ weighted by the sum of the parts. My idea is that we get the following generating ...
- 121
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1
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80
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Integer partitions with summands bounded in size and number
This book says it's easy, but to me, it's not. :(
As for 'at most k summands', in terms of Combinatorics, by using MSET(),
$$ MSET_{\le k}(Positive Integer) = P^{1,2,3,...k}(z) = \prod_{m=1}^{k} \frac{...
- 185
1
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0
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60
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Generating Function for Modified Multinomial Coefficients
The multinomial coefficients can be used to expand expressions of the form ${\left( {{x_1} + {x_2} + {x_3} + ...} \right)^n}$ in the basis of monomial symmetric polynomials (MSP). For example,
$$\...
- 51
4
votes
1
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339
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Why do Bell Polynomial coefficients show up here?
The multinomial theorem allows us to expand expressions of the form ${\left( {{x_1} + {x_2} + {x_3} + {x_4} + ...} \right)^n}$. I am interested in the coefficients when expanding ${\left( {\sum\...
- 51
1
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47
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Congrunces of partitions into distinct parts
Let $P_{d}(n)$ denote the number of partitions of n into distinct parts. The generating function of $P_{d}(n)$ s given by:
$$ \sum_{n \geq 0}P_{d}(n)q^{n}= \prod_{n \geq 0} (1+q^{n}).$$
Now let $P_{2,...
- 21
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1
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73
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Generating function of number of partitions of $n$ into all distinct parts
I am trying to grasp this example from the book A Walk Through Combinatorics:
Show that $\sum_{n \ge 0} p_d(n)x^n = \prod_{i \ge 1}(1+x^i)$ where
$p_d$ stands for partitions of $n$ into all distinct ...
- 309
1
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1
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301
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Generating function of number of partitions of $n$ into parts at most $k$
I am trying to grasp the intuition behind this example.
Show that $\sum_{n \geq 0} p_{\leq k}(n)x^n = \prod_{i=1}^k \frac{1}{1-x^i}$
where $p_{\leq k}(n)$ denotes the number of partitions of the ...
- 309
3
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1
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110
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Finding the number of integer composition using only a specific pair of numbers [closed]
I want to find the number of integer compositions of 19 using numbers 2 and 3.
If I wanted to find the number of integer compositions of 19 using 1 and 2, I could write it as $F(n)=F(n-1)+F(n-2)$ ...
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