All Questions
Tagged with integer-partitions generating-functions
210
questions
0
votes
1
answer
83
views
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 ...
- 47.4k
1
vote
1
answer
49
views
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 ...
- 47.4k
2
votes
1
answer
155
views
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 ...
- 15.8k
2
votes
2
answers
434
views
Gen func "The number of partitions of n where each part occurs 2, 3, 5 times = number of partitions of n..."
The number of partitions of n where each part occurs 2, 3, 5 times = number of partitions of n with parts modulo 2,3,6,9,10 modulo 12
This is from Subbarao 1971 but I don't quite understand the ...
- 4,175
2
votes
1
answer
38
views
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, ...
- 78.3k
0
votes
0
answers
81
views
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)\...
- 47.4k
1
vote
0
answers
34
views
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
0
votes
0
answers
43
views
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 ...
- 21.9k
1
vote
3
answers
73
views
Partitions without repetition
I want to know how many partitions without repetition 19 has. I know I should see the coefficient of $x^{19}$ in $$\prod_{k=1}^\infty(1-x^k),$$
but i'm having trouble finding it. Ay hint?
- 41.3k
1
vote
1
answer
72
views
Proof that two generating function are equals for the sequence which $n$-th number is:
I am not sure I am doing this exercise good
1) $p_n $ | all parts are pairs different
and
2) $p_n $| all parts are not higher than $m$
I found these functions in book, first is:
$$ \prod_{i=1}^\...
- 31
4
votes
1
answer
339
views
Why do Bell Polynomial coefficients show up here?
The multinomial theorem allows us to expand expressions of the form ${\left( {{x_1} + {x_2} + {x_3} + {x_4} + ...} \right)^n}$. I am interested in the coefficients when expanding ${\left( {\sum\...
- 109
0
votes
1
answer
81
views
Integer partitions with summands bounded in size and number
This book says it's easy, but to me, it's not. :(
As for 'at most k summands', in terms of Combinatorics, by using MSET(),
$$ MSET_{\le k}(Positive Integer) = P^{1,2,3,...k}(z) = \prod_{m=1}^{k} \frac{...
- 3,748
1
vote
0
answers
60
views
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,
$$\...
- 51
1
vote
0
answers
47
views
Congrunces of partitions into distinct parts
Let $P_{d}(n)$ denote the number of partitions of n into distinct parts. The generating function of $P_{d}(n)$ s given by:
$$ \sum_{n \geq 0}P_{d}(n)q^{n}= \prod_{n \geq 0} (1+q^{n}).$$
Now let $P_{2,...
- 21
1
vote
1
answer
319
views
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 ...
- 109k