All Questions
21
questions
1
vote
1
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49
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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
votes
1
answer
155
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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 ...
0
votes
0
answers
81
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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
1
vote
1
answer
96
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Representation of number as a sums and differences of natural numbers
Lets consider all the combinations of:
$$1+2+3+4=10,\ \ 1+2+3-4=2,\ \ 1+2-3+4=4,\ \ 1+2-3-4=-4, $$
$$1-2+3+4=6,\ \ 1-2+3-4=-2,\ \ 1-2-3+4=0,\ \ 1-2-3-4=-8,$$
$$-1+2+3+4=8,\ \ -1+2+3-4=0,\ \ -1+2-3+4=2,...
- 1,818
1
vote
3
answers
73
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Partitions without repetition
I want to know how many partitions without repetition 19 has. I know I should see the coefficient of $x^{19}$ in $$\prod_{k=1}^\infty(1-x^k),$$
but i'm having trouble finding it. Ay hint?
- 153
0
votes
1
answer
138
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Infinite product expression of partition function
I'm working on a problem (specifically, I'm using an exam paper without course notes to prepare for a course starting in September),
Define the partition function $P(q)$ and give its infinite product ...
- 2,281
2
votes
1
answer
112
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Integer partitions using generating functions
For each natural number $ n $ we consider the equation
$$x_{1}+2x_{2}+\dots+nx_{n}=n$$
Where $x_{1},\dots,x_{n}$ are nonnegative integers. Prove that this equation has the same number of solutions ...
user1009942
3
votes
1
answer
85
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Showing $\prod_{n\geq 1} (1+q^{2n}) = 1 + \sum_{n\geq 1} \frac{q^{n(n+1)}}{\prod_{i=1}^n (1-q^{2i})}$
I want to show
\begin{align}
\prod_{n\geq 1} (1+q^{2n}) = 1 + \sum_{n\geq 1} \frac{q^{n(n+1)}}{\prod_{i=1}^n (1-q^{2i})}
\end{align}
I know one proof via self-conjugation of partition functions with ...
- 6,490
1
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1
answer
62
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Deriving a partition identity from some given identities
I am trying questions from Apostol Introduction to ANT of Chapter partitions and need help in deducing this identity.
Problem is question 6(a) which will use some information from 2 and 5(b).
...
user775699
2
votes
1
answer
93
views
2 questions related to generating function of partition function in number theory
I am self studying chapter partitions (chapter number-14) from Apostol Introduction to analytic number theory.
I had studied that chapter earlier also and had questions but as I don't have anyone to ...
user775699
2
votes
1
answer
62
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Question about coefficients of generating functions
Theorem: Let $n> 0 \in \mathbb Z.$ Let $p_n$ stand for the number of integer partitions of $n$ and let $k$ be the number of consecutive integers in a partition. Then $p_n + \sum_{k \ge 1}(-1)^k(p_{...
- 23
2
votes
1
answer
379
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Partition function and "Euler function" - what does it mean?
Denote the partition function by $p_k(n)$, and define it as a count of the number of possible sequences of positive integers $a+b+c+...=n$ where the $a,b,c,...$ are not necessarily distinct (so that, ...
1
vote
1
answer
240
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Result on partitions with distinct odd parts
Let $pdo(n)$ be the number of partitions of n into distinct odd parts. Then $p(n)$ is odd if and only if $pdo(n)$ is odd.
I am well aware that a proof of this is available here but I want to do it ...
- 13
0
votes
0
answers
226
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Partitions into non-negative powers of $2$
Let $c(n)$ denote the number of partitions of $n$ into non-negative powers of $2.$ (Thus $c(5)=4$ since $5=4+1=2+2+1=2+1+1+1=1+1+1+1+1).$
(a). Prove that $1+\sum\limits_{n=1}^{\infty} c(n) q^n=\prod\...
user333938
1
vote
0
answers
130
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Generating functions and integer partitioning [duplicate]
Show that the number of partitions of a positive integer n where no
summand appears more than twice is equal to the number of partitions
of n where no summand is divisible by 3
So I begin by ...
- 6,294