# Tag Info

1

A skew diagram is not a partition, so you cannot write it as a tuple. I would just write "$\lambda/\mu$ where $\lambda = 4444$ and $\mu = 421$." You need to keep both $\lambda$ and $\mu$ around to describe $\lambda/\mu$. Of course, if you're interested in some special situation you could make up your own notation.

1

$$\ln\frac{(k+1)(k+2)}{k(k+3)}=\ln\frac{k^2+3k+2}{k^2+3k}=\ln\left(1+\frac2{k^2+3k}\right)\lt\frac2{k^2+3k}\;.$$ Thus \begin{align} \sum_{k=0}^\infty\left(-\ln(j+4k)+\ln(j+4k+1)+\ln(j+4k+2)-\ln(j+4k+3)\right) &\lt\sum_{k=0}^\infty\frac2{(j+4k)^2+3(j+4k)} \\ &\le\sum_{k=0}^\infty\frac2{(1+4k)^2+3(1+4k)} \\ &= \frac\pi{12}+\frac{\ln2}2 \\ ...

1

It appears you need to compute the limit of the lower Riemann sum using some analytical tools -- as opposed to declaring simply that it converges to the integral. The following identity is useful $$\sum_{j=1}^n \sin (jx) = \frac{\sin\left( \frac{nx}{2}\right)\sin\left( \frac{(n+1)x}{2}\right)}{\sin\left( \frac{x}{2}\right)},$$ and is proved easily by ...

2

This is the stars and bars problem. Imagine a line of $64$ stars and $63$ candidate bars between them. Pick $15$ of those bars to be real and read off the groupings. There are ${64 \choose 15}=122131734269895$ To make the list, there are many algorithms on the web to generate the combinations of $15$ choices out of $63$. For example, you can use the fact ...

0

You can use a recursive algorithm to iterate through the solutions, but there is an easy way to count them. The amount of non negative solutions to $$x_1 + x_2 + \cdots + x_n = a$$ is $a + n - 1 \choose n - 1$ or equivalently $a + n - 1 \choose a$. Since we want integer solutions, let $y_k = x_k + 1$. Each $y_k$ is now an ...

0

function partitionCombination(N,m,p) { if (N == 0) print(p); else { L = p.length; for (i = m; i >= 1; i--) { p[L] = i; partitionCombination(N-i,i,p); } p.length = L; } } function partitionOrderCounts(N,p) { if (N == 0) print(p); else { L = p.length; for (i = N; i >= 1; i--) { ...

2

This is the subset-sum problem -- it is NP-Complete, hence although there are certainly algorithms for it, none are necessarily "fast" in general (possibly for special cases of the problem though). The most common approach is dynamic programming. EDIT: while Wikipedia is unnecessarily general, as usual, the subset sum problem usually refers to OP's problem ...

1

lets make the partition equally spaced. $x_i = \frac in$ if you really wanted to you could say $x_i = \frac {(b-a)i}n+a$ but since b = 1 and a = 0, it really isn't necessary. $\int_0^1 x \,dx = \lim_\limits{n\to\infty} \frac 1n\sum_\limits{i=1}^n x_i = \lim_\limits{n\to\infty} \frac {\frac12 (n)(n+1)}{n^2}$ Now, it is usually sufficient to say at this ...

1

The left-sided Riemann sum $S$ for $x$ on $[0,1]$ is expressed as $$S=\lim_{N\to \infty}\sum_{n=0}^{N-1} x_n(x_{n+1}-x_n) \tag 1$$ where $x_0=0$ and $x_N=1$. Using Summation by Parts in $(1)$, we find $$S=\lim_{N\to \infty}\left(x_N^2-x_0^2-\sum_{n=0}^{N-1}x_{n+1}(x_{n+1}-x_n)\right) \tag 2$$ It is straightforward to show that the sum on the right-hand ...

0

One of the references I can recommend is the book of Tom Apostol (1990) on Modular functions and Dirichlet series in number theory. In chapter $5$ the proof is explained in all detail, together with a "plan of the proof", which is very helpful. On $17$ pages we can follow an amazing proof, illustrated with several highly interesting drawings (related to ...

0

Just search for "rademacher partition formula". A discussion with a number of references is here: https://en.wikipedia.org/wiki/Partition_(number_theory)#Partition_function_formulas

3

You have missed seven ways. Mainly, $$2+2+2=6$$$$1+1+2+2=6$$$$1+2+1+2=6$$$$1+2+2+1=6$$$$2+1+2+1=6$$$$2+2+1+1=6$$$$2+1+1+2=6$$So there are $32$ ways. However, one thing worth mentioning is that your definition strays from the normal definition of partitions, where two sums that differ only in the order of their summands are considered the same partition. In ...

1

Let's recall the standard generating function case, where $f(i,r)=x^{ir}$ for a parameter $x$: in that case, $\prod_{j=1}^l f(i_{j_k},r_{j_k}) = x^{\sum_{j=1}^l i_{j_k}r_{j_k}} = x^N$, and so $$\sum_{N=1}^\infty\sum_{k=1}^{p(N)}\prod_{j=1}^lf(i_{j_k},r_{j_k}) = \sum_{N=1}^\infty\sum_{k=1}^{p(N)} x^N = \sum_{N=1}^\infty p(N) x^N,$$ a generating function ...

0

If $x\sim y$, then by symmetry $y\sim x$ so by transitivity $x\sim x$. Thus an element related to any other element is also related to itself. The set therefore decomposes into two subsets; on one the partial equivalence relation is empty and on the other it is an equivalence relation. Thus the partial equivalence relations on $[n]$ are in bijection with ...

0

If you are interested in partitions, then I would suggest you to study the concept further by studying about Ramanujan. S Ramanujan researched heavily in the topic which would help you. Though, there is no such direct way but still you would get a lot of insight. Hope it helps.

1

As the OP said in his comment the answer is the following list of partitions of $S = \{1,2,3\}$: $\left\{ \{1,2,3\} \right\}$ (one set, just $S$ itself ) $\left\{ \{1,2\}, \{3\} \right\}$ (two sets, one of two elements , one singleton) $\left\{ \{1,3\}, \{2\} \right\}$ $\left\{ \{2,3\}, \{1\} \right\}$ $\left\{ \{1\}, \{2\}, \{3\} \right\}$ (all ...

1

Let $P(n,k)$ be the number of partitions of $n$ into exactly $k$ parts. This function satisfies the recurrence $$P(n,k)=P(n-1,k-1)+P(n-k,k)\;:$$ the first term on the right is the number of $k$-partitions of $n$ that have at least one part of size $1$, and the second is the number that have no parts of size $1$. (This is most easily seen by considering ...

0

Yes. The same idea can be applied: Choose $i$ of $n$ elements and partition the remaining $n-i$ elements into $k-1$ subsets, yielding $$\binom niS(n-i,k-1)\;.$$

2

You can do it inductively in the "naive" way: Set $A_1=\{1,12\}$, $A_2=\{2,13^2-2\}$,$\ldots$, $A_{11}=\{11,13^{11}-11\}$, $A_{12}=\{13,13^{13}-13\}$, $A_{13}=\{14,13^{14}-13\}$,... You have, however, to show that this sequence satisfies the properties you want. I think one elementary way to do it is the following: We want to define $A_i$ inductively with ...

4

Note that any $k_j = 1$ is wasted: better to combine it with some other $k_i$, since $1 \cdot k_i < k_i + 1$. Any value $4$ or higher, we might as well split off a $2$: $x < 2 (x-2)$ if $x \ge 4$. So we're left with $2$'s and $3$'s. (Actually a $4$ is interchangeable with two $2$'s, but let's think in terms of $2$'s and $3$'s) $2^3 < 3^2$, so ...

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