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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 ...


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 ...


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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 ...


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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.


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$$ \ln\frac{(k+1)(k+2)}{k(k+3)}=\ln\frac{k^2+3k+2}{k^2+3k}=\ln\left(1+\frac2{k^2+3k}\right)\lt\frac2{k^2+3k}\;. $$ 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 \\ ...


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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 ...


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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 ...


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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 ...


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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 ...


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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 ...



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