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If we fix $z$, we have $x + 2y = N-Kz$. If $N-Kz > 1$, we can always choose the parity of $x+y$ by choosing between $y=0$ and $y=1$. If $N-Kz \le 1$ we are forced to have $x = N-Kz$, $y=0$. Therefore …

$f_{\lambda,n}(t)$ is the generating function for the set of numbers $\lambda_i + n - i$ with $1 \leq i \leq n$. $t^{m+n-1}f_{\lambda',n}(t^{-1})$ is the generating function for the set of numbers $n …

How about induction over $|\lambda|$? Clearly it holds for the unique partition of $1$, whose cell has hook $1$ and content $0$. Suppose it holds for a partition $\lambda$ and consider the partition …

The generating function for integer partitions (see section "generating function") can be defined as: $$\sum_{n=0}^{\infty}P(n)\,x^n := \prod_{k=1}^{\infty}\left(\frac{1}{1-x^k}\right)=(1+x+x^2 …

The Representation Theory of the Symmetric Group, James and Kerber, doi:10.1017/CBO9781107340732, section 2.7. A $q$-hook is defined at the start of the section as a hook of length $q$. The following …

Call the zero-padded vectors $u$ and $v$, so in your example $u = (1,2,3)$ and $v = (1,5,7)$. Observe that $\max_i(u_i) + \min_j(v_j)$ is an upper bound on your max-min: in every permutation, $\min_j …

There's an easy bijection between the partitions of $n$ into $k$ parts of at least $b$ and the partitions of $n - k(b-1)$ into $k$ parts of at least $1$, i.e. partitions of $n - k(b-1)$ into $k$ parts …

The concept here, although I'm not sure how far I can formalise it, is to extend the diagonal as far as the bottom of the Ferrers diagram. Then consider only the lower triangle. So your example $$\be …

I think the bit you're missing is $$\frac{1}{1 - z} = 1 + z + z^2 + \cdots$$ Therefore $$\frac{1}{(q^7;q^7)_\infty} = \prod_{k=1}^\infty \sum_{j=0}^\infty q^{7jk}$$ which clearly only has non-zero coe …

There's no known easy way (for a technical meaning of "easy") and there's a million dollar prize for finding one. This is a variant on the subset sum problem, which is NP complete. If you don't know …

The definition of algorithm is not completely precise, so there is sometimes ambiguity as to whether two computer programs (or other processes) are implementing the same algorithm or related but disti …

Your $sf_2(n)$ is OEIS A071068, Number of ways to write n as a sum of two unordered squarefree numbers. The comments note that (with notation marked up slightly and $a(n)$ replaced with $sf_2(n)$ for …

It's natural to work with the case split. The sums $\sum_{i=1}^{N} x_{i} (i - 1)$ with an odd number of $x_i$s are the sums $0x_1 + \sum_{i=2}^{N} x_{i} (i - 1)$: since we can adjust $x_1$ to fix the …

Since the primary technique applied in the book is manipulation of generating functions, that seems likely to be the intended approach here too. I'm sure there's a more elegant proof than this; if you …

Let's define some $q$-series notation: $(a;q)_n = \prod_{i=0}^{n-1} (1 - aq^i)$ with shorthand notation $(q)_n = (q;q)_n$. E.g. the left-hand side of Euler's pentagonal theorem as stated in the questi …