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For $n > 0$ let $\ell(n) = \lfloor \log_2(n) \rfloor$ so that $n = 2^{\ell(n)} + r_n$ where $0 \le r_n < 2^{\ell(n)}$. Note that $\ell(n) = T(n, 1) - 1$. Then by partitioning the numbers $k$ up to $n$ …

There is some $N(k, p)$ such that $n \ge N(k, p) \implies s(n, k) \equiv 0 \pmod p$. Proof is straightforward by fixing $p$ and using induction on $k$ via the recurrence $$s(n, k) = s(n-1, k-1) - (n-1 …

Yes. Take the complete graph on $n+1$ vertices and delete one edge.

$P_n$ is given as $$P_n(f) = \sum_{\lambda \,\vdash\, n} (-1)^{\lambda_1} \prod \binom{\lambda_j}{\lambda_{j+1}} f_{j}^{\lambda_j}$$ where the sum is over partitions $\lambda = \lambda_1 \ge \lambda_ …

In a comment, Brendan McKay suggested searching for quintuples which are one-free and inverse-free. There are no such quintuples, but there are quadruples, and by repeating elements from one such I pr …

$$\begin{eqnarray*} \textrm{LHS} &=& \exp\left(\frac{a^2+b^2}2\right) \left(\frac{\partial^2}{\partial a^2} + \frac{\partial^2}{\partial b^2}\right)^m \exp\left(-\frac{a^2+b^2}2\right) \\ &=& \exp\lef …

Generalise to $$b(2^m(2k+1)) = \sum\limits_{j=0}^{m}C_{m+1,j} \, b(2^jk), \\ b(0) = 1$$ Consider the infinite matrices: $$M_0 = \begin{pmatrix} 0 & 1 & 0 & 0 & \cdots \\ 0 & 0 & 1 & 0 & \cdots \\ 0 & …

Not a complete answer, but too long for a comment and addressing the conjecture which I take to be the most important part of the question. The double-loop transformation process seems familiar to me …

It's not entirely clear to me how much data your guesses are based on, so I present a table with calculated data and guessed polynomials based on that data and the assumption that $f(1) = f(-1) = 1$. …

In GraphViz notation, digraph { rankdir="BT"; edge [arrowhead=none]; 0 -> a; 0 -> b; 0 -> c; a -> ab; a -> ac; a -> d; b -> ab; b -> bc; b -> d; c -> ac; c -> bc; c -> d; a …

If we take $n=4,j=4$ then $k=1$ is possible (e.g. permutation $2431$). Looking at permutations with $j−1$ inversions we get $S_\beta=\{1432,2341,2413,3142,3214,4123\}$ and $S_\delta=\{2341,2413,3142,4 …

Using a similar approach to LeechLattice's extended comment, consider $A = G \setminus \{0, 1, 3\}$, $B_1 = G \setminus \{0, 1, 2\}$, $B_5 = G \setminus \{0, 1, 6\}$. Over general (i.e. not necessaril …

The optimal value for $k$ is $n$. An example construction for $k = n$ is $W = [x_0, x_1, \ldots, x_{n-1}, x_0 + 1, x_1 + 1, \ldots, x_{n-1} + 1]$. Then the products which must be non-zero are $\prod_i …

$\det \begin{pmatrix} 1 \end{pmatrix} = 1$ works for any $p$. $\det \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix} = -1$ similarly. For $n=3$ we require $p \ge 5$. By exhaustion there's no solution for …

Counterexample: consider $\ell = 3$, $m = 1$. The LHS is $$\frac{B_{7,3}({\color{red} 1}, {\color{green} 0}, {\color{blue} 1}, 0, 9)} {B_{5,3}({\color{red} 1}, {\color{green} 0}, {\color{blue} 1})} = …