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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 …

To be unambiguous about the order of multiplication, let $B(n) = A_1 A_2 \cdots A_n$. We have the D-finite recurrences $B(n)_{r,1} = (\frac{n}{n-1})^k B(n-1)_{r,1} + n^k B(n-2)_{r,1}$ $B(n)_{r,2} = B …

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 …

Following user523984's suggestion in the comments: From triangle = repdigit $$\frac{k(k+1)}2 = \frac{d(10^j-1)}9$$ we get $$k = \frac{-1 \pm \frac13 \sqrt{9 + 8d(10^j-1)}}{2}$$ so we require $9 + 8d(1 …

$Q_n = P_n$ iff there are no carries when multiplying $p_{1}^{m_{1}}\cdot p_{2}^{m_{2}}\cdots p_{k}^{m_{k}}$ in binary. Consider $P_a P_b$: if there are no carries in the binary multiplication of $ab$ …

It turns out to be more straightforward than I expected. Let $C(z) = \sum_{i \ge 0} c_i z^i$ be the g.f. for $c_i$, excluding $c_{-1}$ since that doesn't show up in your recurrence. Starting with $$\f …

Suppose this is known for all Egyptian fractions with minimal representation in $k$ fractions. Then if there's a counterexample in $k+1$ fractions $\frac{1}{n_1} + \frac{1}{n_2} + \cdots \frac{1}{n_{k …

The $n_i$ are any sequence of distinct integers, indexed by $i$. The point is that if $R = r^{kl}$ then $R$ is a power of $p$, so any sum of distinct powers of $R$ is a 0-1 number in base $p$; and sym …

I'm going to use $\operatorname{msb}$ (for most significant bit) as an alias of $f$. Since $q_i$ is a permutation, the property that $q_i(n)<2^k$ iff $n < 2^k$ is equivalent to $\operatorname{msb}(q_i …

$\ell(0)$ is problematic, so I will assume that you actually mean to restrict to positive integers rather than natural numbers. We can rephrase the construction of $a(n)$ to emphasise the increment: l …

QUESTION. Is this true? Or, can you provide a reference to it. $$\mathbf{F}_a(q)=\frac1{(2a-1)!}\mathbf{G}(\mathbf{G}^2-1^2)(\mathbf{G}^2-2^2)(\mathbf{G}^2-3^2)\cdots(\mathbf{G}^2-(a-1)^2);$$ where w …

We assume $F(0) \neq 0$, since otherwise we don't satisfy the assumptions for the series reversion. Let $G = G(0)$ be the fixpoint of the recurrence given: $$G(x) = F\left(\frac{x}{G(x)}\right)$$ Mult …

The review Bounds on Factors of $\mathbb{Z}[x]$ by John Abbott gives various such bounds and shows that none of them is strictly better than the others.