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Just another instance of the (second) Strong Law of Small Numbers: We have A157656(n) = A059100(n-1) for all known terms (i.e., $n\leq 6$), but it's also known that A157656(29) > A059100(28). So, the …

It is easier to describe non-good (bad) numbers with respect to a given prime $p$. For each such number $k$, there exists a prime $q$ such that $q\mid k$ and $q\mid (p^k - 1)$. It follows that $k$ is …

Since $j(n)\equiv 19\pmod{100}$, we have $\varphi(p_n+1-n)\equiv 18\pmod{100}$ meaning that $\nu_2(\varphi(p_n+1-n))=1$. That is, $p_n+1-n$ is $q^k$ or $2q^k$ for a prime $q\equiv 3\pmod4$. If $k=1$, …

Let one sequence be randomly sampled from $\{2,4\}$, and the other obtained similarly after setting the first element to 1. Then these sequences are not periodic and have pairwise distinct partial sum …

Prime factors below $10^{10}$ of $a_n$ can be found in OEIS A007996, and I've tested that none of them divides $a_n$ when squared. Same was reported by Andersen earlier for primes below $2^{32}$. In …

Denote the sum in question by $f(b,n)$, then $$\sum_{b,n\geq 0} f(b,n) y^b z^n = \frac{\log(1+y)\log(1-\frac{yz}{1-z})}{1-z-yz}.$$

I think there is no simple formula here, although we can get some recurrence relations and related identities for generating functions as explained below. Similarly to odious numbers, we have evil nu …

Below I will prove that the proposed test is necessary, that is, if $k\in\text{A106483}$ then $b(2k+1)=6k$. Following the simplification proposed by Will Sawin in the comments, the test for a given od …

The sequence OEIS A260695 can be seen as the interweaving of nonzero Hultman numbers $\mathcal{H}(n,k)$ for $k=1$ and $k=2$. Namely, $$a(n) = \mathcal{H}(n,1+(n\bmod 2)).$$ It is known that $\mathcal{ …

We have $$T(n,k) = [x^ny^{n-k}]\ \frac{2 - (1+y)x}{1-(1+y)x-x^2}=[y^{n-k}]\ L_n(1+y),$$ where $L_n$ is the $n$-th Lucas polynomial. For $k<n$, we have an explicit formula: \begin{split} T(n,k) &= \sum …

While no composite terms of A128465 are known, here is a proof that an odd prime $p$ belongs to A128465 if and only if $b(p)\equiv 2(-1)^{\tfrac{p+1}2}\pmod{p}$. First notice that for an odd prime $p$ …

We have $$\frac{1}{x}f(\frac1x) = \sum_{n\geq 1} (F_{2n-3}-1)x^n,$$ where $F_{2n+1}$ are Fibonacci numbers. Per answers to your previous question, it follows that $$c_{n+1} = \sum_{i=0}^n s(n,i) (F_{2 …

As it was noted in another answer, we have $$f(n,m,i) =[\tfrac{x^n}{n!}\tfrac{y^{m+1}}{(m+1)!}]\ \frac{\big(-\log(1+e^x(e^{-y}-1))\big)^i}{i!}.$$ The linked answer essentially establishes the same rec …

Using the notation from my answer to the previous question, we have $$D(n+1)−(D(n+2)+1)\not\equiv 0\pmod{p}$$ if and only if $g_0(n+2) = 2 g_0(n+1) - p$ and $g_1(n+2) \equiv 2 g_1(n+1) + 1\pmod{p}$, i …

Notice that $P(n-1)$ counts the number of partition of $n$ that contain $1$, while $P(n-2)$ counts the number of partition of $n$ that contain $2$. It follows that $P(n-1)+P(n-2)-P(n)$ equals the di …