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Q1 is too vague, so let me answer Q2 and Q3. Assume that a prime $p$ and a primitive root $g$ modulo $p$ are fixed. Let $g_0(n) := g^n\bmod p$, $g_1(n) := \frac{g^n - g_0(n)}{p}\bmod p$, and $c := g_1 …

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 …

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 …

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 …

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 $k=1$, …

This is just to show that in sextuples of interest $2x^2 - 3y^2 = -1$, while difference $2$ is not possible. First note that neither $|2x^2-2y^2|<6$ nor $|3x^2-3y^2|<6$ is soluble in distinct positive …

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

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 $p$ …

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 …

I've extended OEIS A335976 with many terms. The numerical data so far is in favor of the conjecture, although I think it may be hard to prove it rigorously. Still, we can note a few major factors that …

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 …

This is just an extended comment. There is no need to invoke the algebraic dependency search. The recurrence can be found by directly constructing Groebner basis $B$ under any term order, in which $n$ …

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{ …