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For questions about sequences of integers. References are often made to the online resource oeis.org.
10
votes
Accepted
Tower-of-squares sequence divides linear recurrent A001921 sequence?
The elements of your sequence are
$$a_n=\left(\frac{\alpha^n-\beta^n}{2\sqrt{3}}\right)\left(\frac{\alpha^{n+1}+\beta^{n+1}}{2}\right)$$
where $\alpha=2+\sqrt{3}$ and $\beta=2-\sqrt{3}$. Notice that b …
10
votes
Mod sequences that seem to become constant; and the number 316
Before this becomes another forgotten open problem on MO, let me record here a comment. An equivalent way to state the sequence is as follows: Let $x(1)=2s-1$ and look at the recurrence equation:
$$x( …
4
votes
Accepted
Curious sequences of polynomials
Recurrences like these often times have conserved quantities. For your particular case the quantity
$$A_n=\frac{s_{n+1}+qs_n+qs_{n-k+1}+s_{n-k}}{s_ns_{n-1}\cdots s_{n-k+1}}$$ remains constant for any …
23
votes
Accepted
Zeroes of the random Fibonacci sequence
In response to Mark's comment, it is possible to determine that the probability of $X_n=0$ decays exponentially directly, and this is in fact easier than the theorems about the growth of these random …
19
votes
A possibly surprising appearance of $\sqrt{2}.$
Let's define two auxiliary sequences $c_n=a_{n+2}-a_{n+1}-2$ and $d_n=b_{n+2}-b_{n+1}$ for $n\geq 1$. One can prove with an induction argument that the sequence $c_n$ takes values in $\{0,1,2,3\}$ and …
23
votes
Accepted
Up to $10^6$: $\sigma(8n+1) \mod 4 = OEIS A001935(n) \mod 4$ (Number of partitions with no e...
Let's call A001936(n) by $a(n)$. Here is a sketch of why $$a(n)\equiv \sigma(4n+1)\pmod{4}$$
Firs note that the generating function of $a(n)$ is
$$A(x)=\sum_{n\geq 0}a(k)x^n=\prod_{k\geq 1}\left(\fra …
10
votes
Accepted
Maximal number of edges and triangular cells for n points in a triangular lattice
The following was conjectured by D. Reutter in problem 664A, Elemente der mathematik 27 and proved by H. Harborth in Solution to problem 664A, Elemente der mathematik 29, 14-15
The maximum number …