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For questions about sequences of integers. References are often made to the online resource oeis.org.

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 …
Gjergji Zaimi's user avatar
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 …
Gjergji Zaimi's user avatar
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 …
Gjergji Zaimi's user avatar
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 …
Gjergji Zaimi's user avatar
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( …
Gjergji Zaimi's user avatar
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 …
Gjergji Zaimi's user avatar
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 …
Gjergji Zaimi's user avatar