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

5 votes

Is there any number other than 109 whose reciprocal contains the Fibonacci sequence?

Consider the related decimal $x=\sum_{n=0}^{\infty}\frac{f_n}{10^n}=0.11235955056179775280898876404\cdots.$ Since $$f_n=\frac{\tau^n+(\frac{-1}{\tau})^n}{\sqrt{5}}$$ for $\tau=\frac{1+\sqrt{5}}2,$ W …
Aaron Meyerowitz's user avatar
2 votes

Can these sequences stay integer-valued as many times as we want and then fail?

A better question might be for which $m$ is there a sequence which fails at $c_{m+1}=\frac{c_m(c_m+m+d)}{m}.$ $2,3,12$ are possible but I don't think any other numbers $2^i3^j$ are. All primes up to …
Aaron Meyerowitz's user avatar
1 vote

Integer Recursion Reference Request

I don't have a complete solution but I do have some conjectures and insights which might be useful if someone wants to look into this more deeply. I will assuming here that the $c_i > 0.$ Also, I w …
Aaron Meyerowitz's user avatar
1 vote

Number of squares in a grid under certain conditions

It is not clear to me what you mean by intersecting the sides of unit squares but based on your comment I think it is just what Gerry says: Your grid makes up $n^2$ unit squares and $E(n)$ counts the …
Aaron Meyerowitz's user avatar
6 votes
Accepted

Is the sequence $a_n=c a_{n-1} - a_{n-2}$ always composite for $n > 5$?

Here is another approach to show that $a_n$ is not prime when $c \gt 2$ and $n \gt 2$ We have (proof at end) $$a_{n+m}=a_na_{m+1}-a_{n-1}a_{m} \tag{*}$$ So, by induction on $j \ge 1$, $$a_{n+jn}=a_{ …
Aaron Meyerowitz's user avatar
35 votes

Longest coinciding pair of integer sequences known

The number of divisions of $\mathbb{R}^3$ by $k \ge 0$ planes in general position starts 1,2,4,8, then 15, etc. For $\mathbb{R}^6$ it is 1,2,4,8,16,32,64 then 127. In general for $\mathbb{R}^N$ it is …
Aaron Meyerowitz's user avatar
4 votes

Can an integer or rational sequence satisfy some bounded order recurrence $\mod \ $ almost a...

YES, define $a_k$ by $a_0=0$, $a_k=a_{-k}$ for $k<0$ and for $k>0$ $a_k=m!$ where $m\ge 1$ is as large as possible subject to $k$ being a multiple of $m!$. Then $$a_{k+m!} \equiv a_k \mod m.$$ However …
Aaron Meyerowitz's user avatar