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Summary

Is there a function $Z(a,b)$ or how would one find such a function so that for $a,b\in \mathbb N$, it would produce $0$'s on for each $a$th step for each $b$th value?

For example:
$a=2$, and for $b=1,2,3,4,5...$ the $Z(a,b)$ is: $0,1,0,1,0,1,0,1,0,1....$
$a=3$, and for $b=1,2,3,4,5...$ the $Z(a,b)$ is: $0,1,1,0,1,1,0,1,1....$
$a=4$, and for $b=1,2,3,4,5...$ the $Z(a,b)$ is: $0,1,1,1,0,1,1,1,0...$

And so on.


Introduction

Consider a sequence: $(n-1)(n-2)(n-3)...(n-k)$ where $n,k \in \mathbb{N}$

First we set $k$ to define the length of the sequence; then calculate $n$ like:

1) If $(n\le k)$ then it is equal to $0$
2) If $(n\gt k)$ then it is equal to $\frac{(n-1)!}{(n-k-1)!}$

The sequence can also be written using the product notation as: $\prod ^k _{x=1} (n-x)$


Then we can consider a sequence: $(n-1)(n-3)(n-5)...(n-k)$
Where $k$ is $2m-1$, $m$ being number of the term in the sequence.

I have found the following identities for choosing $k$ :

1) If $(n\lt k)$ then is is equal to: $(n-1)!!\times[-(k-n)!!]\times \frac{1}{2}[1+(-1)^n]$
Using Double Factorial.

2) If $(n=k)$ then it is $0$

3) If $(n\gt k)$ then it is equal to: $\frac{(n-1)!!}{(n-k-2)!!}$

Also write the sequence with: $\prod ^m _{x=1} (n-2x+1)$

Note, the last term in the 1) case can be written like $\frac{1}{2}(1+e^{i \pi n})$ since $n\gt 0$


Now I've tried to generalize this for sequences: $(n-1)...(n-k)$
Where $k= am-a+1$ Where $m$ is the number of the term, and $a$ the "Step" between the terms.

In our first two sequences we have $a=1$ and $a=2$

Note, sequence can also be written like: $\prod ^m _{x=1} (n-ax-a+1)$

First we define the sequence by choosing $a,k \in \mathbb{N}$ then we can claculate values for $n \in \mathbb{N}$ following, using Multi-Factorial:

1) When $(n\lt k)$ then: $$(n-1)!^a\times [-(k-n)!^a] \times Z(a,n)$$

Where $Z(a,n)$ would be a function that will output $0$ for every $a$th value and $1$ otherwise based on value of $n$.

For example; $a=2$ then for: $n=1,2,3,4...$, Function will be: $Z(a,n)= 0,1,0,1...$
Or; $a=3$ for: $n=1,2,3,4,5,6...$, Function will be: $Z(a,n)= 0,1,1,0,1,1...$

2) When $(n= k)$ then it is 0.

3) When $(n\gt k)$ then: $$\frac{(n-1)!^a}{(n-k-a)!^a}$$


Question

Now, how would one find such a function as $Z(a,n)$ ?

In addition, if there is a better way to write some of these things that would be good to know.

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  • $\begingroup$ Relevant: mathworld.wolfram.com/RisingFactorial.html $\endgroup$ Apr 9 '16 at 12:05
  • 2
    $\begingroup$ It seems you have already defined the function $Z(a,n)$. Are you looking for ways to write $Z(a,n)$ as an expression involving other known functions? $\endgroup$
    – David K
    Apr 9 '16 at 12:30
  • $\begingroup$ Yes, like for example $\frac{1}{2}(1+e^{i \pi n})$ is just a "shortened" expression for $Z(2,n)$. How would a expression for the whole $Z(a,n)$ function look like is my main point. $\endgroup$
    – Vepir
    Apr 9 '16 at 12:32
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My comments from the other question apply here as well. But we actually can define $Z(a,n)$ even with such a mild class of functions as one including the ceiling function, addition, subtraction, and division.

$$Z(a,n) = \left\lceil \frac {n-1}a - \left\lceil \frac{n-1}{a} \right\rceil \right\rceil.$$

Actually, the example you gave in the comments extends as well, if you allow yourself interval-type $\sum$-schema, smooth functions, and the complex numbers. $$Z(a,n) = 1-\frac1a\sum_{k=1}^{a-1}\exp\left(2\pi i\frac{k(n-1)}{a}\right)$$ It's not as clear why this one works, but try it out :)

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    $\begingroup$ Eric probably knows this, but for anyone new to the complex number solution, every periodic sequence can be written using those same exponentials (the roots of unity) but with different coefficients. The technical reason is the linear independence of the sequences of the first n powers of the n nth roots of unity. $\endgroup$
    – Mark S.
    Apr 9 '16 at 13:19
  • $\begingroup$ @MarkS.: Eeek, I did know that, but I completely wasn't thinking about it that way; nice catch :D $\endgroup$ Apr 9 '16 at 13:29
  • $\begingroup$ I actually found a simple expression I belive, when considering $0^0=1$, we can write: $Z(n,a)= 1-0^{[(n-1)\mod a]}$ $\endgroup$
    – Vepir
    Apr 9 '16 at 14:41

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