# General function that represents a series of 1 and 0

1. Nov 4, 2011

### Emilijo

How can I get a general function that represents a series of 1 and 0, for example :
1,0,1,0,1,0,1,0...; but also
1,1,0,1,1,0,1,1,0...;
1,1,1,0,1,1,1,0,1... and so on; the first one is not difficult: a = n mod 2; where a=1 or 0; n=1,2,3...
but if n=1 than a=2 (for the first one), and 2 is not solution
That s one of the other difficulties in that problem.

So I just need a general function for an x number of 1 (1,1,1,1...1,0,1,1,1,1,...0...). Always after a certain period of number 1 comes 0
The function must be valuable for all n; as you can see in my example when I put n=1, a=2, and that isn t solution

2. Nov 4, 2011

### Deveno

Re: one,zero,one,zero...

well if p is a prime number, then

f(n) = np-1 mod p will work.

but if we want a function where the 0's come every six places, this approach won't work:

f(n) = n5 mod 6 gives:

(1,2,3,4,5,0,1,2,3,4,5,0,.....)

3. Nov 4, 2011

### Emilijo

Re: one,zero,one,zero...

I want a general formula for all terms, including a function where the 0 s come every six places, and if its possible without primes.
The function must be strictly given for ALL TERMS and proven

Last edited: Nov 4, 2011
4. Nov 4, 2011

### D H

Staff Emeritus
Re: one,zero,one,zero...

What's wrong with divides? (Or not divides in this case.)

5. Nov 5, 2011

### Emilijo

Re: one,zero,one,zero...

Can you give an example including that function f(n) = np-1 mod p for
the series 1,1,1,0,1,1,1,0...; what is n in your case, which prime I have to put instead p... Explain this function.

Last edited: Nov 5, 2011
6. Nov 5, 2011

### disregardthat

Re: one,zero,one,zero...

Let w be a primitive m+1'th root of unity, say $w = e^{i\frac{2\pi}{m+1}}$.

Look at $$x_n = \frac{1+w^n+w^{2n}+...+w^{mn}}{m+1}.$$

We will have that $x_0 = 1, x_1 = x_2 = ... = x_m = 0$. And then it repeats itself.

If we take $y_n = 1-x_{n+1}$ we will get the series $y_1,y_2,... = 1,1,...,1,0,1,,1,...,1...$ where there are m 1's for each repeat.

The reason for this is that for each n $w^n$ is a k'th root of unity for some divisor k of m+1. k = 1 only when n is a multiple of m+1, i.e. n=0,m+1,2(m+1),... and so on. For these n $x_n$ will of course be 1.

When n is not a multiple of m+1, then k will not be equal to 1, and $1+w^n + (w^n)^2 +...+(w^n)^{k-1} = 0$. So the series $1+w^n+w^{2n}+...+w^{mn} = (1+w^n + (w^n)^2 + ... + (w^n)^{k-1}) + (w^n)^k(1+w^n + (w^n)^2 + ... + (w^n)^{k-1}) + ... + (w^n)^{k(\frac{m+1}{k})-1)}(1+w^n + (w^n)^2 + ... + (w^n)^{k-1}) = 0$, and thus $x_n = 0$. Note that for prime m+1, k will be equal to m+1 for all n except multiples of m+1.

For your first example we will get $y_n = 1-\frac{1+(-1)^{n+1}}{2} = \frac{1-(-1)^{n+1}}{2}$, as -1 is a primitive 2nd root of unity.

Generally $$y_n = \frac{m-(w^{n+1}+w^{2(n+1)}+...+w^{m(n+1)})}{m+1},$$ or $$\frac{m-(e^{i\frac{2\pi(n+1)}{m+1}}+e^{i\frac{2\pi 2(n+1)}{m+1}}+...+e^{i\frac{2\pi m(n+1)}{m+1}})}{m+1}.$$

Last edited: Nov 5, 2011
7. Nov 5, 2011

### D H

Staff Emeritus
Re: one,zero,one,zero...

I'll repeat, what's wrong with "divides"? The sequence {a divides n} for a given a and n comprising the positive integers is the opposite of what you want. Take the logical negation and voila! there is your set of sequences.

By the way, you have a missing sequence, the trivial sequence, in your set of sequences. It is the sequence 0,0,0,... So let's start with this in terms of "divides". 1 divides every positive integer, so the sequence given by {dn;1} where dn;1 is 1 divides n is the sequence of all ones (or all true). The logical negation of this sequence yields the trivial series 0,0,0,... .

Now look at using 2 as the divisor. 2 does not divide 1, it divides 2, it does not divide 3, and so on. Thus {dn;2} = {2 divides n} yields 0,1,0,1,... . The logical negation of this sequence yields 1,0,1,0,... : your first example.

Now look at using 3 the divisor. 3 divides n results in the sequence 0,0,1,0,0,1,... . The logical negation yields 1,1,0,1,1,0 : your second example. Doing the same with 4 yields 1,1,1,0,1,1,1,0,...

Divides (or does not divide) is exactly what you want here.

8. Nov 5, 2011

### fleem

Re: one,zero,one,zero...

D H is right. And to add to that, if you want complex sequences (repeating pulse train rather than a simple rectangular wave) then just Boolean-OR multiple divide terms.

9. Nov 5, 2011

### disregardthat

Re: one,zero,one,zero...

D H's solution is perfectly fine, it's more or less the definition of the sequences in question.

10. Nov 5, 2011

### streber

Re: one,zero,one,zero...

disregardthat s function is fine but what if there is a huge number of 1
1,1,1,1,1,1,1,1,1,1,1,1,1,...,1,0,1,...; than there is a lot of calculating;
does exist a summation of
∑_(k=1, to m) e^((2iπk(n+1))/(m+1))

11. Nov 5, 2011

### disregardthat

Re: one,zero,one,zero...

Well, you don't have to do it manually. You already know what the number is anyway.

12. Nov 5, 2011

### Emilijo

Re: one,zero,one,zero...

Yes, but I need that function for sth else and I have to get a general function
Is there exists a solution of summation that gave streber?

13. Nov 5, 2011

### Emilijo

Re: one,zero,one,zero...

Oh, and I can t understand deveno s function; can u explain me
f(n) = n^(p-1) mod p; what if I had a composite number?

14. Nov 5, 2011

### Emilijo

Re: one,zero,one,zero...

I tried to solve the summation by wolfram alpha, but the program doesn t give me a solution. Have you another function but with mod function or sth else, as if H D s function, but his function works just for primes.

15. Nov 5, 2011

### pwsnafu

Re: one,zero,one,zero...

Sorry this doesn't make sense.

When you say
1,1,1,1,...,1,0,1,1,...
you need to know beforehand where that first 0 is. Otherwise you can't write down the problem. As soon as you fix where that first 0 is, D H's formula gives the answer immediately.

So if your problem is "first zero is at 105" then "negation of 105 divides x" is the answer.
Similarly if "first zero is at m" then "negation of m divides x" is the answer.

16. Nov 5, 2011

### Xitami

Re: one,zero,one,zero...

$1-\left\lfloor\frac{n}{6}\right\rfloor + \left\lfloor\frac{n-1}{6}\right\rfloor$

17. Nov 6, 2011

### pwsnafu

Re: one,zero,one,zero...

Xitami, n = 13.

Edit: Scratch that. I mucked my calculations.

Last edited: Nov 6, 2011
18. Nov 6, 2011

### disregardthat

Re: one,zero,one,zero...

Exactly pwsnafu. To properly define your sequences you already have the means to calculate each term.

The solution n (mod 2) to the sequence 1,0,1,0,... is more or less a repetition of the definition itself. I thought you wanted an algebraic solution to the sequences.

19. Nov 6, 2011

### Kael42

Re: one,zero,one,zero...

Just suggestion, and I have only completed high-level maths at a high-school standard so far, so I apologize if I get this messed up, but...
Pehaps a sine function of another sine function with a different, complicated period formed by a polynomial? This could be adjusted so that on the integers 1, 2, 3, 4, etc. give the resulting values 1, 1, 0, 1, etc.
I'm unsure as to how to how to prove this, but it makes sense (I think).

20. Nov 6, 2011

### disregardthat

Re: one,zero,one,zero...

Good idea with trigonometric functions, what about this: $$x_n = \lceil \sin(\frac{2 \pi n}{m+1}) \rceil$$?

x_n from for n from 1 to m will be 1, x_(m+1) will be 0, and so the sequence repeats.

The problem with polynomials of higher degree is that they cannot be periodic, so I think it would be hard to accomplish that which you describe.