# General function that represents a series of 1 and 0

by Emilijo
Tags: function, represents, series
 P: 36 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
 Sci Advisor P: 906 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,.....)
 P: 36 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
 Mentor P: 15,147 General function that represents a series of 1 and 0 What's wrong with divides? (Or not divides in this case.)
 P: 36 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.
 Sci Advisor P: 1,793 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 $$\frac{m-(e^{i\frac{2\pi(n+1)}{m+1}}+e^{i\frac{2\pi 2(n+1)}{m+1}}+...+e^{i\frac{2\pi m(n+1)}{m+1}})}{m+1}.$$
 Mentor P: 15,147 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.
 P: 461 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.
 Sci Advisor P: 1,793 D H's solution is perfectly fine, it's more or less the definition of the sequences in question.
 P: 2 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))
P: 1,793
 Quote by streber 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))
Well, you don't have to do it manually. You already know what the number is anyway.
 P: 36 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?
 P: 36 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?
 P: 36 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.
 P: 129 $1-\left\lfloor\frac{n}{6}\right\rfloor + \left\lfloor\frac{n-1}{6}\right\rfloor$