# General formula for nth element of the sequence 0, 1, 0, 1, …

The sequence is $$f = 0, 1, 0, 1, \ldots$$

I want to find a general formula for the $$n$$th element. The sequence starts at $$n = 0$$ (the $$0$$ here is not the first element $$0$$ but rather denotes the $$0$$th position).

One easy and obvious solution is: $$n$$th $$f = n \bmod 2$$. This works because even positions have $$0$$ and odd positions have $$1$$.

However, this question is part of a homework and modulus has not been discussed (or part of the syllabus or even a prerequisite). And so I am hesitant to use it.

Is there another way to solve this problem using only basic arithmetic operations (one that a beginning high schooler knows of)?

• Do beginning high-schoolers know about Wolfram Alpha? Go to that website and type Boole[OddQ[Range[0, 99]]] in the box. – Bill Thomas Sep 25 '18 at 21:05
• Not sure if you accept recurrence formulas, but that then you would have simple one $a_{n+1}=1-a_{n}$ with $a_0=0$. – Sil Oct 12 '18 at 8:08

$$a_n=(1/2)(1+(-1)^{n+1})$$, $$n=0,1,2,.....$$

The expression for the $$n$$th element of that sequence $$f$$ should take account of the products of $$(-1)$$.

As every number $$n$$ that appears in your sequence is just $$n = \frac{1}{2} + k,$$ where $$k = \pm\frac{1}{2},$$ then the general expression for the $$n$$th term $$f_n$$ would be the sum of the term $$k$$ to the $$n$$th: $$f_n = \frac{1}{2} + (-1)^n \left(\frac{1}{2}\right).$$

Considering that your sequence starts with the value $$0$$ in the first term. $$(-1)^1 = -1, (-1)^2 = 1$$, and so on.

Consider whether you can adjust the sequences from either of:

• $$(-1)^n$$
• $$\cos(\pi n)$$

to get what you are looking for, for example by adding a constant and/or multiplying by a constant

Have you learned about exponents yet, including the special cases $$x^0$$ and $$x^1$$? Here's a slightly more complicated definition using those exponents: $$f(0) = 0$$, $$f(1) = 1$$ (that's what software developers would call "initialization") then $$f(n) = f(n - 2)^{f(n - 1)}$$ for $$n > 1$$.

You could also do $$f(n) = f(n - 2)$$, though you could come across a smart-aleck with that one.

First, some notes:

• If you were to explain the functions you use (same goes for modulus), maybe your homework would get accepted anyway.
• $$\lceil{x}\rceil$$ denotes the ceiling function which takes as input the real number $$x$$ and gives as output the least integer greater than or equal to $$x$$, i.e. $$\lceil{5.3}\rceil=6$$.
• $$\lfloor{x}\rfloor$$ denotes the floor function which takes as input the real number $$x$$ and gives as output the greatest integer less than or equal to $$x$$, i.e. $$\lfloor{5.3}\rfloor=5$$.
• When $$x$$ is an integer, then $$\lceil{x}\rceil=\lfloor{x}\rfloor=x$$.

With this mind:

Your sequence $$\{f_n\}=0,1,0,1,0,1,\dots$$ can be generated by the function $$f_n=\left\lceil\frac{n}{2}\right\rceil-\left\lfloor\frac{n}{2}\right\rfloor,\;\ n\in\mathbb{Z}^{\ge0}$$ For each even positive integer $$n$$ (including $$0$$), the ceiling and floor function evaluate to equal values, canceling each other out; $$f_n=0$$.

For each odd positive integer $$n$$, the ceiling and floor function evaluate to values differing by $$1$$; $$f_n=1$$.

Is the sequence just an eternal alternation of 0 and 1? Then all you need to do is put in a couple dozen alternations in the OEIS, find https://oeis.org/A000035, then just sit back and read until you find a formula you like.

Least significant bit of $$n$$, lsb(n).

This works even if $$n$$ is negative, but it gets a little bit into computer science.

Also decimal expansion of $$\frac{1}{99}$$.

Yeah... if you ignore the integer 0 in 0.01010101010101010101010101010101...

Also the binary expansion of $$\frac{1}{3}$$.

Though I'm not sure this applies under the IEEE 754 floating point format specification.

There's a whole bunch more to look at. I personally like the recurrence relation $$a(n) = 1 - a(n - 1)$$, which is of course initialized with $$a(0) = 0$$.

• Thanks for the amazing resource. – Abhishek Kumar Sep 25 '18 at 15:50

You could compact the alternating sign formula with $$f(n) = \sin \left( \frac{n \pi}{2} \right)^2$$

Applying $$(\sin(x))^2= \left( \frac{1}{2} \right)(1 - \cos(2x))$$

So you get $$0, 1, 0, 1, 0, 1, \ldots$$

We can apply the sign function:

$$a_n=\mathrm{sgn}(1-(-1)^n)\qquad\qquad n=0,1,2,\ldots$$