1
$\begingroup$

If you were given a sequence that went something like $$1,1,3,2,5,3,7,4,9,5....$$ You would notice that there are two patterns alternating, one being 1,3,5,7,9... and the other being 1,2,3,4,5...

How would you define this sequence with one single function, $k(n)$,without the use of words or extra functions such as "If n is even, apply the function $f(n)=\frac{n}{2}$ and if n is odd, apply the function $f(n) = n$"

Is there an elegant way to output such a sequence? I tried to do it and have my own way using a lot of terms but want to know if there is a much simpler solution

Furthermore, if your particular equation works for this equation, would it work for $p=3,4,5,6...$ patterns within a pattern. In addition, the patterns rotate uniformly meaning if there were 4 patterns, every 4 terms the same function is used to output a number based on the term number: An example of 4 patterns in a pattern that follow the following functions $f(n)=n,f(n)=n^2,f(n)=3n,f(n)=\lfloor{\sqrt{n}}\rfloor$, a certain function $k_p(n)$ where $p=4$ would output $$1,4,9,2,$$$$5,36,21,2$$$$9,100,33,3...$$ (The spacing is just for visual purposes and makes it easier to see how each pattern rotates every 4 terms)

Whatever solution that outputs such a pattern should also be able to change or be manipulated easily to accommodate more or less patterns with functions for each pattern varying in complexity

Update: A common way that people are trying to use to solve the first example with only two simple patterns is through the use of the properties of $(-1)^n$ which outputs -1 or 1, an alternating sequence between two values, perfect for the first example. I hope that further answers to this question should somehow expand to the next part of it as I know there are more than enough equations that output the first pattern.

My Work: I tried using modulus arithmetic to output the values I wanted that would sort of turn off or turn on certain functions depending on the term $n$. For the first example, I made the equation $k_2(n)= (mod(n,2))*n + (mod(n+1,2))*(\frac{n}{2})$ I think further work can be done in improving the modulus method as there is a common number between the mod sections of $k_2(n)$ and the number of patterns $p$ which is 2

$\endgroup$
  • $\begingroup$ I read some time ago about this topic, I think it was in the book Concrete mathematics of Graham and Knuth. If not it was in Generatingfunctionology. $\endgroup$ – Masacroso Apr 4 '17 at 8:38
  • $\begingroup$ Thank you @Masacroso, I will look into those two $\endgroup$ – Stone Apr 4 '17 at 8:39
  • $\begingroup$ This feels like code golf. en.wikipedia.org/wiki/Code_golf $\endgroup$ – Elizabeth S. Q. Goodman Apr 4 '17 at 8:46
2
$\begingroup$

Note that

$$f(x)=\frac12((-1)^x+1)$$ is zero when $x$ is odd and $1$ when $x$ is even, similarly, we can make such function that is zero when $x$ is even and $1$ when odd.

Thus,

$$1,1,3,2,5,3,7,4,9,5,...$$

Could be written as

$$k(x)=(x+1)\left(\frac12(1-(-1)^x)\right)+\frac{x}{2}\left(\frac12(1-(-1)^x)\right)$$

Which simplifies to

$$k(x)=\frac12+\frac{3x}4-\left(\frac12+\frac{3x}4\right)(-1)^x$$


In fact, you can do this with every power of $2$; simply take that first function $f$, and apply it on $f$, so that we have a function that is $1$ only if $x$ is divisible by $4$, etcetera. If we wish to do this with $k=3$, then it already becomes more complicated. We need a number that cycles with period $3$ when raising it to a power; well, an easy one would be $\omega_3=e^{\frac23i\pi}$, a cube root of $1$. Then we see

$$f_3(x)=\frac13\left(\omega_3^x+\omega_3^{2x}+\omega_3^{3x}\right)$$

has the exact property we want; it is $1$ only if $x$ is a multiple of $3$, and $0$ otherwise. We can in fact do this for every prime number $p$; define $\omega_p=e^{\frac 2pi\pi}$ such that

$$f_p(x)=\frac1p\left(\omega_p^x+\omega_p^{2x}+\cdots+\omega_p^{px}\right)$$

and then $f_p$ is $1$ only if $x$ is divisible by $p$, and $0$ otherwise. We can now create such function for every $n$, since if $n=p_1p_2\cdots p_k$, then $f_n(x)=(f_{p_1}f_{p_2}\cdots f_{p_k})(x)$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I in fact noticed this common pattern that I believe is sometimes used in computer science when alternating between two values or in this case, alternating between two functions. However, we can't exactly define new values for $(-1)^3$ or $(-1)^k$ that output something besides -1 and 1. Is there perhaps a better way that accommodates for more patterns? In addition, I prefer if we don't just map values to our liking, the solution would be much more elegant but if needed then please do so. $\endgroup$ – Stone Apr 4 '17 at 8:48
  • $\begingroup$ "I prefer if we don't just map values to our liking..." what do you mean? Am I misunderstanding the question? $\endgroup$ – vrugtehagel Apr 4 '17 at 8:54
  • $\begingroup$ So unless I don't know what mapping means, but things like mapping $(-1)^3$ to some other value that might help us solve much more complicated values. $\endgroup$ – Stone Apr 4 '17 at 9:03
  • $\begingroup$ Does this method only work for a prime p? Can this work for composite numbers as well? It's quite a nice idea but is limited to prime numbers if I'm understanding you correctly $\endgroup$ – Stone Apr 4 '17 at 9:10
  • $\begingroup$ Read the last line: "We can now create such function for every $n$..." $\endgroup$ – vrugtehagel Apr 4 '17 at 9:11
0
$\begingroup$

For this particular question : $$T_n=\frac{3n}{4} + (-1)^{n-1}\frac{n}{4}$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I'm pretty sure there are quite a few ways to solve the first part, but would this be possible to expand on? I noticed that this seems to be limited to just 2 patterns. Also i'm not quite sure what to do with n = 1, should we treat $(-1)^0 = 1$? $\endgroup$ – Stone Apr 4 '17 at 8:39
  • $\begingroup$ Yes : $(-1)^{0}=1$, or you may replace $n-1 \rightarrow n+1$, would work the same.Frankly, I too don't know can this be expanded to more than $2$ terms or not. $\endgroup$ – Jaideep Khare Apr 4 '17 at 8:41
0
$\begingroup$

Using trigonometric functions:

$$f(n) = |\sin(\frac{n\pi }{2})|n + |\cos(\frac{n\pi }{2})| \frac{n}{2}$$

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

$$f(n)=\frac{n}{2-n+2\lfloor\frac{n}2\rfloor}$$ Not very elegant, though.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

$$n\cdot\frac{n\bmod2+1}{2}.$$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.