Express the sequence which alternates three positive ones and three negative ones $1,1,1,-1,-1,-1,\dots$

Question

But when I see that it was closed because of being unclear I decided to make it better and ask again.

First I want to give a link to make it more clear first what is alternating series. What you see is that you can generate $1,-1,1,-1,\dots$ using this formula: $(-1)^n$.

Now if you want to have two positives and two negatives together $\{1,1,-1,-1,1,1,-1,-1,...\}$ you can explain it with floor function $(-1)^{\lfloor \frac{n}{2} \rfloor}$ or you can use triangular numbers: $(-1)^{\frac{n(n+1)}{2}}$. See other possible ways here and here.

Now when we want to have three positives and three negatives together:$\{1,1,1,-1,-1,-1,1,1,1,-1,-1,-1,...\}$ We could explain with floor function: $(-1)^{\lfloor \frac{n}{3} \rfloor}$ but because of we didn't learn floor function yet we don't have the permission to use it.

Also we didn't learn trigonometry then we cannot use it either. (Here is a solution using trigonometry may help.) Note that we want a single formula solving it with different formula is very easy.

Now what I want is a simple solution using simple algebra and without using floor function or trigonometry. This is the main sequence that if we find the previous sequence will solved easily.

$$A_n=\{3,7,11,-15,-19,-23,...\}$$

We should only add $4n+3$ next to it.

The news that I should tell you is that our teacher told us is that using absolouting value is not allowed then we can know $sign$ is also not allowed our teacher told us it should be in the form of an algebric formula like the other one that we used for two negative once and two positive once. He told us for solving this we should use degree $3$ algebra but how?

Answer 1

The formula is

$$A_n=(-1)^{F_{n+1}+n}\,(4n-1)$$

with first term $A_1=3$ and where $F_n$ is the $n$th Fibonacci number. This formula involves neither $sign$ nor $floor$.

I believe you meant $4n-1$ instead of $4n+3$, since otherwise the first term would be $4+3=7$ instead of $3$.

Explanation

Fibonacci numbers are defined as follows:

• The first and second Fibonacci numbers are both $1$
• For $n\ge3$, the $n$th Fibonacci number equals the sum of the $(n-1)$th and $(n-2)$th Fibonacci number.

Consider that

• $F_1$ is odd, and
• $F_2$ is odd.

Therefore,

• $F_3$ is even
• $F_4$ is odd
• $F_5$ is odd.

If this is continued, the pattern will repeat itself indefinitely after every 3 numbers:

• $F_6$ is even
• $F_7$ is odd
• $F_8$ is odd, and so forth.

Expressing this in the form $F_{n+1}$, excluding $F_{0+1}$, yields

• $F_{1+1}$ is odd
• $F_{2+1}$ is even
• $F_{3+1}$ is odd
• $F_{4+1}$ is odd
• $F_{5+1}$ is even
• $F_{6+1}$ is odd.

Consider that

• 1 is odd
• 2 is even
• 3 is odd
• 4 is even
• 5 is odd
• 6 is even.

This pattern, obviously, repeat itself after every 2 numbers. Adding it to the previous in order yields

• $F_{1+1}+1$ is even
• $F_{2+1}+2$ is even
• $F_{3+1}+3$ is even
• $F_{4+1}+4$ is odd
• $F_{5+1}+5$ is odd
• $F_{6+1}+6$ is odd.

This pattern repeats itself after every $lcm(3,2)=6$ numbers. Then, we raise $-1$ to each term of this pattern, obtaining

• $(-1)^{F_{1+1}+1}=1$
• $(-1)^{F_{2+1}+2}=1$
• $(-1)^{F_{3+1}+3}=1$
• $(-1)^{F_{4+1}+4}=-1$
• $(-1)^{F_{5+1}+5}=-1$
• $(-1)^{F_{6+1}+6}=-1$

Finally, we multiply every $n$th term by $4n-1$.

• $(-1)^{F_{1+1}+1}(4*1-1)=1*3=3$
• $(-1)^{F_{2+1}+2}(4*2-1)=1*7=7$
• $(-1)^{F_{3+1}+3}(4*3-1)=1*11=11$
• $(-1)^{F_{4+1}+4}(4*4-1)=(-1)*15=-15$
• $(-1)^{F_{5+1}+5}(4*5-1)=(-1)*19=-19$
• $(-1)^{F_{6+1}+6}(4*6-1)=(-1)*23=-23$

The multiplication produces the desired sequence, which is $\{3,7,11,-15,-19,-23\}$.

Answer 2

Here's one function which, starting at $n=1$ produces the sequence $1,\,1,\,1,\,-1,\,-1,\,-1,\ldots$.: $$(-1)^{\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}+\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}+n}$$ If we're not using trigonometry or complex numbers, I suspect the techniques that I used to find this would be inscrutable, but I have detailed them below the horizontal line.

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I found this by playing around with generating functions. In particular, we want $$a_{n+3}\equiv 1-a_n\pmod 2.$$ In terms of generating functions functions, if we let $A(x)=a_0+a_1x+a_2x^2+a_3x^3+\ldots$ then we get $$A\equiv x^3\left(\frac{1}{1-x}-A\right)\pmod 2$$ which solves as $$A\equiv \frac{x^3}{(1-x)(1+x^3)}\pmod 2.$$ Since we want the denominator to factor as much as possible, we subtract $0\equiv 2x^2\pmod 2$ from the appropriate place giving $$A\equiv \frac{x^3}{(1-x)(1-2x^2+x^3)}\equiv \frac{x^3}{(1-x)^2(1+x-x^2)}\pmod 2.$$ At this point, we are happy since our denominator can only have real roots, of algebraic degree $2$, so we can definitely find a closed form for the power series of the right hand side in terms of the given functions. Finding this closed form is easy if you know about generating functions, and would yield a suitable answer. One could use similar techniques to find various other similar formulae.

However, at this point, I realized that, since the recurrence relation $L_{n+2}=L_{n+1}+L_n$ has a period of $3$ mod $2$ when started with $0,\,1,\ldots$ mod $2$, we can basically build the answer off of it. In particular, the Lucas sequence starts this way and has the simple closed form $L_n=\varphi^n+(-\varphi)^{-n}$. It turns out that, if one observes the form that came out of the generating function, and matches the parities up properly, we get that, starting with $n=1$, the sequence $L_{n+1}+n$ alternates between even and odd parity every $3$ terms. Thus, we just write $(-1)^{L_{n+1}+n}$ as our answer.

If one can solve cubic and quartic equations (whose general formulae require complex numbers), then one can at least get alternations of length $5$. Solving general polynomials gives alternations of any length. However, it seems that things are doomed to get more complicated or even impossible without using trigonometry or complex numbers.

Answer 3

You can find an exponent of $-1$ that fits by solving the recurrence

$$e_{n+3}=e_n+1,\\e_0=e_1=e_2=0.$$

The general solution is

$$e_n=c_0+c_1\omega^{n}+c_2\omega^{-n}+\frac n3$$ where $\omega$ is a cubic root of $1$. To get a real solution, you must have $c_2=c_1^*$.

By plugging the initial conditions, you determine the constants, giving

$$e_n=\frac{n-1}3-\frac2{3\sqrt3}\sin\left(\frac{2\pi}3(n-1)\right).$$

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This solution directly generalizes to alternating sequences of length $k$.

Answer 4

One way would be to expand the floor function, $$a_n=\begin{cases}(-1)^\frac{n}3(4n+3)&\text{ if }n\equiv0\pmod3\\(-1)^\frac{n-1}3(4n+3)&\text{ if }n\equiv1\pmod3\\(-1)^\frac{n-2}3(4n+3)&\text{ if }n\equiv2\pmod3\end{cases}$$