Describe a list of numbers that has alternating functions without using trigonometric functions? The list is: $ \{5,6,10,12,15,18,20,24,25...\}$
Is there a way to describe this list of numbers mathematically without using $\sin(x)
, \cos(x), \cos(nx)$, etc?
There is a way using the trigonometric function $\sin$. Simply use the jumping between $0$ and $1$ of $$\frac{1+\sin(n\pi)}{2}$$ and $$\frac{1+\sin\left(n\pi+\frac{3\pi}{2}\right)}{2 }$$ which instead oscillates between $1$ and $0$. Then you can get the answer by describing $A = \{5, x, 10, y, 15, z, 20\}$ as well as $ B = \{x, 6, y, 12, z, 18\}$. Every other number in the lists $A$ and $B$ will either be numbers or they will be zero, since they're multiplied with said $\sin$ function. $A + B$ is therefore a list containing two different methods of sorting, i.e. $A$'s function could be $ x^2 $ and $B$'s function could be $x^3-x^2+2x $ where every item in the list $C$, where $C = A + B$, is based on a different equation than the one before and after that said item.
Now, again, is there a way to create this said list $C$, containing the pattern of two very different functions that create the list $C $ where every item in the slot $ 2n$ is connected to one function but that function is different from the item in the slot $2n + 1$?
 A: What you have is three sequences
$$ a_1,a_2,a_3,\cdots $$
$$ b_1,b_2,b_3,\cdots$$
with a third sequence which is a 'shuffle' of the first two
$$ a_1,b_1,a_2,b_2,a_3,b_3,\cdots=c_1,c_2,c_3,c_4,c_5,c_6\cdots$$
Define two functions on $\mathbb{N}$:


*

*$k(n)=\left\lfloor\dfrac{n+1}{2}\right\rfloor$

*$h(n)=\dfrac{1+(-1)^{n+1}}{2}$
Then $k$ defines a sequence $1,1,2,2,3,3,\cdots$ and
$h$ defines a sequence $1,0,1,0,1,0,\cdots$
Then the sequence $c_n$ can be defined
$$ c_n=a_{k(n)}h(n)+b_{k(n)}h(n+1) $$
Then
\begin{eqnarray}
c_1&=&a_{k(1)}h(1)+b_{k(1)}h(2)\\
&=&a_1\cdot1+b_1\cdot0\\
&=&a_1
\end{eqnarray}
\begin{eqnarray}
c_2&=&a_{k(2)}h(2)+b_{k(2)}h(3)\\
&=&a_1\cdot0+b_1\cdot1\\
&=&b_1
\end{eqnarray}
etc.
Substituting in your particular sequences will result in
$$c_n=\left(\frac{11+(-1)^n}{2}\right)k_n$$
A: If you just want to define a sequence so that the terms come from one function and then another, alternately, a simple way is:
$$
c_n = \begin{cases}
f(n) & \text{if $n \equiv 0 \pmod 2$}, \\
g(n) & \text{if $n \equiv 1 \pmod 2$}.
\end{cases}
$$
