How to best express this series as a statement?

[Endy0816]

{(1,1), (2,0.1), (3,0.1), (4,0.01), (5,0.01)}

assuming that this pattern continues on forever?

[joigus]

(n,10^1-n) would almost do the job you need. But you need the powers of ten to be "out of step" and catch up every two steps, so to speak.

If you want a simple recipe the simplest one I can think of is with the floor function:

floor(m)=floor(n+p/q)=n

where n is the integer part of rational number m and p/q<1 is the rational part you need to add to n to get m.

Then,

(n,10^-C(n))

where C(n) := floor((n-1)/2)

https://en.wikipedia.org/wiki/Floor_and_ceiling_functions

Quote

The floor of x is also called the integral part, integer part, [...] and was historically denoted [x] (among other notations).

A good thing about floor is that you have it built in in Mathematica, Wolfram, and the like.

You probably can do this with trigonometric functions too.

Edited August 9 by joigus
minor correction

[Ghideon]

3 hours ago, Endy0816 said:

{(1,1), (2,0.1), (3,0.1), (4,0.01), (5,0.01)}

assuming that this pattern continues on forever?

Idea: Use floor*?  

[math]10^{-\lfloor  \frac{n-1}{2} \rfloor }[/math]

So that the pairs of numbers are expressed as [math](n, 10^{-\lfloor  \frac{n-1}{2} \rfloor }), n \in \mathbb{N}[/math]

*) https://en.wikipedia.org/wiki/Floor_and_ceiling_functions#:~:text=In mathematics%2C the floor function,⌋ or floor(x).

edit: x-post with @joigus

Edited August 9 by Ghideon
x-post, fixing latex

[Endy0816]

@joigus @Ghideon

Thanks guys! This will work great