# Compute a sequence of number whose with fixed min distance and max distance

Is there a way to generate n numbers whose space between them grows incrementally and which that space varies between a min value and a max value? It's not important the domain of these numbers.

Is there a way to generate n numbers whose space between them grows incrementally and which that space varies between a min value and a max value? It's not important the domain of these numbers.

I immagine to create a procedure like this:

const serie = computeSerie(n, minSpace, maxSpace)

// domain is not important, for example [1, +infinity] but also [0, 1], what you prefer

const serie1 = computeSerie(5, 1, 1) // [1, 2, 3, 4, 5]
const serie2 = computeSerie(5, 2, 2) // [1, 3, 5, 7, 9]
const serie3 = computeSerie(5, 1, 4) // [1, ...] I don't know, I suppose to use a pow math function (?)
const serie4 = computeSerie(7, 1, 6) // [1, 2, 4, 8, 13, 18, 24] I don't know if it is the right sequence


I need a sequence that visually is similar to:

Visually:

serie1: |-|-|-|-|
serie2: |--|--|--|--|
serie3: |-|???|--|
serie4: |-|--|---|----|-----|------|


as you can see, serie4 is something that grows and the first distance is 1, the last one is 6 and in total it includes 7 numbers.

I've no idea what is the logic behind. Which is the best way (or simply a way) to create a sequence of number whose distance between them grows and with a fixed min and max distance?

This is not hard if you break it down step by step. First of all, let's name the values we need to work with.

Let $$a_1 ... a_n$$ be the sequence you want to generate.

Let $$b$$ be the smallest interval, and $$c$$ be the amount that each interval is incremented. So, the largest interval will be $$b + (n-2)c$$.

[edit:] If you have the largest interval, say $$d$$, then you can find the increment:
$$d = b +(n-2)c \\ c = (d-b)/(n-2)$$

And then we just put it all together:

$$a_2 = a_1 + b \\ a_3 = a_2 + b + c \\ ... \\ a_{i+1} = a_i + b + (i-1)c \\ ... \\ a_n = a_{n-1} + b + (n-2)c \\$$

• Thank you Caledon, bit it's a bit different. I need to fix the smallest and largest interval, not the the amount that each interval is incremented Oct 12, 2021 at 16:31
• You're right. I used the increment because it makes the notation much simpler. If you know what the largest interval is, then it's pretty easy to calculate the increment. Oct 12, 2021 at 16:33
• Last question: can you explain me why the largest interval is 𝑏+(𝑛−2)𝑐? Oct 13, 2021 at 6:33
• the first interval is b. The 2nd interval is b + c. The 3rd is b + 2c. Continue like this. There are n-1 intervals; the final one will be b + (n-2)c. Oct 13, 2021 at 13:47