(a working semi-related decrement function)
[T]500 = [T]500 * [S]0.156048361
= [N1] (134.8206296 * [M]0.85)
= [N2] (114.5975351 * [M]0.85)
= [N3] (97.40790487 * [M]0.85)
= [N4] (82.79671914 * [M]0.85) = 70.37721127
In this case T = 500 , N = 4, M = 0.85 , S = 0.156048361 = (0.85-1)/0.85^(4+1)-1)
T = Total
N = is how many times you want to decrement.
M = is the multiplier
S = a number you start to decrement = (M-1)/(M^(N+1)-1)
This is possible through the sheer helpfulness of other on stackexchange. I'm an indie game dev. And this is awesome for any kind of numerical progression you want to have a boundary. Such as damage scales, loot scales, expenses etc.
The S calculation assured my results were accurate regardless of how much i fiddled with T, N and M.
Now the goal is:
Increment a number that gets added X(always the same amount), N times and then multiplied N times(the same N) via linearly decrementing M multiplier and keep the SUM total a specified number. The S calculation for the decrement formula/point provides that kind of preservation of the sum total.
As far as I understand the variable that needs to be calculated for this is the how much the multiplier decrements since i want to be able to adjust the initial value.
This way if i have contraption that produces stuff with a flat value + % of the stuff it has i can tweak the base criteria enough so that it produces half its limit for the third of its time table and have it diminished later.
I can make up the numbers manually and have them add up but I've always wanted a way to do this kind of thing, with actual math.
I assume this looks something like this (you can see i'm not great at it) :
[T]100000 = ( [S]0 + [F]500) * [MS]3 ) = ( [R1]1500 + [F]500 ) * [MS]3 * [MD]?
Where MS*MD = MR
Continue by doing N Times:
(R2 + F) * MR1
(R3 + F) * MR2
T = Total
S = Start
F = Flat value to add every time
MS = Starting multiplier
MD = Multiplier decriment
MS * MD = MR
............ Even if you don't calculate it an assessment how doable and/or hard it is - works.