Help on a regression

[Galandros]

Data:
a b
0 10000
1 20000
2 40000
3 75000
4 140000
5 255000
6 470000
7 865000
8 1590000
9 2920000
10 5355000
11 9820000
12 18005000
13 33005000
14 60510000
15 110925000
16 203350000
17 372785000
18 683385000
19 1252785000
20 2296600000
EDIT: correct values thanks do DarkerLine

I want to find the formula that gives this. Dependent variable is a. (of course)

I have tried the most usual regressions. Linear (no hopes on it at start though), Power, exponential, polynom, etc..
Hint: a big *maybe*: it looks something like: b = 100000 + 50000 * (RoundUp(1.6^a)-1)

I worked on trying to figure if the formula is like that one with a spreadsheet a this online tool: http://www.xuru.org/rt/TOC.asp
No big results. I made everything I could... Maybe I need more serious tools.

Let the fun begin!
PS: why a need this formula? It's a non TI-project of mine.
PS2: I learned a bit with this So I am happy even if we don't find it.

[DarkerLine]

That data looks pretty clearly exponential to me. Is there any reason you don't like that model?

Edit: oh, and the fit becomes much better if you ignore the first few terms (which are a bit off the model, presumably because they have less significant digits).

[Galandros]

[DarkerLine]

Perfect results are generally not the idea. For actual data, perfect results are impossible since the data depends on a bunch of other factors including measurement error. In your case, it looks like every value of B is rounded to the nearest 5K, so any formula that gives a correct result up to that tolerance is as good as perfect.

Of course, knowing where the data comes from generally helps make such a judgment.

[Galandros]

[Weregoose]

Verify that b=1,297,589,000 is correct in your list; it's a little strange.

5000int(2.5(11/6)^a-[font="verdana"]³√([font="times new roman"]e

[DarkerLine]

That's a little too complex compared to the other ones.

It's definitely exponential, plus some bit of rounding that got thrown in there. You'd have to experiment to figure out the exact form of the formula (the rounding could happen anywhere), but this tells us that the exponent is around 1.84 (the formula is probably fairly simple, so I'm guessing it's either 1.8 or, less likely, 1.85). Then again, the values might have been hard-coded -- maybe they were originally calculated by a formula, but then altered.

[Galandros]

[DarkerLine]

My guess is that the formula is simply iterated, and the rounding is done at each step. In that case, the formula is more or less 11/6 (or maybe an approximation to it) times the previous value, rounded to the nearest 5000. There's some trickiness with the rounding here, but this is much closer to the truth than anything else.

Also, here are several corrections to your values:
1: 20000 not 10000
2: 40000 not 20000
3: 75000 not 40000
19: 1252785000 not 1297589000

[Galandros]

[DarkerLine]

The value of 1297589000 was a typo that one person made and everyone else copied. I searched Google for 683.385.000 but not 1.297.589.000 and found a table with a better number.

[Galandros]

DarkerLine words to coding:
10000
FOR(1, 20)
store = CEILING( (11/6) * ans ;0,5)
END

The problem is that the step is around 11/6 like this:
1 2
2 2
3 1,875
4 1,866666667
5 1,821428571
6 1,843137255
7 1,840425532
8 1,838150289
9 1,836477987
10 1,83390411
11 1,833800187
12 1,833503055
13 1,833101916
14 1,833358582
15 1,833168071
16 1,833220645
17 1,833218589
18 1,833188031
19 1,833205294
20 1,83319564
I tried some roundings (up, down, normal before the ceiling but no perfect results).

This steps make me remember some iterative algorithms like with the Greek method to find a square root (it gets close of the square root with fractions).

[DarkerLine]

The reason for that convergence is the rounding. As the numbers get larger, the rounding stops mattering as much, and therefore the ratio converges to the true value of around 1.8332.