Self referencing formula

[uziel]

Hello, I have a tough question. I am unsure if it fits in this category; if not, I apologize.

Given three numbers, how would a future average get calculated based only on the initial three numbers and the number of periods in the future?

In the example below the first three values are the basis for all future calculations. Is this possible?

 

[Harry_the_cat]

Nice problem!
If the initial 3 values are a, b and c:

Average 1	(a + b + c)/3	(a + b + c)/3
Average 2	(b + c + (a + b + c)/3)/3	(a + 4b + 4c)/9
Average 3	(c + (a+4b+4c)/9 + (a+b+c)/3)/3	(4a + 7b + 16c)/27
Average 4		(16a + 28b + 37c)/81
Average 5		(37a + 85b + 121c)/243
Average 6		(121a + 232b + 376c)/729

 

[Harry_the_cat]

Now it's a matter of finding a pattern in the last column.
1. The denominator is 3ⁿ.
2. The coefficients of a, b and c add up to 3ⁿ.
3. The "old" coefficient of c becomes the "new" coefficient of a.

It is also intuitive that the further down the list you go, these averages will squeeze in on one another and approach a limiting value.

So there's a start. I need to think some more. In the meantime, any ideas??

 

[uziel]

Thank you for your help! What you did is very helpful, but I cannot determine a way to calculate that programmatically. For example, variable 'a' is 16 in #4 and 37 in #5. Is there a way to calculate that or must we do the math by hand?

#	a	b	c	div
1	1	1	1	3	1,228	3
2	1	4	4	9	1,250	9
3	4	7	16	27	1,279	27
4	16	28	37	81	1,252	81
5	37	85	121	243	1,260	243
6	121	232	376	729	1,264	729

 

[Harry_the_cat]

Because of the way the previous 3 terms are added and then averaged (see unfinished column 2 in Post #2)

note that \(\displaystyle a_4 = 16 = 9*1 +3*1 + 1*4 = 9*a_1 +3*a_2 +1*a_3\)

Also \(\displaystyle a_5 = 37 = 9*1 +3*4 +1*16 = 9*a_2 + 3*a_3 + 1* a_4\)

This works all the way through for a, b and c

So \(\displaystyle a_n = 9*a_{n-3} + 3*a_{n-2} + 1*a_{n-1}\)

This still relies on knowing the last 3 coefficients, not just a function of n. Some more thinking needed!

 

[uziel]

Actually, I think this might solve the question. I do not think you can go any further because a+b+c=div, which seems obvious now. Thank you so much!