Self referencing formula

Harry_the_cat

Senior Member
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

Senior Member
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

New member
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

Senior Member
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

New member
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!