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?

- Thread starter uziel
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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?

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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 |

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1. The denominator is 3

2. The coefficients of a, b and c add up to 3

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??

# | 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 |

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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!