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joemath314159Best ResponseYou've already chosen the best response.4
where do people get these sequences from?
 2 years ago

joemath314159Best ResponseYou've already chosen the best response.4
i cant say ive seen a sequence on this site that has made any sense in the least bit.
 2 years ago

Akshay_BudhkarBest ResponseYou've already chosen the best response.0
sri is it an olypiad question?
 2 years ago

joemath314159Best ResponseYou've already chosen the best response.4
if its a polynomal in n, then its at least a cubic >.>
 2 years ago

joemath314159Best ResponseYou've already chosen the best response.4
I showed last time i can make an infinite number of formulas fitting that pattern using 6 random numbers <.<
 2 years ago

joemath314159Best ResponseYou've already chosen the best response.4
we get questions like these a lot. someone had posted something similar, and just like this one, we couldnt find a pattern. So using the fact that we never know how to the tail of a sequence acts, i made up a tail, and found a formula. if its correct or not, we will never know...but it does work for the first couple of numbers.
 2 years ago

joemath314159Best ResponseYou've already chosen the best response.4
give me 6 numbers between 5 and 5, and not 0.
 2 years ago

joemath314159Best ResponseYou've already chosen the best response.4
its going to look ugly btw, but im 100% sure it works.
 2 years ago

joemath314159Best ResponseYou've already chosen the best response.4
k, give me a couple of mins, gotta do calculations.
 2 years ago

joemath314159Best ResponseYou've already chosen the best response.4
ok, the formula is: \[\frac{3}{10}(3)^n+\frac{5}{504}(2)^n\frac{335}{9}(1)^n+\frac{1107}{40}(2)^n\frac{101}{36}(4)^n+\frac{9}{14}(5)^n\] for n = 1,2,3...,6 it equals those numbers up top. http://www.wolframalpha.com/input/?i=%E2%88%923%2F10%28%E2%88%923%29^n%2B5%2F504%28%E2%88%922%29^n%E2%88%92335%2F9%281%29^n%2B1107%2F40%282%29^n%E2%88%92101%2F36%284%29^n%2B9%2F14%285%29^n%2C+n+%3D+6 where you type in the equation, if you scroll to the right you can change the value of n to make sure it works.
 2 years ago
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