Recurrence Relation Question

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PeterRobinson44

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This question confuses me very much. Any insight would be greatly appreciated. I don't really know where to start with answering the question.


a(0) = 45

Find a(6)



a(n) =

( a(n) / 3 | n = 0 mod 3
( a(n) - 1 / 3 | n = 1 mod 2
( a(n) + 1 / 3 | n = 2 mod 3


I put those barriers ( | ) up there to not have everything so compacted.


I tried my best to write out the question but kind of difficult with the characters available to write it out. Also, the question is written out correctly except however if anything the n = 0 mod 3, n = 1 mod 2, n = 2 mod 3 values might be off a bit. I apologize for that. I saw a question like this a while back and can't find where it is exactly but I remember it confused me when I first saw it so I wanted to understand it better. Any insight would be appreciated. Once again sorry and if you have any questions about format or values please reach out to me and I will answer those questions.
 
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Just to be sure that we have the correct problem can you upload a picture of the exact problem.

I suspect that you meant to write (twice!) n=1 mod 3

Should that very first a(n)= be a(n+1) = ??????

Whenever you divide an integer by 3 the remainder is either 0, 1 or 2. They are telling you the rule for a(n) depending on the remainder when you divide n by 3. For example, if n= 23, then n = 2 mod 3, since when you divide 23 by 3 the remainder is 2. So you use the rule a(n) + 1/3
 
You are remembering this? I suspect you are not remembering correctly!

For one thing, numbers that satisfy "n= 0 (mod 3)" are simply the multiples of 3- 0, 3, -3, 6, -6, 9, -9 etc.- while numbers that satisfy "n= 1 (mod 2)" are numbers that are one more than 2- 3, 5, 7, 9, etc. Some numbers, 3, 7, etc., are on both lists- they satisfy both conditions so, if what you wrote is correct, both formulas would have to apply. I suspect they were all are "mod 3", no mod 2 (or was that a typo by you?)
Also a "recurrence relation" wouldn't have "a(n)=" a function of a(n). Finally, while "a(n)- 1/3" and "a(n)+ 1/3" are possible I am concerned that this might be "(a(n)- 1)/3" and "(a(n)+ 1)/3".
I suspect what you are remembering was
a(n+1)= ( a(n) / 3 | n = 0 mod 3
( (a(n) - 1)/ 3 | n = 1 mod 3
( (a(n) + 1)/ 3 | n = 2 mod 3

You are asked to find a(6). Since we are given that a(0)= 45, we need to work our way up to n= 6. "2" of course satisfies n= 2 (mod 3) so a(2)= (a(1)+ 1)/3= (45+1)/3= 46/3. Then "3" is 0 mod 3 so a(3)= a(2)/3= (46/3)/3= 46/9. "4" is 1 more than 3 so a(4)= (a(3)- 1)/3= (46/9+ 1)/3= 54/27. "5" is 2 more than 3 so a(5)= (a(4)+ 1)/3= (54/27+ 1)/3= 81/81= 1. Finally, "6" is a multiple of 3 so a(6)= a(5)/3= 1/3!
 
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