Solving Sequence Problem: Explicit Sequence for 1, 1, 2, 2...

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In summary, it is possible to find an explicit sequence for the given pattern without using the greatest integer function or a recurrence formula. The odd terms are (n+1)/2, n = 1, 3, 5, 7, ... and the even terms are n/2, n = 2, 4, 6, 8, ... By using the alternating (-1)^n, the sequence can be described as a, b, a, b, a, b, a, b, ... where a = (n+1)/2 and b = n/2. This can be extended for sequences with repeating terms of width n by using \lfloor \frac{m}{n} \rf
  • #1
Kopake
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I was wondering, is it possible to find an explicit sequence for: 1, 1, 2, 2, 3, 3, 4 ,4 ,5 ,5 ,6 ,6 etc.
without using the greatest integer function, or a recurrency formula? Any help would be appreciated.
 
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  • #2
whats wrong with recurrence?
 
  • #3
nothing wrong with recurrence, its just that my professor said its possible to find a_n without using it and I can't figure it out
 
  • #4
are you in calc 2?
 
  • #5
The odd terms are (n+1)/2, n = 1, 3, 5, 7, ...
The even terms are n/2, n = 2, 4, 6, 8, ...

All you have to do is to figure out how to alternate between the two functions and that can be achived by using the alternating (-1)^n

To warm up, figure out how to describe the sequence
a, b, a, b, a, b, a, b, ...
using the alternating (-1)^n.

After you figure that, you can just plug in the two formulas from your original sequence instead of a, b.
 
  • #6
smallphi said:
The odd terms are (n+1)/2, n = 1, 3, 5, 7, ...
The even terms are n/2, n = 2, 4, 6, 8, ...

All you have to do is to figure out how to alternate between the two functions and that can be achived by using the alternating (-1)^n

To warm up, figure out how to describe the sequence
a, b, a, b, a, b, a, b, ...
using the alternating (-1)^n.

After you figure that, you can just plug in the two formulas from your original sequence instead of a, b.

oooooooooooooooooooooooooooooo
 
  • #7
yep, I'm currently in calc 2, and thanks a bunch smallphi!
 
  • #8
smallphi said:
The odd terms are (n+1)/2, n = 1, 3, 5, 7, ...
The even terms are n/2, n = 2, 4, 6, 8, ...

All you have to do is to figure out how to alternate between the two functions and that can be achived by using the alternating (-1)^n

To warm up, figure out how to describe the sequence
a, b, a, b, a, b, a, b, ...
using the alternating (-1)^n.

After you figure that, you can just plug in the two formulas from your original sequence instead of a, b.

both of those series yield even numbers
 
  • #9
ice109 said:
both of those series yield even numbers

Eh? Look again.
 
  • #10
JohnDuck said:
Eh? Look again.

what? for n=1,3,5,7,9 the first sequence yields even numbers.
for n=2,4,6,8,10 the second sequence yields even numbers?
 
  • #11
Sloan's is your friend

Many possibles http://www.research.att.com/~njas/sequences/?q=1%2C1%2C2%2C2%2C3%2C3%2C4%2C4%2C5%2C5%2C6%2C6%2C7%2C7%2C8%2C8&language=english&go=Search"
 
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  • #12
ice109 said:
what? for n=1,3,5,7,9 the first sequence yields even numbers.
for n=2,4,6,8,10 the second sequence yields even numbers?

For the odd numbers:
1 -> (1+1)/2 = 1
3 -> (3+1)/2 = 2
5 -> (5+1)/2 = 3

Even numbers:

2 -> 2/2 = 1
4 -> 4/2 = 2
6 -> 6/2 = 3
 
  • #13
Not sure if the floor function is legal for your purposes but if so, this works

m = {1, 2, 3, 4, ...}

[tex]a = \lfloor \frac{m}{2} \rfloor + 1[/tex]

Added later:

Oops! Sorry didn't see this on first read: "without using the greatest integer function"
 
Last edited:
  • #14
This works without using the floor or ceil functions

m = {0, 1, 2, 3, ... }

[tex]a = \frac{2 * m + (-1) ^ m - 1}{4}[/tex]
 
  • #15
In an earlier post, I came up with this to answer the OP

m = {0, 1, 2, 3, ... }

[tex]a = \frac{2 * m + (-1) ^ m - 1}{4}[/tex]

But I was wondering whether this can be extended to produce sequences with repeating terms of width n? The [tex]-1^m[/tex] trick doesn't seem to help.

It's trivially easy to do it with floor

m = {0, 1, 2, ... }
n = {1, 2, 3, ...}

[tex]a_m = \lfloor \frac{m}{n} \rfloor[/tex]

But what about without floor?
 

FAQ: Solving Sequence Problem: Explicit Sequence for 1, 1, 2, 2...

What is an explicit sequence?

An explicit sequence is a sequence where the rule or formula for finding the value of each term is explicitly stated. This means that you can easily determine the value of any term in the sequence without having to rely on previous terms.

What is the rule for finding the value of each term in the sequence 1, 1, 2, 2...

The rule for finding the value of each term in this sequence is to repeat the previous term twice. So the first term is 1, the second term is also 1, then 2, 2, and so on.

What is the nth term in this sequence?

The nth term in this sequence is 1 if n is odd, and 2 if n is even. This can also be written as a piecewise function: f(n) = 1 if n is odd, and f(n) = 2 if n is even.

How can I use this sequence to solve problems?

This sequence can be used to solve problems involving patterns, counting, and probability. By understanding the rule for finding the value of each term, you can use it to predict future terms in the sequence or determine missing values.

Are there other ways to generate this sequence?

Yes, there are different ways to generate this sequence. For example, you can use a recursive formula where each term is equal to the previous term plus 1. So the first term is 1, the second term is 1 + 1 = 2, the third term is 2 + 1 = 3, and so on.

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