How Many Distinct Numbers Are in This Mathematical Sequence?

  • MHB
  • Thread starter MarkFL
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    2016
In summary, distinct numbers in a list are unique and have no duplicates. To determine them, each number in the list must be compared with all other numbers. A list can have only one distinct number if all numbers are the same. Finding distinct numbers can be useful in identifying patterns and outliers in the data. They can be in any order, as long as they are unique.
  • #1
MarkFL
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MHB
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anemone has asked me to fill in for her this week.

Here is this week's POTW:

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How many distinct numbers are in the list

\(\displaystyle \frac{1^2-1+4}{1^2+1},\,\frac{2^2-2+4}{2^2+1},\,\frac{3^2-3+4}{3^2+1},\,\cdots,\frac{2011^2-2011+4}{2011^2+1}\)

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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  • #2
Congratulations to the following members for their correct solutions:

  • kaliprasad
  • lfdahl

Here is the solution anemone provided:

We can rewrite the general term of this sequence that contains for a total of \(\displaystyle 2011\) terms as

\(\displaystyle a_n=\frac{n^2-n+4}{n^2+1}=1-\frac{n-3}{n^2+1}\), where $n\in \mathbb{Z}^+.$

We want to look for the terms, says, \(\displaystyle a_n\) and \(\displaystyle a_m\) that have the same value and see how many such pairs are there in the given sequence.

Since \(\displaystyle a_n\) and \(\displaystyle a_m\) are equal, we get:

\(\displaystyle 1-\frac{n-3}{n^2+1}=1-\frac{m-3}{m^2+1}\)

Upon simplifying gives:

\(\displaystyle \frac{n-3}{n^2+1}=\frac{m-3}{m^2+1}\)

\(\displaystyle (n-3)(m^2+1)=(m-3)(n^2+1)\)

\(\displaystyle nm^2+n-3m^2-3=mn^2+m-3n^2-3\)

\(\displaystyle nm^2-mn^2-m+n-3m^2+3n^2=0\)

\(\displaystyle nm(m-n)-(m-n)-3(m^2-n^2)=0\)

\(\displaystyle nm(m-n)-(m-n)-3(m-n)(m+n)=0\)

\(\displaystyle (m-n)(nm-1-3(m+n))=0\)

We know $n=m$ leads to the trivial case, we will assume that $n\ne m$.

Hence we get $nm-1-3(m+n)=0\implies m(n-3)=3n+1$. From here we can tell $n\ne 3$ or it leads us to $3n+1=0$, which is a contradiction.

Thus, \(\displaystyle m=\frac{3n+1}{n-3}=3+\frac{10}{n-3}\). Since $m$ is an integer, we know the denominator $n-3$ has to divide $10$, we have only four possibilities:

$n-3=1,\,\implies n=4,\,m=13$

$n-3=2,\,\implies n=5,\,m=8$

$n-3=5,\,\implies n=8,\,m=5$

$n-3=10,\,\implies n=13,\,m=4$

We therefore find there are only two pairs of terms that are equal, namely $a_4=a_{13}=\dfrac{16}{17}$ and $a_5=a_{8}=\dfrac{12}{13}$.

We can conclude by now that there are $2011-2=2009$ distinct numbers in the given sequence.
 

FAQ: How Many Distinct Numbers Are in This Mathematical Sequence?

What is the definition of distinct numbers in a list?

Distinct numbers in a list refer to numbers that are unique and do not have any duplicates in the given list.

How do you determine the distinct numbers in a list?

To determine the distinct numbers in a list, you need to compare each number in the list with all the other numbers and remove any duplicates. This process can be done manually or by using a computer program.

Can a list have only one distinct number?

Yes, a list can have only one distinct number if all the numbers in the list are the same. In this case, the distinct number would be the only number in the list.

What is the significance of finding distinct numbers in a list?

Finding distinct numbers in a list can help identify patterns or outliers in the data. It can also be useful in data analysis and statistics.

Are distinct numbers in a list always in a specific order?

No, distinct numbers in a list can be in any order. The only requirement is that they are unique and do not have any duplicates.

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