nth term of a sequence

**p vs np** · September 9th 2008, 01:59 AM

Any suggestions for an approach to proving that the nth term of the following sequence:

1,2,2,3,3,3,4,4,4,4,5,5,5...

is [ ((2*n)^(1/2) + (1/2)) ], where [x] denotes the floor value of x and ^ denotes to the power of

**NonCommAlg** · September 9th 2008, 06:12 AM

Originally Posted by p vs np

Any suggestions for an approach to proving that the nth term of the following sequence:

1,2,2,3,3,3,4,4,4,4,5,5,5...

is [ ((2*n)^(1/2) + (1/2)) ], where [x] denotes the floor value of x and ^ denotes to the power of

let $A_n$ be the n-th term of the sequence. it's easy to see that your sequence can be written as: $A_{\frac{m(m-1)}{2} + k}=m, \ \ m \geq 1, \ 1 \leq k \leq m.$ now let:

$\frac{m(m-1)}{2}+k=n,$ where $m \geq 1$ and $1 \leq k \leq m.$ so $A_n=m,$ and: $8n=4m^2-4m+8k=(2m-1)^2+8k-1.$ thus : $8n > (2m-1)^2. \ \ \ (1)$

also we have: $8n=(2m+1)^2-8(m-k)-1.$ thus: $8n < (2m+1)^2. \ \ \ \ (2)$

taking square root in (1) and (2) will give us: $m < \sqrt{2n} + \frac{1}{2} < m+1,$ which means: $\left \lfloor \sqrt{2n} + \frac{1}{2} \right \rfloor=m=A_n.$

**Plato** · September 9th 2008, 07:57 AM

Here is another way to generate that sequence: $a_n = \left\lceil {\frac{{ - 1 + \sqrt {8n + 1} }}{2}} \right\rceil$ .
That is of course using the ceiling function.
Notice the sequence changes on the triangular numbers, 1,3,6,10,…; $\frac{{n(n + 1)}}{2}$ .

**bkarpuz** · September 9th 2008, 11:31 AM

First of all, I have to say that this is a very good exercise.

$ \begin{array}{cccccccccccc} n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & \cdots \\ a_{n} & 1 & 2 & 2 & 3 & 3 & 3 & 4 & 4 & 4 & 4 & \cdots\\ & * & & * & & & * & & & & * & \end{array} $

Consider the terms marked with $*$ , for these numbers, we have $n=\sum\limits_{i=1}^{a_{n}}i=\frac{a_{n}(a_{n}+1)} {2}.$

Now, let $a_{n_{0}}\leq a_{n}\leq a_{n^{0}}$ , where $n^{0}\geq n$ is the smallest number which is not less than $n$ and satisfies $n^{0}=\frac{a_{n^{0}}(a_{n^{0}}+1)}{2}$ (in short, lets say $*$ property) and $n_{0}\leq n$ is the greatest number which is less than $n$ and satisfies $*$ property.
Therefore, solving this parabola with, we have $a_{n^{0}}=\frac{-1\pm\sqrt{1+8n^{0}}}{2}$ . Note that the desired term is the positive one, hence $a_{n^{0}}=\frac{-1+\sqrt{1+8n^{0}}}{2}$ . Thus, the solution is completed when $n$ has $*$ property.

If $n$ does not have $*$ property, then we see that $n_{0}<n<n^{0}$ , you can easily show that $a_{n}=\left\lceil\frac{-1+\sqrt{1+8n}}{2}\right\rceil$ , because of the definitions of $n_{0},n^{0}$ , being $a_{n_{0}}+1=a_{n^{0}}=a_{n}$ and the sequezing $n_{0}=\frac{a_{n_{0}}(a_{n_{0}}+1)}{2}\leq n\leq\frac{a_{n^{0}}(a_{n^{0}}+1)}{2}=n^{0}.$

$\therefore$ Plato's answer is right!

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