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Thread: nth term of a sequence

  1. #1
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    nth term of a sequence

    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
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  2. #2
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    Quote Originally Posted by p vs np View Post
    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 $\displaystyle A_n$ be the n-th term of the sequence. it's easy to see that your sequence can be written as: $\displaystyle A_{\frac{m(m-1)}{2} + k}=m, \ \ m \geq 1, \ 1 \leq k \leq m.$ now let:

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

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

    taking square root in (1) and (2) will give us: $\displaystyle m < \sqrt{2n} + \frac{1}{2} < m+1,$ which means: $\displaystyle \left \lfloor \sqrt{2n} + \frac{1}{2} \right \rfloor=m=A_n.$
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  3. #3
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    Here is another way to generate that sequence: $\displaystyle 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,; $\displaystyle \frac{{n(n + 1)}}{2}$.
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  4. #4
    Senior Member bkarpuz's Avatar
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    Cool

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

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

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

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

    $\displaystyle \therefore$ Plato's answer is right!
    Last edited by bkarpuz; Sep 9th 2008 at 09:02 PM. Reason: [math] n=\sum\limits_{i=1}^{a_{n}}i=\frac{a_{n}(a_{n}+1)}{2}. [/math] is corrected.
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