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, where[ ((2*n)^(1/2) + (1/2)) ][x]denotes thefloor value of xand^denotesto the power of

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- Sep 9th 2008, 01:59 AMp vs npnth 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, where*[ ((2*n)^(1/2) + (1/2)) ]**[x]*denotes the*floor value of x*and*^*denotes*to the power of* - Sep 9th 2008, 06:12 AMNonCommAlg
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: $\displaystyle \left \lfloor \sqrt{2n} + \frac{1}{2} \right \rfloor=m=A_n.$ http://www.mathhelpforum.com/math-he...ags/Canada.gif - Sep 9th 2008, 07:57 AMPlato
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,…; $\displaystyle \frac{{n(n + 1)}}{2}$. - Sep 9th 2008, 11:31 AMbkarpuz
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! (Nod)