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Prove that $n$th Term of the sequence $1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,\cdots$ is given by

$$T_n =\left[\sqrt{2n+\frac{1}{2}}\right]$$ where $[.]$ is Floor function

My Try:

Its clear that $1$st Term is $1$, $3$rd Term is $2$, $6$th term is $3$ and so on

We have Triangular numbers as $1,3,6,10,...$ whose $m$th term is given by $\frac{m(m+1)}{2}$

Hence for the given sequence

$$T_{\frac{m(m+1)}{2}}=m$$

Letting $$\frac{m(m+1)}{2}=n$$ we get Quadratic in $m$ as

$$m^2+m-2n=0$$ So $m$ takes $\frac{-1\pm \sqrt{1+8n}}{2}$ and since $m$ is a positive integer we have

$$m=\frac{-1+\sqrt{1+8n}}{2}$$

hence

$$T_n=\frac{-1+\sqrt{1+8n}}{2}$$

how to proceed further?

Umesh shankar
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    Your expression for $T_n$ is correct. It is confusing to reuse $n$ as a variable from the question and $T_n$ often represents the triangular numbers which makes your expression backwards.. As a MathJax hint we use \lfloor and \rfloor, which render as $\lfloor$ and $\rfloor,$ to represent the floor function. – Ross Millikan Jan 06 '18 at 03:51
  • However your $T_n$ is only correct when $n$ is a triangular number. – Ross Millikan Jan 06 '18 at 03:58
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    @J.-E.Pin: I don't think this is a duplicate. It is the same sequence, but this question asks to justify a particular formula that is not one of the ones mentioned in that question. – Ross Millikan Jan 06 '18 at 04:02
  • This is more or less a shot in the dark, but have you considered induction? – gen-ℤ ready to perish Jan 06 '18 at 04:20
  • The interest of the question refered to by @J.-E. Pin by is its reference to http://oeis.org/A002024 – Jean Marie Jan 06 '18 at 05:47
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    Beware: your formula should be $\lfloor \sqrt{2n}+\tfrac12\rfloor$, i.e., with1/2 term outside the square root. – Jean Marie Jan 06 '18 at 05:56

1 Answers1

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Note that $$T_n=\frac{-1+\sqrt{1+8n}}{2}$$ is true only if $$n=m(m+1)/2.$$ Now you have to define $T_n$ for other values of $n$. Since terms repeat between two such triangular numbers and you know the value of $T_n$ for $n=m(m+1)/2$ the rest of the solution should be obvious.

Mohammad Riazi-Kermani
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