Given two constants, a and b, and a sequence constructed as follows:
1, 1, 1, 1…(a – 1 times), 2, 2, 2, 2,…(a-2) times,……a-1
For example, for a = 6,
1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5
Ques: Determine the bth term in the sequence
- This is opposite to the sequence given here: Finding the nth element of a sequence
I am not working for a proof here, any pointers to the explanation would help loads. Thank you
Best Answer
Let's flip the sequence to give,
5, 4, 4, 3, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1
Now if we perform, $a-t_n$ for all the terms it gives us,
1, 2, 2, 3, 3, 3.....5
We know the term $t_n = \lfloor \sqrt{2n} + \frac{1}{2} \rfloor$. Our term for the flipped sequence is just $t_n = a-\lfloor \sqrt{2n} + \frac{1}{2} \rfloor$.
The index of the original sequence w.r.t the flipped sequence is $\frac{a(a-1)}{2} - n$. So putting it together, $$ t_{\left(\frac{a(a-1)}{2} - n\right) } = a-\lfloor \sqrt{2n} + \frac{1}{2} \rfloor $$
Replacing $n$ with $\frac{a(a-1)}{2} - n$, $$ t_{n} = a-\lfloor \sqrt{2\left(\frac{a(a-1)}{2} - n\right) } + \frac{1}{2} \rfloor $$