Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The recurrence series is given by $$ t_{n+1} = t_n + 2\left\lfloor{n -1 \over 4} +1\right\rfloor , \; t_0 = 1$$

What would be the closed form of $t_n$ in terms of $n$ and it's sum up to $n$ terms? The sequence generated can be found on here on OEIS.

share|cite|improve this question
What's the initial value, say $t_0$ or $t_1$? – Patrick Li Sep 29 '12 at 13:13
@PatrickLi sorry ... the value of $t_0 = 1$ – Monkey D. Luffy Sep 29 '12 at 13:15
up vote 1 down vote accepted

First note that the sequence of $2\lfloor \frac{n-1}{4} + 1 \rfloor$ is $$2, 2, 2, 2, 4, 4, 4, 4, 6, 6, 6, 6, \dots$$

For any integer $n$, let $k$ be the largest integer such that $4k \leq n$. In other words, $k=\lfloor \frac{n}{4}\rfloor$.

First we sum over the first $4k$ terms which gives us $$ 4\times 2 + 4\times 4 + \dots + 4\times 2k = 8(1+2+\dots + k) = 4k(k+1)$$

Next let's add the remaining terms from $4k+1$ to $n$. There are $n-4k$ terms,each of them being $2(k+1)$. So the final result is $$4k(k+1)+2(n-4k)(k+1)+1$$ where the last value of $1$ comes from the initial value $t_0=1$.

share|cite|improve this answer

Ignore $t_0$ and $t_1$, the first terms $-1$ are: $\;(1\cdot2)\;(2\cdot2)\;(3\cdot2)\;(4\cdot2)\;(3\cdot4)\;(4\cdot4)\;(5\cdot4)\cdots$ which suggests:$$t_{n}=2{\lfloor\frac{n-1}{4}+1\rfloor}\cdot \left(n-1-2\lfloor\frac{n-1}{4}\rfloor\right)+1$$

(this looks more complicated than it actually is)

Note that when $n$ is a multiple of $4$ we have $\lfloor\frac{n-1}{4}\rfloor=\lfloor\frac{n}{4}-1\rfloor$ and otherwise $\lfloor\frac{n-1}{4}\rfloor=\lfloor\frac{n}{4}\rfloor$

$\bullet\;\;n$ multiple of $4:$

$$t_{n+1}=2{\lfloor\frac{n}{4}+1\rfloor}\cdot \left(n-2\lfloor\frac{n}{4}\rfloor\right)+1=2{\lfloor\frac{n}{4}\rfloor}\cdot \left(n-2\lfloor\frac{n}{4}\rfloor\right)+n+1$$

$$=2{\lfloor\frac{n}{4}\rfloor}\cdot \left(n-2\lfloor\frac{n}{4}\rfloor\right)+2{\lfloor\frac{n}{4}\rfloor}\cdot 2+1=2{\lfloor\frac{n}{4}\rfloor}\cdot \left(n+2-2\lfloor\frac{n}{4}\rfloor\right)+1$$

$$=2{\lfloor\frac{n}{4}\rfloor}\cdot \left(n-1-2\lfloor\frac{n}{4}-1\rfloor\right)+2\lfloor\frac{n}{4}\rfloor+1$$$$=2{\lfloor\frac{n-1}{4}+1\rfloor}\cdot \left(n-1-2\lfloor\frac{n-1}{4}\rfloor\right)+2\lfloor\frac{n-1}{4}+1\rfloor+1$$ $$=t_n+2\lfloor\frac{n-1}{4}+1\rfloor$$

$\bullet\;\;n$ not a multiple of $4:$

$$t_{n+1}=2{\lfloor\frac{n}{4}+1\rfloor}\cdot \left(n-2\lfloor\frac{n}{4}\rfloor\right)+1=2{\lfloor\frac{n-1}{4}+1\rfloor}\cdot \left(n-2\lfloor\frac{n-1}{4}\rfloor\right)+1$$

$$=2{\lfloor\frac{n-1}{4}+1\rfloor}\cdot \left(n-1-2\lfloor\frac{n-1}{4}\rfloor\right)+2\lfloor\frac{n-1}{4}+1\rfloor+1=t_n+2\lfloor\frac{n-1}{4}+1\rfloor$$

This verifies our assumption. The sum was nicely done by Patrick Li so I'll leave it there.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.