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Re: [Unsolved] How many "magic square" (combinatorial)? MY SOLUTION

My solution

$\boxed 1$ How many ways writing positive integer $n\;$ as the sum of the positive integers different each pairs? (no permutation) Example: $6=6=1+5=2+4=1+2+3 \quad $ (4 ways) $\boxed 2$ How many ways writing positive integer $n\;$ as the sum of the positive integers? (no permutation)...

Re: What to do when can't get Latex formatted correctly? I want show codecogs by formula $\displaystyle \sum_{k=1}^{n-1}\sin\left(\frac{\left(2\left\lfloor\sqrt{kn}\right\rfloor+1\right)\pi}{2n}\right)=\cot\left(\frac{\pi}{2n}\right)\cos\left(\frac{\pi}{2n}\right)$ but no preview :(...

Re: [Unsolved] How many "magic square" (combinatorial)? $n^2$ non-negative integers is not necessarily different! $\displaystyle S_3(0)={0+2\choose 4}+{0+3\choose 4}+{0+4\choose 4}=1$ $$\begin{matrix}0&0&0\\0&0&0\\0&0&0\end{matrix}$$ $\displaystyle S_3(1)={1+2\choose 4}+{1+3\choose...

A "magic square" is an $n\times n$ table includes $n^2$ non-negative integers satisfying conditions sum on each row and each column are equal and equal to $r$. Define $S_3(r)$ is number of all $3\times 3$ magic square with row sum is $r$ Prove that: $\displaystyle S_3(r)={r+2\choose...

Your result is absolute correct!(Clapping) Hint: $(1-x^2)^m(1+x)^{2m}=(1-x)^m(1+x)^{3m}$ Good luck! (Sun)

Evaluate sum: $\displaystyle S=\sum_{k=0}^{2n}(-1)^k{2n\choose k}{4n\choose 2k}$

very nice solution! Some similar results $a_n=1+\left\lceil\dfrac{n(n-2)}{3}\right\rceil$ $a_n=1+(2n-2)\left\lfloor\dfrac{n}{3}\right\rfloor-3\left\lfloor\dfrac{n}{3}\right\rfloor^2$ Also, by induction show that $a_n-a_{n-1}=\left\lfloor\dfrac{2(n-1)}{3}\right\rfloor$

$a_0$ is'nt importan! because $a_1=a_0+\left\lfloor\dfrac{1^2-2.1+2-a_0}{1}\right\rfloor=1$

Define $\{a_n\}$ is integer sequences (all term are integers) satisfy condition $a_n=a_{n-1}+\left\lfloor\dfrac{n^2-2n+2-a_{n-1}}{n}\right\rfloor $ for $n=1,2,...$ *note: $\left\lfloor x\right\rfloor$ is a greatest integer number less than or equal $x$ Find general term of sequences.

Re: Find the Sum My solution

Put $1\le n\in\mathbb Z$ Find the Sum: $S_n=\displaystyle \sum_{k=1}^n\dfrac{2k+1-n}{(k+1)^2(n-k)^2+1}$

Prove that: $\displaystyle\sum_{k=0}^n \frac{\cos(k x)}{\cos^kx} = \frac{1+(-1)^n}{2\cos^nx} + \dfrac{2\sin\big(\lfloor\frac{n+1}{2}\rfloor x\big) \cos\big(\lfloor\frac{n+2}{2}\rfloor x\big)} {\sin x\cos^n x} \qquad\qquad (\frac{2x}{\pi}\not\in \mathbb Z)$ *note: $\lfloor x\rfloor$ is floor...