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Why does the following pattern hold?

$$1=1=2^0$$

$$\frac{2(2^2-1)}{3}=2=2^1$$

$$\frac{3^2(3^2-1)(3^2-2^2)}{3^2 \times 5}=8=2^3$$

$$\frac{4^2(4^2-1)^2(4^2-2^2)(4^2-3^2)}{3^3 \times 5^2\times 7}=64=2^6$$

$$\frac{5^3(5^2-1)^2(5^2-2^2)^2(5^2-3^2)(5^2-4^2)}{3^4 \times 5^3 \times 7^2 \times 9}=1024=2^{10}$$

$$ n^{\lceil n/2\rceil}\prod_{k=1}^{n-1}\frac {(n^2-k^2)^{\lfloor (n-k+1)/2\rfloor}}{(2k+1)^{n-k}}=2^{n(n-1)/2}.$$

The motivation for the question comes from the following two questions:

Coefficients of binomial continued fractions

Determinants of products of binary matrices and binomial coefficients

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a partial answer: $$ \frac{5^3(5^2-1)^2(5^2-2^2)^2(5^2-3^2)(5^2-4^2)}{3^4 \times 5^3 \times 7^2 \times 9} $$

may be re-written as

$$ \frac{\frac{9!}{0!}\frac{7!}{2!}\frac{5!}{4!}}{\frac{9!}{4!2^4}\frac{7!}{3!2^3}\frac{5!}{2!2^2}\frac{3!}{1!2^1}} $$

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Here we show the claim is valid for odd positive integers $n$. For even $n$ it can be shown analogously.

We start with the denominator, the somewhat easier part.

We obtain for odd integers $n>1$ \begin{align*} \color{blue}{\prod_{k=1}^{n-1}(2k+1)^{n-k}}&=3^{n-1}\cdot5^{n-2}\cdots(2n-3)^2(2n-1)\\ &=\prod_{k=1}^{n-1}\frac{(2n+1-2k)!}{(2n-2k)!!}\tag{1}\\ &=\prod_{k=1}^{n-1}\frac{(2k+1)!}{(2k)!!}\tag{2}\\ &\,\,\color{blue}{=\prod_{k=1}^{n-1}\frac{(2k+1)!}{2^kk!}}\tag{3} \end{align*}

Comment:

  • In (1) we rearrange the right-hand side of the first line by using factorials and double-factorials $(2p)!!=(2p)(2p-2)\cdots4\cdot2$.

  • In (2) we change the order of the factors of the numerator by setting $k\to n-k$.

  • In (3) we use the identities $(2p)!=(2p)!!(2p-1)!!$ and $(2p)!!=2^pp!$.

Now we take a look at the numerator \begin{align*} \color{blue}{n^{\lceil \frac{n}{2}\rceil}}&\color{blue}{\prod_{k=1}^{n-1}\frac {(n^2-k^2)^{\lfloor \frac{n-k+1}{2}\rfloor}}{(2k+1)^{n-k}}}\\ &=\left(\prod_{k=1}^{n-1}\frac {k^{\lfloor \frac{n-k+1}{2}\rfloor}}{(2k+1)^{n-k}}\right)n^{\lceil \frac{n}{2}\rceil} \left(\prod_{k=1}^{n-1}\frac {(n-k)^{\lfloor\frac{n-k+1}{2}\rfloor}}{(2k+1)^{n-k}}\right)\tag{4}\\ &=1^1\cdot2^1\cdot3^2\cdot4^2\cdots(n-2)^{\frac{n-1}{2}}(n-1)^{\frac{n-1}{2}}n^{\frac{n+1}{2}}\\ &\qquad\cdot (n+1)^{\frac{n-1}{2}}(n+2)^{\frac{n-1}{2}}\cdots(2n-4)^2(2n-3)^2(2n-2)^1(2n-1)^1\tag{5}\\ &=(2n-1)!\cdot\frac{(2n-3)!}{2!}\cdot\frac{(2n-5)!}{4!}\cdots\frac{n!}{(n-2)!}\tag{6}\\ &\,\,\color{blue}{=n!\prod_{k=1}^{\frac{n-1}{2}}\frac{(2n+1-2k)!}{(2k)!}}\tag{7} \end{align*}

Comment:

  • In (4) we use $n^2-k^2=(n-k)(n+k)$ and put the factor $n^{\lceil \frac{n}{2}\rceil}$ in the middle. We also change the order of the factors of the numerator at the left-hand product by $k\to n-k$.

  • In (5) we use a more descriptive style for the expression (4).

  • In (6) we collect terms to factorials starting with the right-most factor in steps of two terms. We compensate the missing factors at the left-hand side by multiplying with factors $\frac{1}{(2k)!}$ accordingly.

  • In (7) we write the expression (6) more compactly using the product symbol.

We can now take the results (3) and (7) and obtain \begin{align*} \color{blue}{n^{\lceil \frac{n}{2}\rceil}}&\color{blue}{\prod_{k=1}^{n-1}\frac {(n^2-k^2)^{\lfloor \frac{n-k+1}{2}\rfloor}}{(2k+1)^{n-k}}}\\ &=n!\prod_{k=1}^{\frac{n-1}{2}}\frac{(2n+1-2k)!}{(2k)!}\prod_{k=1}^{n-1}\frac{2^kk!}{(2k+1)!}\tag{8}\\ &=2^{\binom{n}{2}}\prod_{k=1}^{\frac{n-1}{2}}\frac{1}{(2k)!}\prod_{k=1}^{\frac{n-3}{2}}\frac{1}{(2k+1)!}\prod_{k=1}^{n-1}k!\tag{9}\\ &\,\,\color{blue}{=2^{\binom{n}{2}}}\tag{10} \end{align*} and the claim follows.

Comment:

  • In (8) we take the result (3) and multiply it with the reciprocal of (7).

  • In (9) we cancel $\prod_{k=1}^{\frac{n-1}{2}}(2n+1-2k)!$ with $\prod_{k=\frac{n-1}{2}}^{n-1}\frac{1}{(2k+1)!}$ and factor out $\prod_{k=1}^{n-1}{2^k}=2^{\sum_{k=1}^{n-1}k}=2^{\binom{n}{2}}$.

  • In (10) we cancel everything besides the power of $2$.

Note: In order to show the case when $n$ is even we can adapt this answer starting from (5).

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