Proving that $u_{n}\in\mathbb{Z}\forall n\ge 0$, such that $\det \begin{bmatrix}u_{n} & u_{n+1} \\ u_{n+2}&u_{n+3}\end{bmatrix}=n!\forall n\ge 0$

Question

Define a sequence $u_n\forall n\ge 0$ such that $u_{0},u_{1},u_{2}=1$ and $\det\begin{bmatrix}u_n & u_{n+1} \\ u_{n+2} & u_{n+3}\end{bmatrix}=n! \forall n\ge 0$. Prove that $u_n\in\mathbb{Z}\forall n\ge 0$.

As of now I have just gone on to find the values of the sequence to see if they form a pattern. The first few terms come out to be $1,1,1,2,3,5,8,15, 48, 105$. No obvious pattern is visible. Also induction is used, will have to be multi counter induction, and I am not familiar with its application. Also if we try to replace $n$ by $n-1$, we get following expressions.

$$\begin{aligned}u_{n}u_{n+3}-u_{n+1}u_{n+2}&=n!\\ u_{n-1}u_{n+2}-u_{n}u_{n+1}&=(n-1)!\end{aligned}$$

On subtracting we get the equality $u_{n}(u_{n+3}-u_{n+1})-u_{n+2}(u_{n+1}-u_{n-1})=(n-1)^{2}(n-2)!$. I can not see how to proceed. Any hints are appreciated. Thanks.

Answer 1

The pattern is $$ u_2 = 1 = 1 u_0 \\ u_3 = 2 = 2 u_1 \\ u_4 = 3 = 3 u_2 \\ u_5 = 8 = 4 u_3 \\ u_6 = 15 = 5 u_4 \\ u_7 = 48 = 6 u_5 $$ and generally, $u_n = (n-1) u_{n-2}$ for $n \ge 2$. It follows that $$ u_n = (n-1)!! = (n-1)(n-3)(n-5) \cdots $$ for $n \ge 1$ where $!!$ denotes the “double factorial.” It is now easy to verify that $$ \det\begin{bmatrix}u_n & u_{n+1} \\ u_{n+2} & u_{n+3}\end{bmatrix} = u_n u_{n+1}\det\begin{bmatrix}1 & 1 \\ {n+1} & {n+2}\end{bmatrix} = u_n u_{n+1} = n! \, . $$ Apart from the initial term $u_0 = 1$ this is A006882 in the On-Line Encyclopedia of Integer Sequences®.

Answer 2

An idea is from $u_{n}u_{n+3} - u_{n+1}u_{n+2}= n!$, see if $u_{n}| n!+u_{n+1}u_{n+2}$ for small values of $n$ and induct on that Too. Let $A(n)=\begin{bmatrix}u_n & u_{n+1} \\ u_{n+2} & u_{n+3}\end{bmatrix}$

Suppose that for all $k \leq n$: $A(n)$ has integer entries and that $A_{11}(k)| k!+ A_{2,1}(k)A_{1,2}(k)$.

Then prove that this still hold for $k=n+1$.