Factorial sequence

Algebra Level 5

Consider the sequence $(u_{n})_{n}$ defined by $u_{0}=u_{1}=u_{2}=1$ and $\displaystyle \det \begin{pmatrix} u_{n+3} & u_{n+2} \\ u_{n+1} & u_{n} \end{pmatrix} = n!$ for $n \ge 0$ .

Find the remainder when $u_{17}$ is divided by 17.

The answer is 13.

1 solution

Ajinkya Shivashankar
Feb 14, 2017

Relevant wiki: Double Factorials and Multifactorials

Expand the determinant to get reccurence relation for $u_n$ as $u_{n+3} u_n-u_{n+2}u_{n+1}=n!$ . Note that for small numbers computation is easy so we shall go ahead and find the first few terms as 1,1,1,2,3,8,15,48,105,384,945 which is the Double Factorial and its 17th term is 10321920 which on dividing by 17 gives answer as 13

Now for a proof by induction, Let $u_n=(n-1)!!$ As double factorial either contains only even or only odd terms, multiplying 2 of different parity gives the factorial.

$u_{n+3} \times u_{n}=(n+2)(n!)$ and $u_{n+2} \times u_{n+1}=(n+1)(n!)$ (written this way to simplify the next step, even though it is just (n+1)!)

Subtracting the 2 equations to obtain LHS $=(n!)=$ RHS.

All that remains is to check the base case And we are done.

Factorial sequence

The answer is 13.

1 solution

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