Cómo encontrar esta secuencia $a_{n}$ forma cerrada si tal condición extraña

Question

Dejemos que $\{a_{n}\}$ tal $$a_{0}=a_{1}=a_{2}=1,a_{3}=2,a_{4}=3,a_{5}=4,a_{6}=10,a_{7}=18$$ y tal $$\begin{vmatrix} a_{n+8}&a_{n+7}&a_{n+6}\\ a_{n+5}&a_{n+4}&a_{n+3}\\ a_{n+2}&a_{n+1}&a_{n} \end{vmatrix}=2\cdot (n+1)!$$ Encuentre el $a_{n}$ formulario colsed

o $$a_{n+8}a_{n+4}a_{n}+a_{n+5}a_{n+1}a_{n+6}+a_{n+2}a_{n+3}a_{n+7} -a_{n+6}a_{n+4}a_{n+2}-a_{n+7}a_{n+5}a_{n}-a_{n+8}a_{n+1}a_{n+3}=2\cdot(n+1)!$$

si encontramos $a_{n}$ forma colsed, creo que fácil de inducción matemática resolverlo?

Answer 1

Los primeros términos calculados son : $1,1,1,2,3,4,10,18,28,80,162,280,880$

Así que la fórmula cerrada parece ser $\bbox[5px,border:2px solid]{\begin{cases}a_0=1\\a_1=1\\a_2=1\\a_n=(n-1)a_{n−3}\quad n\ge3\end{cases}}$

Veamos cómo es el determinante bajo la hipótesis de recurrencia hasta $a_{n+7}$ .

$D=\begin{array}{l} +a_{n+8}a_{n+4}a_{n}\\ +a_{n+5}a_{n+1}a_{n+6}\\ +a_{n+2}a_{n+3}a_{n+7}\\ -a_{n+6}a_{n+4}a_{n+2}\\ -a_{n+7}a_{n+5}a_{n}\\ -a_{n+8}a_{n+1}a_{n+3}\end{array} = \begin{array}{l} +(n+3)a_{n+8}a_{n+1}a_{n} \\ +(n+4)(n+5)(n+2)a_{n+2}a_{n+1}a_{n} \\ +(n+2)(n+6)(n+3)a_{n+2}a_{n}a_{n+1} \\ -(n+5)(n+2)(n+3)a_{n}a_{n+1}a_{n+2} \\ -(n+6)(n+3)(n+4)a_{n+1}a_{n+2}a_{n} \\ -(n+7)(n+4)(n+2)a_{n+2}a_{n+1}a_{n} \end{array}$

$=(n+3)a_{n+8}a_{n+1}a_{n}+[-n^3-14n^2-61n-82]\,a_{n+2}a_{n+1}a_{n}$

$=(n+3)a_{n+8}a_{n+1}a_{n}-[(n+3)(n+7)(n+4)-2]\,a_{n+2}a_{n+1}a_{n}$

$=(n+3)a_{n+1}a_{n}\times[a_{n+8}-(n+7)(n+4)a_{n+2}]+2a_{n+2}a_{n+1}a_{n}$

$=(n+3)a_{n+1}a_{n}\times[a_{n+8}-(n+7)a_{n+5}]+2a_{n+2}a_{n+1}a_{n}$

Examinemos para $n=3k,\ 3k+1,\ 3k+2$ el producto $a_{n+2}a_{n+1}a_{n}$

$a_{n+2}a_{n+1}a_{n}=(n+1)(n)(n-1)a_{n-1}a_{n-2}a_{n-3}=\ldots=(n+1)(n+0)(n-1)\ldots(4)(3)(2)a_2a_1a_0=(n+1)!\times 1\times 1\times 1=(n+1)!$

$a_{n+2}a_{n+1}a_{n}=(n+1)(n+0)(n-1)\ldots(5)(4)(3)a_3a_2a_1=\frac{(n+1)!}{2}\times 2\times 1\times 1=(n+1)!$

$a_{n+2}a_{n+1}a_{n}=(n+1)(n+0)(n-1)\ldots(6)(5)(4)a_4a_3a_2=\frac{(n+1)!}{2\times 3}\times 3\times 2\times 1=(n+1)!$

Así que $D=(n+3)a_{n+1}a_{n}\times[a_{n+8}-(n+7)a_{n+5}]+2\,(n+1)!$

Esto permite concluir que $a_{n+8}-(n+7)a_{n+5}=0$ que pone fin a la recurrencia.