Recurrence relation system?

Question

Given the following recurrence relation for $a_n, b_n, c_n, d_n,e_n,n \in \Bbb{N}$:

\begin{cases} a_{n+1} = a_n + b_n + c_n+d_n, \\ b_{n+1} = -2a_n -\frac{5}{2}b_n-3c_n-3d_n-e_n, \\ c_{n+1} = \frac{5}{2}a_n +3b_n+\frac{7}{2}c_n+3d_n+2e_n \\ d_{n+1} = -\frac{5}{2}a_n -\frac{3}{2}b_n-\frac{3}{2}c_n-d_n-e_n \\ e_{n+1} = (9a_n+5b_n+3c_n+d_n+e_n)/2 \\ \end{cases}

Is there a way to get something like

$a_{n+1} = p a_n + qa_{n-1} + ra_{n-2}+....$

$b_{n+1} = e b_n + fb_{n-1} + gb_{n-2}+....$

...

Thanks in advance!

Edit: accidentally I found:

$a_{n+1}=-a_{n-1}$

Answer 1

This is a sequence given by the linear recursion implemented by the matrix $$ A= \begin{bmatrix} 1 & 1 & 1 & 1 & 0 \\ -2 & -\frac{5}{2} & -3 & -3 & -1 \\ \frac{5}{2} & 3 & \frac{7}{2} & 3 & 2 \\ -\frac{5}{2} & -\frac{3}{2} & -\frac{3}{2} & -1 & -1 \\ \frac{9}{2} & \frac{5}{2} & \frac{3}{2} & \frac{1}{2} & \frac{1}{2} \end{bmatrix} $$ Its characteristic polynomial is, computed with sage,

sage: A = matrix( QQ, 5, 5,
....:     [ 1, 1, 1, 1, 0, 
....:       -2, -5/2, -3, -3, -1, 
....:       5/2, 3, 7/2, 3, 2, 
....:       -5/2, -3/2, -3/2, -1, -1, 
....:       9/2, 5/2, 3/2, 1/2, 1/2] )
sage: A.charpoly()                                                                                                                                                                                    
x^5 - 3/2*x^4 + 7/4*x^3 - 13/8*x^2 + 3/4*x - 1/8
sage: A.charpoly().factor()                                                                                                                                                                           
(x - 1/2)^3 * (x^2 + 1)

This implies that, determined by the first five values of each of the sequences $(a_n)$, $(b_n)$, $(c_n)$, $(d_n)$, $(e_n)$, all other values are a linear combination of the expressions $$ (1/2)^n\ ,\ n(1/2)^n\ ,\ n^2(1/2)^n\ ;\ i^n\ ,\ (-i)^n\ . $$ In particular, each of the five sequences satisfies a linear recursion given by (the coefficients of) the characteristic polynomial of $A$.

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Note: The "accidental relation" satisfied by $(a_n)$ depends on the initial values.

Answer 2

general method
This is a linear homogeneous system with constant coefficients. Write it using vectors $\mathbf{x}_n = [a_n, b_n, c_n, d_n, e_n]^T$ using a $5\times 5$ matrix $E$ of coefficients: $$ \mathbf{x}_{n+1}=E\,\mathbf{x}_n $$ General solution... $\mathbf{x}_n = E^n \, \mathbf{x}_0$.
Using the minimal polynomial $p(x)$ for the matrix $E$, we get $p(E) = 0$. So using the coefficients of that polynomial, we get linear homogeneous recurrence equation (of order at most $5$, with constant coefficients) for $a_n, b_n, c_n, d_n, e_n$.

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this case
The matrix is $$ E = \left[ \begin {array}{ccccc} 1&1&1&1&0\\ -2&-5/2&-3 &-3&-1\\ 5/2&3&7/2&3&2\\ -5/2&-3/2 &-3/2&-1&-1\\ 9/2&5/2&3/2&1/2&1/2\end {array} \right] $$ with minimal polynomial $$ p(x) = {x}^{5}-\frac{3}{2}\,{x}^{4}+\frac{7}{4}\,{x}^{3}-{\frac {13\,}{8}}{x}^{2}+\frac{3}{4}\,x-\frac{1}{8} $$ The matrix $E$ satisfies $$ {E}^{5}= \frac{3}{2}\,{E}^{4}-\frac{7}{4}\,{E}^{3}+{\frac {13\,}{8}}{E}^{2}-\frac{3}{4}\,E+\frac{1}{8} I $$ Multiply by $E^{n-5}$ $$ {E}^{n}= \frac{3}{2}\,{E}^{n-1} -\frac{7}{4}\,{E}^{n-2} +{\frac {13\,}{8}}{E}^{n-3}- \frac{3}{4}\,E^{n-4}+\frac{1}{8} E^{n-5} $$ Apply to the vector $\mathbf{x}_0$ $$ {E}^{n}\,\mathbf{x}_0 = \frac{3}{2}\,{E}^{n-1}\,\mathbf{x}_0 -\frac{7}{4}\,{E}^{n-2}\,\mathbf{x}_0 +{\frac {13\,}{8}}{E}^{n-3}\,\mathbf{x}_0 -\frac{3}{4}\,E^{n-4}\,\mathbf{x}_0 +\frac{1}{8} E^{n-5}\,\mathbf{x}_0 $$ or $$ \mathbf{x}_{n}= \frac{3}{2}\,\mathbf{x}_{n-1} -\frac{7}{4}\,\mathbf{x}_{n-2} +{\frac {13\,}{8}}\mathbf{x}_{n-3} -\frac{3}{4}\,\mathbf{x}_{n-4} +\frac{1}{8} \mathbf{x}_{n-5} $$ The five coordinates of this vector equation are \begin{align} a_{n}&= \frac{3}{2}\,a_{n-1} -\frac{7}{4}\,a_{n-2} +{\frac {13\,}{8}}a_{n-3} -\frac{3}{4}\,a_{n-4} +\frac{1}{8} a_{n-5} \\ b_{n}&= \frac{3}{2}\,b_{n-1} -\frac{7}{4}\,b_{n-2} +{\frac {13\,}{8}}b_{n-3} -\frac{3}{4}\,b_{n-4} +\frac{1}{8} b_{n-5} \\ c_{n}&= \frac{3}{2}\,c_{n-1} -\frac{7}{4}\,c_{n-2} +{\frac {13\,}{8}}c_{n-3} -\frac{3}{4}\,c_{n-4} +\frac{1}{8} c_{n-5} \\ d_{n}&= \frac{3}{2}\,d_{n-1} -\frac{7}{4}\,d_{n-2} +{\frac {13\,}{8}}d_{n-3} -\frac{3}{4}\,d_{n-4} +\frac{1}{8} d_{n-5} \\ e_{n}&= \frac{3}{2}\,e_{n-1} -\frac{7}{4}\,e_{n-2} +{\frac {13\,}{8}}e_{n-3} -\frac{3}{4}\,e_{n-4} +\frac{1}{8} e_{n-5} \end{align}