### 15.7.10 Finding linear recurrences

The reverse_rsolve command finds a linear recurrence relation given the first few terms.

- reverse_rsolve takes v=[v0,…,v2n−1], a vector made of the first 2n terms of a sequence (vn) which is supposed to verify a linear recurrence relation xnvn+k+⋯+x0vk=0 of degree smaller than n, where the xj are n+1 unknowns.
- reverse_rsolve(v) returns the list x=[xn,…,x0] (if xn≠ 0 then xn is reduced to 1).

In other words, reverse_rsolve solves the linear system of n equations of the form

xnvn+k+⋯+x0vk=0,k=0,1,…,n−1.

The matrix A of the system has n rows and n+1 columns:

A=

⎡
⎢
⎢
⎢
⎢
⎢
⎣

vn | ⋯ | v0 | 0
⋮ |  | ⋮ | ⋮
vn+k | ⋯ | vk | 0
⋮ |  | ⋮ | ⋮
v2n−1 | ⋯ | vn−1 | 0

⎤
⎥
⎥
⎥
⎥
⎥
⎦

reverse_rsolve returns the solution x=[xn,…,x1,x0] of Ax=0 which satisfies xn=1.

#### Examples

Find a sequence satisfying a linear recurrence of degree at most 2 whose first elements are 1, −1, 3, 3.

reverse_rsolve([1,-1,3,3])

| ⎡ ⎣ | 1,−3,−6 | ⎤ ⎦

Hence x0=−6, x1=−3, x2=1, and the recurrence relation is vk+2 −3vk+1 −6 vk=0.

Find a sequence satisfying a linear recurrence of degree at most 3 whose first elements are 1, −1, 3, 3,−1, 1.

reverse_rsolve([1,-1,3,3,-1,1])

⎡
⎢
⎢
⎣

1,−

1
2

,

1
2

,−1

⎤
⎥
⎥
⎦

Hence, x0=−1, x1=1/2, x2=−1/2, x3=1, and the recurrence relation is

vk+3−

1
2

vk+2+

1
2

vk+1−vk=0.