$10$ Equations in $10$ variables

Question

Define,

$$F_k = xp^k + yq^k + zr^k + us^k + vt^k$$

Let,

$$F_0 = 2$$

$$F_1 = 3$$

$$F_2 = 16$$

$$F_3 = 31$$

$$F_4 = 103$$

$$F_5 = 235$$

$$F_6 = 674$$

$$F_7 = \color{blue}{1667}$$

$$F_8 = 4526$$

$$F_9 = 11595$$

Solve for $x, y, z, u, v, p, q, r, s, t$

Can anyone post a solution if you have solved it? Thanks.

$\color{blue}{Edit}$:

It seems there is a typo. Compare to OEIS A072684:

$$2, 3, 16, 31, 103, 235, 674, \color{blue}{1669}, 4526, 11595,\dots$$

Answer 1

I don't think there is closed form solution in $\mathbb R$. I fed the equation into Mathematica and the solution involves some simultaneous quintic equations.

The numerical solution is:

p =  1.0950321639628354969
q = -1.0576149539068368954
r = -1.6352442469443802722
s =  2.6240480141706460457
t =  1.7260360724137917277

x = -1.0539164346585613344
y = -0.43941289847517655928
z =  1.2243270533315074403
u =  1.9766293143625121080
v =  0.29237296543971834535

Answer 2

Up to permutation of following $5$ pairs of parameters, We have $$\begin{array}{ccrrr} (x,p) &\approx& (&1.224327053331513,&-1.63524424694438),\\ (y,q) &\approx& (&-0.4394128984751902,&-1.057614953906837),\\ (z,r) &\approx& (&-1.053916434658547,& 1.095032163962836),\\ (u,s) &\approx& (&0.2923729654397071,&1.726036072413793),\\ (v,t) &\approx& (&1.976629314362518,&2.624048014170645) \end{array}$$

and $(p,q,r,s,t)$ are roots of the polynomial

$${\lambda}^{5}-\frac{167973}{61031}\lambda^4-\frac{222201}{61031}\lambda^3 +\frac{649807}{61031}\lambda^2 + \frac{165745}{61031}\lambda - \frac{523491}{61031} = 0$$

The coefficients $$(\alpha_0,\alpha_1,\alpha_2,\alpha_3,\alpha_4) = \left( \frac{523491}{61031},-\frac{165745}{61031},-\frac{649807}{61031},\frac{222201}{61031},\frac{167973}{61031} \right)$$ are solution for following matrix equation

$$\begin{bmatrix} 2 & 3 & 16 & 31 & 103\\ 3 & 16 & 31 & 103 & 235\\ 16 & 31 & 103 & 235 & 674\\ 31 & 103 & 235 & 674 & 1667\\ 103 & 235 & 674 & 1667 & 4526 \end{bmatrix} \begin{bmatrix} \alpha_0\\ \alpha_1\\ \alpha_2\\ \alpha_3\\ \alpha_4 \end{bmatrix} = \begin{bmatrix} 235\\ 674\\ 1667\\ 4526\\ 11595 \end{bmatrix} \tag{*1} $$

The basic idea goes like this. Instead of solving $p,q,r,s,t$, we look at following polynomial first:

$$(\lambda -p)(\lambda - q)(\lambda - r)(\lambda - s)(\lambda - t) = \lambda^5 - (\alpha_4 \lambda^4 + \alpha_3\lambda^3 + \alpha_2\lambda^2 + \alpha_1\lambda + \alpha_0)$$

It is easy to see the list of power sums

$$u_k = x p^k + y q^k + z r^k + u s^k + v t^k$$

satisfies a recurrence relation of the form:

$$u_{n+5} = \alpha_4 u_{n+4} + \alpha_3 u_{n+3} + \alpha_2 u_{n+2} + \alpha_1 u_{n+1} + \alpha_0 u_{n}$$

This means one can backout the parameters $\alpha_0,\ldots,\alpha_4$ using the supplied power sums. This is the rationale behind equation $(*1)$.

Once we have $\alpha_0, \ldots, \alpha_4$, we can use whatever means to locate the roots of the quintic polynomial. The roots of the quintic polynomial will be our variables $(p, q, r, s, t)$.

After we obtain $(p,q,r,s,t)$, it is trivial to setup linear equations to compute the corresponding $(x,y,z,u,v)$. I will skip the details here.