Find an explicit formula for $f(n)=f(n-1)+f(n-2)/2+f(n-3)/6$ when $f(1)=1, f(2)=2, f(3)=3$

Question

I was trying to solve the set of equations $a+b+c=1, a^2+b^2+c^2=2, a^3+b^3+c^3=3$ for $a^n+b^n+c^n=x$ I took $x=f(n)$ and found that $f(n)=f(n-1)+f(n-2)/2+f(n-3)/6$ when $f(1)=1, f(2)=2, f(3)=3$ I then went on to try to find an explicit formula but I don’t know how to get started, can anyone please help?

Answer 1

What do you want to find? the values of $a, b,c $ or the general equation for $f_n$? I'm not sure your requirement, but two questions are related.

You have explicit formula, $f(n) = a^n + b^n + c^n$. So it is enough to find each values for $a, b, c$.

Let $a+b+c = A$, $ab + bc + ca = B$, $abc = C$. Then $a, b, c$ are three roots of $x^3 - A x^2 +Bx - C = 0$.

We have $A = f_1 = 1$. Note that $f_1^2 = f_2 + 2B$, i.e. $B = -1/2$. Also one can deduce that \begin{align*} f_1^3 &= a^3 + 3 a^2 b + 3 a^2 c + 3 a b^2 + 6 a b c + 3 a c^2 + b^3 + 3 b^2 c + 3 b c^2 + c^3\\ & = f_3 + 3ab(a+b) + 3bc(b+c) + 3ca(c+a) + 6C \\ & = f_3 + 3ab(A - c) + 3bc(A-a) + 3ca(A-b) + 6C \\ &= f_3 + 3AB - 9 C + 6C\\ 1& = 3 -3/2 - 3C \\ C& = 1/6 \end{align*} so $a, b, c$ satisfies $x^3 - x^2 - x/2 -1/6 =0$. Actually, this is exactly the characteristic polynomial of the recurrence formula you found. It dose not have rational solutions, so let three solutions be $\alpha, \beta, \gamma$, then we have $$ f(n) = \alpha^n+ \beta^n + \gamma^n$$

Numerically, $\alpha \approx 1.4308$, $\beta \approx -0.21542 - 0.26471 i$, $\gamma =\overline{\beta}$.

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I'm not sure if i understood your question. Are you interested in the following general situation? :

$x_n = a_1^n + \cdots + a_k^n$, $x_1 = 1, x_2 = 2, \cdots, x_k = k$

In this case, if you proceed as above, this would be helpful.