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

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?

• It might helps you. math.stackexchange.com/questions/3521010/… – dust05 Jun 3 '20 at 6:04
• Actually I understood that but I quite couldn’t relate it much with my question, I would be glad if you can explain me further – Asv Jun 3 '20 at 7:14
• I think I misunderstood your point. – dust05 Jun 3 '20 at 9:48

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

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.

• Thanks a lot, actually I wasn’t able to see that a, b and c would be the roots of a cubic – Asv Jun 3 '20 at 10:32