递归数列

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#递归

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求 $\begin{cases}k(n+3)=k(n)+k(n+1)+k(n+2)\\k(1)=k(2)=k(3)=1\end{cases}$ 的通项公式 $k(n)$ 。

RSolve[{k[n+3]==k[n+2]+k[n+1]+k[n],k[1]==k[2]==k[3]==1},k[n],n]

⎧ ⎩ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ a = 1 3 (1 + 19 - 3 33 - - \sqrt - - - - - - - - - \sqrt 3 + 19 + 3 33 - - \sqrt - - - - - - - - - \sqrt 3) b = 1 3 - 1 6 (1 + i 3 \sqrt) 19 - 3 33 - - \sqrt - - - - - - - - - \sqrt 3 - 1 6 (1 - i 3 \sqrt) 19 + 3 33 - - \sqrt - - - - - - - - - \sqrt 3 c = 1 3 - 1 6 (1 - i 3 \sqrt) 19 - 3 33 - - \sqrt - - - - - - - - - \sqrt 3 - 1 6 (1 + i 3 \sqrt) 19 + 3 33 - - \sqrt - - - - - - - - - \sqrt 3 d = - 1 3 - 1 66 (1 + i 3 \sqrt) 847 - 33 33 - - \sqrt - - - - - - - - - - \sqrt 3 - ( 1 - i 3 \sqrt ) 77 + 3 33 - - \sqrt - - - - - - - - - \sqrt 3 6 11 2 / 3 e = - 1 3 - 1 66 (1 - i 3 \sqrt) 847 - 33 33 - - \sqrt - - - - - - - - - - \sqrt 3 - ( 1 + i 3 \sqrt ) 77 + 3 33 - - \sqrt - - - - - - - - - \sqrt 3 6 11 2 / 3 f = 1 33 (- 11 + 847 - 33 33 - - \sqrt - - - - - - - - - - \sqrt 3 + 11 (77 + 3 33 - - \sqrt) - - - - - - - - - - - - \sqrt 3) k (n) = a n d n + b n f n + c n e n

$\begin{cases} a=\dfrac{1}{3}\left(1+\sqrt[3]{19-3\sqrt{33}}+\sqrt[3]{19+3 \sqrt{33}}\right)\\ b=\dfrac{1}{3}-\dfrac{1}{6}\left(1+i\sqrt{3}\right)\sqrt[3]{19-3 \sqrt{33}}-\dfrac{1}{6}\left(1-i\sqrt{3}\right)\sqrt[3]{19+3\sqrt{33}} \\ c=\dfrac{1}{3}-\dfrac{1}{6}\left(1-i\sqrt{3}\right)\sqrt[3]{19-3 \sqrt{33}}-\dfrac{1}{6}\left(1+i\sqrt{3}\right)\sqrt[3]{19+3\sqrt{33}} \\ d=-\dfrac{1}{3}-\dfrac{1}{66}\left(1+i\sqrt{3}\right)\sqrt[3]{847-33 \sqrt{33}}-\dfrac{\left(1-i\sqrt{3}\right)\sqrt[3]{77+3\sqrt{33}}}{6\ 11^{2/3}} \\ e=-\dfrac{1}{3}-\dfrac{1}{66}\left(1-i\sqrt{3}\right)\sqrt[3]{847-33 \sqrt{33}}-\dfrac{\left(1+i\sqrt{3}\right)\sqrt[3]{77+3\sqrt{33}}}{6\ 11^{2/3}} \\ f=\dfrac{1}{33}\left(-11+\sqrt[3]{847-33\sqrt{33}}+\sqrt[3]{11\left(77+3 \sqrt{33}\right)}\right) \\ {\Huge{k(n)=a^nd^n+b^nf^n+c^ne^n}} \end{cases}$