再来几个求数列之通项公式的难题:

眼光落地

<p><font color="#4d2bd5"><strong><em>&nbsp;再来几个求数列之通项公式的难题:</em></strong></font></p><p><font size="3"><font color="#ee1169"><strong>&nbsp;求数列之通项公式</strong><font size="2">(能用Mathematica 5.2编程吗)</font>
                        </font><strong>:</strong></font></p><p><font color="#000000">&nbsp;&nbsp; 1. 数列f_(n) = f_{(n - 1)} + 2 * f_{(n - 3)} + 1,&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;f₀&nbsp;= 0;&nbsp;f₁ = 1; f₂ = 2 .&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;前几项为{0,1,2,3,6,11,18,31, ... ...}</font></p><p><font color="#000000">&nbsp;&nbsp; 2. 数列f_(n) = f_{(n - 1)} + 3 * f_{(n - 4)} + 1,&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp; f₀&nbsp;= 0;&nbsp;f₁ = 1; f₂ = 2;&nbsp; f₃ = 3 .&nbsp;&nbsp;</font></p><p><font color="#000000">&nbsp;&nbsp; 3. 数列f_{(n) ⁼}f_{(n - 1)} + m _^* f_{(n - (m+1))} + 1,&nbsp;&nbsp; &nbsp;f₀&nbsp;= 0;&nbsp;f₁ = 1; f₂ = 2;&nbsp; f₃ = 3;&nbsp;&nbsp;&nbsp;... ;f_(m) = m .</font>&nbsp;&nbsp;&nbsp;</p>

sightqun

<p>function result=shulie(m,n)<br/>%返回的result便是所求数列的前n项<br/>if n-m&lt;2<br/>&nbsp;&nbsp;&nbsp; error('使用出错');<br/>end</p><p>result=zeros(n+1,1);</p><p>for i=1:m+1<br/>&nbsp;&nbsp;&nbsp; result(i,1)=i-1;<br/>end</p><p>for i=m+2:n+1<br/>&nbsp;&nbsp;&nbsp; result(i,1)=result(i-1,1)+m*result(i-m-1,1)+1;<br/>end</p>

眼光落地

<p><font color="#7963d0"><strong><font style="BACKGROUND-COLOR: #c7ebbe;">看不懂呀,<font size="2">Mathematica 5.2函数首字母应该用大写嘛,请写出完整的程序吗????</font></font></strong></font></p><p><strong><font style="BACKGROUND-COLOR: #c7ebbe;">能写出递推关系及通项公式吗</font></strong></p>

眼光落地

很复杂,公式里很可能有复数,有简明的公式来表示吗?????????????/

眼光落地

谁有办法呀!!!!!!!!!!!!!!!!!!