如何迭代计算此序列?

问题描述

我要迭代计算此序列.

A(0,j)=j+1
A(i,0)=A(i-1,0)
A(i,j)=A(i-1,A(i,j-1))

这是我的尝试

    public function calculsuite1Action($i,$j)
{
    $A = array();
    for ($k = 0; $k <= $i * $i * $i + $j; $k++) {
        $A[0][$k] = $k + 1;
    }
    for ($c = 1; $c <= $i; $c++) {
        for ($k = 0; $k <= $i * $i * $i + $j - $c; $k++) {
            if ($k == 0) {
                $A[$c][$k] = $A[$c - 1][1];
            } else {
                $A[$c][$k] = $A[$c - 1][$A[$c][$k - 1]];
            }
            if ($c == $i && $k == $j) {
                return $A[$i][$j];
            }

        }
    }
}

我正在尝试使用PHP或任何其他编程语言找到解决方案.我该怎么办?

I'm trying to find a solution using PHP or any other programming language.How can I do that?

推荐答案

我尝试为前几个条目寻找模式.如果我没有犯任何错误，则序列非常简单.归结为

I have tried looking for a pattern for the first few entries. If I haven´t done any mistakes the sequence is pretty simple. It boils down to

A(i, j) = j+1

只需使用此JavaScript程序来验证我没有犯任何错误

Just used this JavaScript program to verify I didn't make any mistakes

for (i = 0; i < 5; i++){
  for (j = 0; j < 5; j++){
    console.log("A("+i+", "+j+") = "+calc(i,j));
  }
}

function calc(i, j){
  if(i==0)
    return j+1;
  else if(j == 0)
    return calc(i-1, 0);
  else
    return calc(i-1, calc(i, j-1));
}

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