PPS定义

/**
 * @param {number[]} seq 序列
 * @param {number} FSterm 展开次数 
 */
function expandPPS(seq, FSterm) {
    var len = seq.length;
    var x = len > 0 ? seq[len - 1] : null;
    var parentY = x;
    var rootY = null;
    var b = null;
    var badpart = [];
    var L = 0;
    var flag = false;
    if (parentY >= 1 && parentY <= len) {
        rootY = parentY;
        b = seq[rootY - 1];
        badpart = seq.slice(rootY, len - 1);
        L = len - rootY;
        flag = badpart.some((val) => val === b);
    } else {
        L = len - parentY;
    }
    var goodpart = seq.slice(0, -1);
    var result = goodpart.slice();
    for (var i = 1; i <= FSterm; i++) {
        result.push(flag ? b : x - 1);

        var bad_modified = badpart.map((val) => (val < x ? val : val + L * i));
        result = result.concat(bad_modified);
    }
    return result;
}

PPS的极限表达式为0,1,0,2,2,0,5,5,0,8,8,0,11,11,0,14,14,……

分西

\[0,0 = 1,1\] \[0,1 = 1,2\] \[0,1,0,0,3 = 1,2,1,2\] \[0,1,0,0,3,0,0,6 = 1,2,1,2,1,2\] \[0,1,0,0,3,0,0,6,0,0,9 = 1,2,1,2,1,2,1,2\] \[0,1,0,0,3,3 = 1,2,2\] \[0,1,0,0,3,3,0,0,7 = 1,2,2,1,2\] \[0,1,0,0,3,3,0,0,7,0,0,10 = 1,2,2,1,2,1,2\] \[0,1,0,0,3,3,0,0,7,7 = 1,2,2,1,2,2\] \[0,1,0,0,3,3,0,0,7,7,0,0,11,11 = 1,2,2,1,2,2,1,2,2\] \[0,1,0,0,3,3,3 = 1,2,2,2\] \[0,1,0,0,4 = 1,2,3\] \[0,1,0,0,4,0,0,6 = 1,2,3,1,2\] \[0,1,0,0,4,0,0,6,6 = 1,2,3,1,2,2\] \[0,1,0,0,4,0,0,7 = 1,2,3,1,2,3\] \[0,1,0,0,4,0,0,7,0,0,10= 1,2,3,1,2,3,1,2,3\] \[0,1,0,1= 1,2,3,2\] 这里004的强展开使得×ω重复了ω次，01起到了×ω的作用。 \[0,1,0,1= 1,2,3,2\] \[0,1,0,1,0,0,6= 1,2,3,2,1,2,3\] \[0,1,0,1,0,0,6,0,6= 1,2,3,2,1,2,3,2\] \[0,1,0,1,0,1= 1,2,3,2,2\] \[0,1,0,1,0,1,0,1= 1,2,3,2,2,2\] \[0,1,0,2= 1,2,3,2,3\] 这里第三个0,1是×ω的作用。 \[0,1,0,2,0,0,0,7 = 1,2,3,2,3,1,2\] \[0,1,0,2,0,0,0,7,0,7 = 1,2,3,2,3,1,2,3,2\] \[0,1,0,2,0,0,0,7,0,8 = 1,2,3,2,3,1,2,3,2,3\] \[0,1,0,2,0,0,0,7,0,8,0,0,13,0,14 = 1,2,3,2,3,1,2,3,2,3,1,2,3,2,3\] \[0,1,0,2,0,0,1 = 1,2,3,2,3,2\] 这里001是×ω的作用 \[0,1,0,2,0,0,1,0,0,1 = 1,2,3,2,3,2,2\] 这里的06相当于×ω重复ω次的操作重复ω次（*双层指数+1） \[0,1,0,2,0,0,1,0,6 = 1,2,3,2,3,2,3\] \[0,1,0,2,0,0,1,0,6,0,0,1,0,11 = 1,2,3,2,3,2,3\] \[0,1,0,2,0,0,1,0,6,0,6 = 1,2,3,3\] \[0,1,0,2,0,0,1,0,7 = 1,2,3,4\] 这里07强展开变成06,从而三层指数+1。 \[0,1,0,2,0,0,1,0,7,0,0,1 = 1,2,3,4,2\] \[0,1,0,2,0,0,1,0,7,0,0,1,0,12 = 1,2,3,4,2,3,4\] \[0,1,0,2,0,0,1,0,7,0,0,1,0,12,0,0,1,0,17 = 1,2,3,4,2,3,4,2,3,4\] 02重复的是0102xxx的内容，展开过程中意外把06升级成07，造成了圣诞树效应。 \[0,1,0,2,0,0,2 = 1,2,3,4,3\] \[0,1,0,2,0,0,2,0,0,1 = 1,2,3,4,3,2\] \[0,1,0,2,0,0,2,0,0,1,0,10 = 1,2,3,4,3,2,3,4\] 这里的0010A是09重复。09是001重复，0A自然是×ω^ω^ω。 01020020010(10)00(10)由于和0102002有相似的性质。 \[0,1,0,2,0,0,2,0,0,1,0,10,0,0,0,1 = 1,2,3,4,3,2,3,4,2\] \[0,1,0,2,0,0,2,0,0,1,0,10,0,0,0,1,0,16 = 1,2,3,4,3,2,3,4,2,3,4\] \[0,1,0,2,0,0,2,0,0,1,0,10,0,0,9 = 1,2,3,4,3,2,3,4,3\] \[0,1,0,2,0,0,2,0,0,1,0,10,0,0,9,0,0,0,1 = 1,2,3,4,3,2,3,4,3,2\] \[0,1,0,2,0,0,2,0,0,1,0,10,0,0,9,0,0,0,1,0,19 = 1,2,3,4,3,2,3,4,3,2,3,4\] \[0,1,0,2,0,0,2,0,0,1,0,10,0,0,9,0,0,0,1,0,19,0,0,18 = 1,2,3,4,3,2,3,4,3,2,3,4,3\] \[0,1,0,2,0,0,2,0,0,1,0,10,0,0,9,0,0,9 = 1,2,3,4,3,3\] \[0,1,0,2,0,0,2,0,0,1,0,10,0,0,9,0,0,9,0,0,0,1 = 1,2,3,4,3,3,2\] \[0,1,0,2,0,0,2,0,0,1,0,10,0,0,9,0,0,9,0,0,0,1,0,22 = 1,2,3,4,3,3,2,3,4\] \[0,1,0,2,0,0,2,0,0,1,0,10,0,0,9,0,0,9,0,0,0,1,0,22,0,0,21,0,0,21 = 1,2,3,4,3,3,2,3,4,3,3\] \[0,1,0,2,0,0,2,0,0,1,0,10,0,0,9,0,0,9,0,0,9 = 1,2,3,4,3,3,3\] \[0,1,0,2,0,0,2,0,0,1,0,10,0,0,9,0,14 = 1,2,3,4,3,4\] \[0,1,0,2,0,0,2,0,0,1,0,10,0,0,9,0,14,0,0,9 = 1,2,3,4,3,4,3\] \[0,1,0,2,0,0,2,0,0,1,0,10,0,0,9,0,14,0,0,9,19 = 1,2,3,4,3,4,3,4\] \[0,1,0,2,0,0,2,0,0,1,0,10,0,0,9,0,14,0,14 = 1,2,3,4,4\] \[0,1,0,2,0,0,2,0,0,1,0,10,0,0,9,0,15 = 1,2,3,4,5\] \[0,1,0,2,0,0,2,0,0,1,0,10,0,0,9,0,15,0,0,9,0,20 = 1,2,3,4,5,3,4,5\] 00A和002一样，也意外把090(14)升级成了090(15)，发生了圣诞树效应。 \[0,1,0,2,0,0,2,0,0,1,0,10,0,0,10 = 1,2,3,4,5,4\] \[0,1,0,2,0,0,2,0,0,1,0,10,0,0,10,0,0,1,0,18,0,0,18 = 1,2,3,4,5,4,2,3,4,5,4\] 由于002也意外升级了00(10),所以也发生了圣诞树效应。 \[0,1,0,2,0,0,2,0,0,2 = 1,2,3,4,5,4,3\] 接下来继续分析 \[0,1,0,2,0,0,2,0,0,2,0,0,1 = 1,2,3,4,5,4,3,2\] \[0,1,0,2,0,0,2,0,0,2,0,0,1,0,12 = 1,2,3,4,5,4,3,2,3\] \[0,1,0,2,0,0,2,0,0,2,0,0,1,0,13 = 1,2,3,4,5,4,3,2,3,4\] \[0,1,0,2,0,0,2,0,0,2,0,0,1,0,13,0,0,0,1 = 1,2,3,4,5,4,3,2,3,4,2\] \[0,1,0,2,0,0,2,0,0,2,0,0,1,0,13,0,0,0,1,0,19 = 1,2,3,4,5,4,3,2,3,4,2,3,4\] \[0,1,0,2,0,0,2,0,0,2,0,0,1,0,13,0,0,12 = 1,2,3,4,5,4,3,2,3,4,3\] \[0,1,0,2,0,0,2,0,0,2,0,0,1,0,13,0,0,12,0,0,12 = 1,2,3,4,5,4,3,2,3,4,3,3\] \[0,1,0,2,0,0,2,0,0,2,0,0,1,0,13,0,0,12,0,17 = 1,2,3,4,5,4,3,2,3,4,3,4\] \[0,1,0,2,0,0,2,0,0,2,0,0,1,0,13,0,0,12,0,18 = 1,2,3,4,5,4,3,2,3,4,5\] \[0,1,0,2,0,0,2,0,0,2,0,0,1,0,13,0,0,12,0,18,0,0,12,0,23 = 1,2,3,4,5,4,3,2,3,4,5,2,3,4,5\] \[0,1,0,2,0,0,2,0,0,2,0,0,1,0,13,0,0,13 = 1,2,3,4,5,4,3,2,3,4,5,4\] \[0,1,0,2,0,0,2,0,0,2,0,0,1,0,13,0,0,13,0,0,0,1,0,22,0,0,22 = 1,2,3,4,5,4,3,2,3,4,5,4,2,3,4,5,4\] \[0,1,0,2,0,0,2,0,0,2,0,0,1,0,13,0,0,13,0,0,12 = 1,2,3,4,5,4,3,2,3,4,5,4,3\] \[0,1,0,2,0,0,2,0,0,2,0,0,1,0,13,0,0,13,0,0,12,0,0,12 = 1,2,3,4,5,4,3,3_{\color{Grey} 1,2,3,4,5,4,3,2,3,4,5,4,3,3}\] \[0,1,0,2,0,0,2,0,0,2,0,0,1,0,13,0,0,13,0,0,12,0,20 = 1,2,3,4,5,4,3,4\] \[0,1,0,2,0,0,2,0,0,2,0,0,1,0,13,0,0,13,0,0,12,0,20,0,20 = 1,2,3,4,5,4,3,4,4\] \[0,1,0,2,0,0,2,0,0,2,0,0,1,0,13,0,0,13,0,0,12,0,21 = 1,2,3,4,5,4,3,4,5\] \[0,1,0,2,0,0,2,0,0,2,0,0,1,0,13,0,0,13,0,0,12,0,21,0,0,0,12 = 1,2,3,4,5,4,3,4,5,3\] \[0,1,0,2,0,0,2,0,0,2,0,0,1,0,13,0,0,13,0,0,12,0,21,0,0,0,12 = 1,2,3,4,5,4,3,4,5,3\] \[0,1,0,2,0,0,2,0,0,2,0,0,1,0,13,0,0,13,0,0,12,0,21,0,0,20 = 1,2,3,4,5,4,4\] \[0,1,0,2,0,0,2,0,0,2,0,0,1,0,13,0,0,13,0,0,12,0,21,0,0,20,0,25 = 1,2,3,4,5,4,5\] \[0,1,0,2,0,0,2,0,0,2,0,0,1,0,13,0,0,13,0,0,12,0,21,0,0,20,0,26 = 1,2,3,4,5,6_{\color{Grey}1,2,3,4,5,4,5,6}\] \[0,1,0,2,0,0,2,0,0,2,0,0,1,0,13,0,0,13,0,0,12,0,21,0,0,20,0,26,0,0,20,0,31 = 1,2,3,4,5,6,4,5,6_{\color{Grey}1,2,3,4,5,4,5,6,4,5,6}\] \[0,1,0,2,0,0,2,0,0,2,0,0,1,0,13,0,0,13,0,0,12,0,21,0,0,21 = 1,2,3,4,5,6,5\] \[0,1,0,2,0,0,2,0,0,2,0,0,1,0,13,0,0,13,0,0,13 = 1,2,3,4,5,6,5,4\] \[0,1,0,2,0,0,2,0,0,2,0,0,2 = 1,2,3,4,5,6,5,4,3\] 002造成了圣诞树效应,00(13)造成了圣诞树效应，00(21)，00(20)0(26)也造成了圣诞树效应。 根据上述分析，我们可以得出 \[0,1,0,2,0,0,2,0,0,2,0,0,2,0,0,2 = 1,2,3,4,5,6,7,6,5,4,3\] \[0,1,0,2,0,0,2,0,0,2,0,0,2,0,0,2,0,0,2 = 1,2,3,4,5,6,7,8,7,6,5,4,3\]

\[0,1,0,2,0,3 = 1,3\]

在部分记号可寻找的地方

我们上文说过：“02重复的是0102xxx的内容，展开过程中意外把06升级成07，造成了圣诞树效应。”。 这种意外升级在BMS中有迹可循。 在\((0)(1,1,1)(2,1)(1,1,1)\)的展开过程中，(2,1)被升级成了(3,2)，这和PPS的“意外升级”一样。

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