This task is a variation of the short story by Arthur C. Clarke.

(Solvers should be aware of the consequences of completing this task.)

In detail, to specify what is meant by a   “name”:

The integer 1 has 1 name     “1”.
The integer 2 has 2 names   “1+1”,   and   “2”.
The integer 3 has 3 names   “1+1+1”,   “2+1”,   and   “3”.
The integer 4 has 5 names   “1+1+1+1”,   “2+1+1”,   “2+2”,   “3+1”,   “4”.
The integer 5 has 7 names   “1+1+1+1+1”,   “2+1+1+1”,   “2+2+1”,   “3+1+1”,   “3+2”,   “4+1”,   “5”.


Task
Display the first 25 rows of a number triangle which begins:

                                      1
                                    1   1
                                  1   1   1 
                                1   2   1   1
                              1   2   2   1   1
                            1   3   3   2   1   1

Where row   



n


{\displaystyle n}

   corresponds to integer   



n


{\displaystyle n}

,   and each column   



C


{\displaystyle C}

   in row   



m


{\displaystyle m}

   from left to right corresponds to the number of names beginning with   



C


{\displaystyle C}

.

A function   



G
(
n
)


{\displaystyle G(n)}

   should return the sum of the   



n


{\displaystyle n}

-th   row.

Demonstrate this function by displaying:   



G
(
23
)


{\displaystyle G(23)}

,   



G
(
123
)


{\displaystyle G(123)}

,   



G
(
1234
)


{\displaystyle G(1234)}

,   and   



G
(
12345
)


{\displaystyle G(12345)}

.

Optionally note that the sum of the   



n


{\displaystyle n}

-th   row   



P
(
n
)


{\displaystyle P(n)}

   is the     integer partition function.

Demonstrate this is equivalent to   



G
(
n
)


{\displaystyle G(n)}

   by displaying:   



P
(
23
)


{\displaystyle P(23)}

,   



P
(
123
)


{\displaystyle P(123)}

,   



P
(
1234
)


{\displaystyle P(1234)}

,   and   



P
(
12345
)


{\displaystyle P(12345)}

.



Extra credit
If your environment is able, plot   



P
(
n
)


{\displaystyle P(n)}

   against   



n


{\displaystyle n}

   for   



n
=
1
…
999


{\displaystyle n=1\ldots 999}

.

Related tasks
 Partition function P


 
(function () {
    var cache = [
        [1]
    ];
//this was never needed.
   /* function PyRange(start, end, step) {
        step = step || 1;
        if (!end) {
            end = start;
            start = 0;
        }
        var arr = [];
        for (var i = start; i < end; i += step) arr.push(i);
        return arr;
    }*/ 
 
    function cumu(n) {
        var /*ra = PyRange(cache.length, n + 1),*/ //Seems there is a better version for this
            r, l, x, Aa, Mi;
       // for (ll in ra) { too pythony
       for (l=cache.length;l<n+1;l++) {
            r = [0];
//            l = ra[ll];
//            ran = PyRange(1, l + 1);
//            for (xx in ran) {
            for(x=1;x<l+1;x++){
//                x = ran[xx];
                r.push(r[r.length - 1] + (Aa = cache[l - x < 0 ? cache.length - (l - x) : l - x])[(Mi = Math.min(x, l - x)) < 0 ? Aa.length - Mi : Mi]);
            }
            cache.push(r);
        }
        return cache[n];
    }
 
    function row(n) {
        var r = cumu(n),
//            rra = PyRange(n),
            leArray = [],
            i;
//        for (ii in rra) {
        for (i=0;i<n;i++) {
//            i = rra[ii];
            leArray.push(r[i + 1] - r[i]);
        }
        return leArray;
    }
 
    console.log("Rows:");
    for (iterator = 1; iterator < 12; iterator++) {
        console.log(row(iterator));
    }
 
// PL clearly this was not tested:
//    console.log("Sums")[23, 123, 1234, 12345].foreach(function (a) {
    console.log("Sums");
    [23, 123, 1234, 12345].forEach(function (a) {
        var s = cumu(a);
        console.log(a, s[s.length - 1]);
    });
})()
 