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


 
Red []
 
context [
    sum-part: function [nums [block!] count [integer!]][
        out: 0.0
        loop count [
            out: out + nums/1 
            if empty? nums: next nums [break]
        ]
        out
    ]
    nums: make map! [1 [1] 2 [1 1]]
    sums: make map! [1 1 2 2]
    set 'names function [row /show /all][
        if row < 1 [cause-error 'user 'message "Argument needs to be >= 1"]
        if show [
            unless nums/:row [names row]
            repeat i row [either all [probe reduce [i nums/:i sums/:i]][print nums/:i]]
        ]
        either sums/:row [sums/:row][
            out: clear []
            half: to integer! row / 2
            if row - 1 > last: length? nums [
                repeat i row - last - 1 [names last + i]
            ]
            repeat col row - 1 [
                either col = (half + 1) [
                    append out at nums/(row - 1) half
                    break
                ][
                    append out sum-part nums/(row - col) col
                ]
            ]
            also sums/:row: sum nums/:row: copy out  clear out
        ]
    ]
]
 
print "rows: ^/"
names/show 25
print "^/sums: ^/"
probe names 23
probe names 123 
probe names 1234
 
 