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


V cache = [[BigInt(1)]]
F cumu(n)
   L(l) :cache.len .. n
      V r = [BigInt(0)]
      L(x) 1 .. l
         r.append(r.last + :cache[l - x][min(x, l - x)])
      :cache.append(r)
   R :cache[n]
 
F row(n)
   V r = cumu(n)
   R (0 .< n).map(i -> @r[i + 1] - @r[i])
 
print(‘rows:’)
L(x) 1..10
   print(‘#2:’.format(x)‘ ’row(x))
 
print("\nsums:")
 
V pp = [BigInt(1)]
 
F partitions(n)
   :pp.append(BigInt(0))
 
   L(k) 1 .. n
      V d = n - k * (3 * k - 1) I/ 2
      I d < 0
         L.break
 
      I k [&] 1 != 0
         :pp[n] += :pp[d]
      E
         :pp[n] -= :pp[d]
 
      d -= k
      I d < 0
         L.break
 
      I k [&] 1 != 0
         :pp[n] += :pp[d]
      E
         :pp[n] -= :pp[d]
 
   R :pp.last
 
V ns = Set([23, 123, 1234, 12345])
V max_ns = max(ns)
 
L(i) 1 .. max_ns
   I i > max_ns
      L.break
   V p = partitions(i)
   I i C ns
      print(‘#6: #.’.format(i, p))