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


'print triangle
> n, 1..25
  k = 50-n*2
  #.output(#.str("","<"+k+"<"),#.rs)
  > k, 1..n
    i = p(n,k)
    s = #.str(i,">3<")
    ? k<n, s += " "+#.rs
    #.output(s)
  <
<
p(n,k)=
  ? k=0 | k>n, <= 0
  ? k=n, <= 1
  <= p(n-1,k-1)+p(n-k,k)
.
 
'calculate partition function
#.output()
#.output("G(23) =    ",g(23))
#.output("G(123) =   ",g(123))
#.output("G(1234) =  ",g(1234))
#.output("G(12345) = ",g(12345))
g(n)=
  p[1] = 1
  > i, 2..n+1
    j = 2
    k,p[i] = 0
    > j>1
      k += 1
      j = i-#.lower((3*k*k+k)/2)
      ? j!<1, p[i] -= (-1)^k*p[j]
      j = i-#.lower((3*k*k-k)/2)
      ? j!<1, p[i] -= (-1)^k*p[j]
    <
  <
  <= p[n+1]
.