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


extern crate num;
 
use std::cmp;
use num::bigint::BigUint;
 
fn cumu(n: usize, cache: &mut Vec<Vec<BigUint>>) {
    for l in cache.len()..n+1 {
        let mut r = vec![BigUint::from(0u32)];
        for x in 1..l+1 {
            let prev = r[r.len() - 1].clone();
            r.push(prev + cache[l-x][cmp::min(x, l-x)].clone());
        }
        cache.push(r);
    }
}
 
fn row(n: usize, cache: &mut Vec<Vec<BigUint>>) -> Vec<BigUint> {
    cumu(n, cache);
    let r = &cache[n];
    let mut v: Vec<BigUint> = Vec::new();
 
    for i in 0..n {
        v.push(&r[i+1] - &r[i]);
    }
    v
}
 
fn main() {
    let mut cache = vec![vec![BigUint::from(1u32)]];
 
    println!("rows:");
    for x in 1..26 {
        let v: Vec<String> = row(x, &mut cache).iter().map(|e| e.to_string()).collect();
        let s: String = v.join(" ");
        println!("{}: {}", x, s);
    }
 
    println!("sums:");
    for x in vec![23, 123, 1234, 12345] {
        cumu(x, &mut cache);
        let v = &cache[x];
        let s = v[v.len() - 1].to_string();
        println!("{}: {}", x, s);
    }
}