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


package main
 
import (
	"fmt"
	"math/big"
)
 
func main() {
 
	intMin := func(a, b int) int {
		if a < b {
			return a
		} else {
			return b
		}
	}
 
	var cache = [][]*big.Int{{big.NewInt(1)}}
 
	cumu := func(n int) []*big.Int {
		for y := len(cache); y <= n; y++ {
			row := []*big.Int{big.NewInt(0)}
			for x := 1; x <= y; x++ {
				cacheValue := cache[y-x][intMin(x, y-x)]
				row = append(row, big.NewInt(0).Add(row[len(row)-1], cacheValue))
			}
			cache = append(cache, row)
		}
		return cache[n]
	}
 
	row := func(n int) {
		e := cumu(n)
		for i := 0; i < n; i++ {
			fmt.Printf(" %v ", (big.NewInt(0).Sub(e[i+1], e[i])).Text(10))
		}
		fmt.Println()
	}
 
	fmt.Println("rows:")
	for x := 1; x < 11; x++ {
		row(x)
	}
	fmt.Println()
 
	fmt.Println("sums:")
	for _, num := range [...]int{23, 123, 1234, 12345} {
		r := cumu(num)
		fmt.Printf("%d %v\n", num, r[len(r)-1].Text(10))
	}
}