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


#include <stdio.h>
#include <gmp.h>
 
#define N 100000
mpz_t p[N + 1];
 
void calc(int n)
{
	mpz_init_set_ui(p[n], 0);
 
	for (int k = 1; k <= n; k++) {
		int d = n - k * (3 * k - 1) / 2;
		if (d < 0) break;
 
		if (k&1)mpz_add(p[n], p[n], p[d]);
		else	mpz_sub(p[n], p[n], p[d]);
 
		d -= k;
		if (d < 0) break;
 
		if (k&1)mpz_add(p[n], p[n], p[d]);
		else	mpz_sub(p[n], p[n], p[d]);
	}
}
 
int main(void)
{
	int idx[] = { 23, 123, 1234, 12345, 20000, 30000, 40000, 50000, N, 0 };
	int at = 0;
 
	mpz_init_set_ui(p[0], 1);
 
	for (int i = 1; idx[at]; i++) {
		calc(i);
		if (i != idx[at]) continue;
 
		gmp_printf("%2d:\t%Zd\n", i, p[i]);
		at++;
	}
}