Problem Statement

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”:

Display the first 25 rows of a number triangle which begins: 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)}

.

If your environment is able, plot

P ( n )

{\displaystyle P(n)}

against

n

{\displaystyle n}

for

n

1 … 999

{\displaystyle n=1\ldots 999}

.

Let's start with the solution:

array cumu(int n) {
	array(array(int)) cache = ({({1})});

	for(int l = sizeof(cache); l < n + 1; l++) {
		array(int) r = ({0});
		for(int x = 1; x < l + 1; x++) {
			r = Array.push(r, r[-1] + cache[l - x][min(x, l-x)]);
		}
		cache = Array.push(cache, r);
	}
	return cache[n];
}

array row(int n) {
	array r = cumu(n);
	array res = ({});
	for (int i = 0; i < n; i++) {
		res = Array.push(res, r[i+1] - r[i]);
	}
	return res;
}

int main() {
	write("rows:\n");
	for(int x = 1; x < 11; x++) {
		write("%2d: ", x);
		for(int i = 0; i < sizeof(row(x)); i++) {
			write((string)row(x)[i] + " ");
		}
		write("\n");
	}

	array(int) sum_n = ({23, 123, 1234, 12345});
	write("\nsums:\n");
	for (int x = 0; x < sizeof(sum_n); x++) {
		write((string)sum_n[x] + " " + (string)cumu(sum_n[x])[-1] + "\n");
	}
	return 0;
}