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


 
object Main {
 
  // This is a special class for memoization
  case class Memo[A,B](f: A => B) extends (A => B) {
	  private val cache = Map.empty[A, B]
	  def apply(x: A) = cache getOrElseUpdate (x, f(x))
  }
 
  // Naive, but memoized solution
  lazy val namesStartingMemo : Memo[Tuple2[Int, Int], BigInt] = Memo {
    case (1, 1) => 1
    case (a, n) => 
	    if (a > n/2) namesStartingMemo(a - 1, n - 1)
	    else if (n < a) 0
	    else if (n == a) 1
	    else (1 to a).map(i => namesStartingMemo(i, n - a)).sum
 
  }
 
  def partitions(n: Int) = (1 to n).map(namesStartingMemo(_, n)).sum
 
  // main method
  def main(args: Array[String]): Unit = {
    for (i <- 1 to 25) {
    	for (j <- 1 to i) { 
	      print(namesStartingMemo(j, i));
	      print(' ');
	    }
    	println()
    }
    println(partitions(23))
    println(partitions(123))
    println(partitions(1234))
    println(partitions(12345))
  }
}
 