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:

Proc _NineBillionNames(15)
End

_NineBillionNames
  Param (1)
  Local (3)
  
  @(1*a@ + 1) = 1
 
  For b@ = 2 To a@
    For c@ = 1 To b@
      @(b@*a@ + c@) = @((b@ - 1)*a@ + (c@ - 1)) + @((b@ - c@)*a@ + c@)
    Next
  Next
  
  For b@ = 1 To a@
    d@ = a@ * 2 - 2 * b@ - 2
    For c@ = 1 To b@
      Print Tab(d@ + c@ * 4 + (1 - Len(Str(@(b@*a@ + c@))))); @(b@*a@ + c@);
    Next
  Next
  
  Print
Return