HP 50g Double factorial

[joeres]

(05-01-2019 05:05 PM)Giuseppe Donnini Wrote:  Here's another solution:

Considering that  n!!  is equivalent to  n·(n-2)·(n-4)· ... ·1  for every positive integer  n ≠ 0  (odd or even), all it needs is a simple loop which calculates the next factor and multiplies it with the running product. Such a solution could do entirely without local variables and algebraic objects, requiring only stack manipulations, and very few in number:

1          @ Initialize result.
SWAP 2     @ Loop down from n to 2.
FOR f f *  @ Multiply with current factor.
-2 STEP    @ Decrement factor by 2 ; repeat until factor < 2.

Taking into account the special case where  n = 0 , we finally have:

Code:


@ NAME  : DF (Double factorial)
@ STACK : ( n --> n!! )
\<<
  1                 @ Initialize result.
  IF SWAP DUP 0 ==  @ If n = 0,
  THEN DROP         @   then n!! = 1 (just drop n, 1 already on stack).
  ELSE 2            @ Otherwise: Loop down from n to 2.
  FOR f f *         @   Multiply with current factor.
  -2 STEP           @   Decrement factor by 2 ; repeat until factor < 2.
\>>

Hi Giuseppe,

Thank you, very interesting and nice solution. I tested it (END added). Works fine, but if I use very large numbers, the program runs in an infinite loop. The other programs output the message "Integer too large". But I can easily query this.

Kind regards
Joerg