HP 50g Double factorial

[Giuseppe Donnini]

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.

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