###
 Laguerre function

Example

  laguerre(x,3)

Result

     1   3    3   2
  - --- x  + --- x  - 3 x + 1
     6        2

The computation uses the following recurrence relation.

  L(x,0,k) = 1

  L(x,1,k) = -x + k + 1

  n*L(x,n,k) = (2*(n-1)+1-x+k)*L(x,n-1,k) - (n-1+k)*L(x,n-2,k)

In the "for" loop i = n-1 so the recurrence relation becomes

  (i+1)*L(x,n,k) = (2*i+1-x+k)*L(x,n-1,k) - (i+k)*L(x,n-2,k)
###



Eval_laguerre = ->
  # 1st arg

  push(cadr(p1))
  Eval()

  # 2nd arg

  push(caddr(p1))
  Eval()

  # 3rd arg

  push(cadddr(p1))
  Eval()

  p2 = pop()
  if (p2 == symbol(NIL))
    push_integer(0)
  else
    push(p2)

  laguerre()

#define X p1
#define N p2
#define K p3
#define Y p4
#define Y0 p5
#define Y1 p6

laguerre = ->
  n = 0
  save()

  p3 = pop()
  p2 = pop()
  p1 = pop()

  push(p2)
  n = pop_integer()

  if (n < 0 || isNaN(n))
    push_symbol(LAGUERRE)
    push(p1)
    push(p2)
    push(p3)
    list(4)
    restore()
    return

  if (issymbol(p1))
    laguerre2(n)
  else
    p4 = p1;      # do this when p1 is an expr
    p1 = symbol(SECRETX)
    laguerre2(n)
    p1 = p4
    push(symbol(SECRETX))
    push(p1)
    subst()
    Eval()

  restore()

laguerre2 = (n) ->
  i = 0

  push_integer(1)
  push_integer(0)

  p6 = pop()

  for i in [0...n]

    p5 = p6

    p6 = pop()

    push_integer(2 * i + 1)
    push(p1)
    subtract()
    push(p3)
    add()
    push(p6)
    multiply()

    push_integer(i)
    push(p3)
    add()
    push(p5)
    multiply()

    subtract()

    push_integer(i + 1)
    divide()