Sierpinski Layers

GolfScript (42 bytes)

~,-1%1\{' '*\.2base{'  /\\'2/=}%[email protected]*^}/;;

Online demo

This exploits a well-known relationship between Pascal's triangle and the Sierpinski triangle.

CJam, 48 46 bytes

li(_S*"/\\":T+\{_N@S+TW%/S2**" /"/"\ "f/:+T*}*

Simple recursive approach based on observations 1 and 2 in the question.

Try it online.

How it works

li(         " L := int(input()) - 1            ";
_S*         " A := L * ' '                     ";
"/\\":T+    " A += (T := '/\')                 ";
\{          " do L times:                      ";
  _N        "   push A, '\n'                   ";
  @S+       "   A += ' '                       ";
  TW%/      "   B := A.split(reverse(T))       ";
  S2**      "   A := '  '.join(B)              ";
  " /"/     "   B := A.split(' /')             ";
  "\ "f/    "   C := { X.split('\ ') : X ∊ B } ";
  :+T*      "   A := T.join(sum(C, []))        ";
}*          "                                  ";

---

CJam, 51 bytes

li__2mL,1a\{2_@##)1$f*+}/<f{2b_"  /\\"2/f=@@,-S*\N}

I like this approach better, but it can't compete with the recursive one. Even after eliminating 2mL (which results in at least O(2ⁿ) execution time), I'm still at 48 bytes...

This approach encodes /\'s as 1's and double spaces between them as 0's. Considering the resulting arrays binary numbers, we see that the configuration of the nth row corresponds to the nth integer larger than 1 that can be expressed as a product of different Fermat numbers (integers of the form 2^2k+1).

How it works

li__2mL,1a         " push L := int(input()), L, R := range(log(L)/log(2)), A := [1] ";
\{2_@##)1$f*+}/    " for I in R: A += { a × 2**(2**I) : a ∊ A }                     ";
<                  " A := A[:L]                                                     ";
f{                 " for I in R: push L, I                                          ";
  2b_"  /\\"2/     "   push (B := base(I, 2)), B, S := [ '  ' '/\' ]                ";
  f=               "   for J in I: J := S[J]                                        ";
  @@,-S*\N         "   push (L - len(B)) * ' ', J, '\n'                             ";
}                  "                                                                ";

Python 2 - 140 139 127 122 121 118 116

N=input()
b,V,m,n=' V/\\'
s=b*~-N+m+n
R=str.replace
while N:print s;N-=1;s=R(R(R(R(s+b,b+m,V),n+b,V),n+m,b+b),V,m+n)

Based on temporary string replacements (https://stackoverflow.com/a/8687380/3419103):

1. / > V
2. \ > V
3. \/ > __ (2 spaces)
4. V > /\