A204091 The number of 1 X n Haunted Mirror Maze puzzles with a unique solution ending with a mirror.
1, 2, 10, 46, 210, 958, 4370, 19934, 90930, 414782, 1892050, 8630686, 39369330, 179585278, 819187730, 3736768094, 17045465010, 77753788862, 354678014290, 1617882493726, 7380056440050, 33664517212798, 153562473183890, 700483331493854, 3195291711101490
Offset: 0
Examples
For n=2 we get the following ten boards:
('Z', '/')
('Z', '|')
('V', '/')
('V', '|')
('G', '/')
('G', '|')
('/', '/')
('/', '|')
('|', '/')
('|', '|')
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..1000
- Samples of these types of puzzles can be found at this site.
- Index entries for linear recurrences with constant coefficients, signature (5,-2).
Programs
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Mathematica
Join[{1}, LinearRecurrence[{5, -2}, {2, 10}, 40]] -
PARI
Vec((1-x)*(1-2*x)/(1-5*x+2*x^2) + O(x^30)) \\ Colin Barker, Nov 26 2016
-
Python
def a(n, d={0:1,1:2,2:10}): if n in d: return d[n] d[n]=5*a(n-1) - 2*a(n-2) return d[n] for n in range(25): print(a(n), end=', ') -
Python
from itertools import product def Lprint(n): print("The following generate boards with a unique solution") s = 0 for x in product(["Z", "V", "G", "/", "|"], repeat=n): if x[-1] == "/" or x[-1] == "|": y = list(x) # Splitting x up into a list pieces z = list() while y: if "/" in y or "|" in y: if "/" in y and "|" in y: m = min(y.index("/"), y.index("|")) else: if "/" in y: m = y.index("/") else: m = y.index("|") if y[0] not in ["/", "|"]: z.append(y[:m]) y = y[m + 1 :] else: z.append(y) y = [] goodword = True for w in z: # For each element in the list checking for Z&V together if "Z" in w and "V" in w: goodword = False if goodword: s += 1 print(x) return s print(Lprint(5))
Formula
a(n) = 5*a(n-1) - 2*a(n-2) with a(1) = 2, a(2) = 10.
a(n) = 2 * sum_{i=0}^{i=n-1} a(n)(2^(n-i)-1), a(0) = 1.
G.f.: (1-x)*(1-2*x) / (1-5*x+2*x^2).
a(n) = (2^(1-n) * ((5+sqrt(17))^n - (5-sqrt(17))^n))/sqrt(17) for n>0. - Colin Barker, Nov 26 2016
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