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 A204089 The number of 1 by n Haunted Mirror Maze puzzles with a unique solution ending with a mirror, where mirror orientation is fixed. 7
 1, 1, 4, 14, 48, 164, 560, 1912, 6528, 22288, 76096, 259808, 887040, 3028544, 10340096, 35303296, 120532992, 411525376, 1405035520, 4797091328, 16378294272, 55918994432, 190919389184, 651839567872, 2225519493120, 7598398836736, 25942556360704, 88573427769344 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 COMMENTS Apart from a leading 1, the same as A007070. - R. J. Mathar, Jan 16 2012 Since the uniqueness of a solution is unaffected by the orientation of the mirrors in this 1 by n case, we assume mirror orientation is fixed for this sequence. Dropping the requirement that the board end with a mirror gives A204090. Allowing for mirror orientation gives A204091. Allowing for orientation and dropping the requirement gives A204092. LINKS Vincenzo Librandi, Table of n, a(n) for n = 0..1000 David Millar, Haunted Puzzles, The Griddle Index entries for linear recurrences with constant coefficients, signature (4,-2). FORMULA G.f.: (1-x)*(1-2*x)/(1-4*x+2*x^2). a(n) = Sum_{i=0..n-1} a(i) * (2^(n-i)-1), with a(0)=1. a(n) = 4*a(n-1) - 2*a(n-2), a(1)=1, a(2)=4. G.f.: (x-1)*(2*x-1)/(1 - 4*x + 2*x^2) = 1/(1 + U(0)) where U(k)= 1 - 2^k/(1 - x/(x - 2^k/U(k+1) )); (continued fraction 3rd kind, 3-step). - Sergei N. Gladkovskii, Dec 05 2012 a(n) = ((-(2-sqrt(2))^n+(2+sqrt(2))^n))/(2*sqrt(2)). - Colin Barker, Dec 06 2015 EXAMPLE For M(3) we would have the following possibilities: ('Z', 'Z', '/') ('Z', 'G', '/') ('Z', '/', '/') ('V', 'V', '/') ('V', 'G', '/') ('V', '/', '/') ('G', 'Z', '/') ('G', 'V', '/') ('G', 'G', '/') ('G', '/', '/') ('/', 'Z', '/') ('/', 'V', '/') ('/', 'G', '/') ('/', '/', '/') MATHEMATICA Join[{1}, LinearRecurrence[{4, -2}, {1, 4}, 25]] PROG (Python) def a(n, d={0:1, 1:4}): .if n in d: ..return d[n] .d[n]=4*a(n-1) - 2*a(n-2) .return d[n] (Python) #Produces a(n) through enumeration and also displays boards: def Mprint(n): .print('The following generate boards with a unique solution') .s=0 .for x in product(['Z', 'V', 'G', '/'], repeat=n): ..if x[-1]=='/': ...#Splitting x up into a list pieces ...y=list(x) ...z=list() ...while y: ....#print(y) ....if '/' in y: .....if y != '/': #Don't need to add blank pieces to z ......z.append(y[:y.index('/')]) .....y=y[y.index('/')+1:] ....else: .....z.append(y) .....y=[] ...#For each element in the list checking for Z&V together ...goodword=True ...for w in z: ....if 'Z' in w and 'V' in w: .....goodword=False ...if goodword: ....s+=1 ....print(x) .return s (PARI) Vec((1-x)*(1-2*x)/(1-4*x+2*x^2) + O(x^30)) \\ Michel Marcus, Dec 06 2015 CROSSREFS Cf. A007070, A204090, A204091, A204092. Sequence in context: A127359 A289928 A007070 * A092489 A094827 A094667 Adjacent sequences:  A204086 A204087 A204088 * A204090 A204091 A204092 KEYWORD nonn,easy AUTHOR David Nacin, Jan 10 2012 STATUS approved

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Last modified January 20 05:27 EST 2020. Contains 331067 sequences. (Running on oeis4.)