A204089 -id:A204089 - cp's OEIS frontend

A007070 a(n) = 4a(n-1) - 2a(n-2) with a(0) = 1, a(1) = 4.

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, 302408598355968

Offset: 0

N. J. A. Sloane, Mira Bernstein, Simon Plouffe

Joe Keane (jgk(AT)jgk.org) observes that this sequence (beginning at 4) is "size of raises in pot-limit poker, one blind, maximum raising."
It appears that this sequence is the BinomialMean transform of A002315 - see A075271. - John W. Layman, Oct 02 2002
Number of (s(0), s(1), ..., s(2n+3)) such that 0 < s(i) < 8 and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n+3, s(0) = 1, s(2n+3) = 4. - Herbert Kociemba, Jun 11 2004
a(n) = number of distinct matrix products in (A+B+C+D)^n where commutators [A,B]=[C,D]=0 but neither A nor B commutes with C or D. - Paul D. Hanna and Joshua Zucker, Feb 01 2006
The n-th term of the sequence is the entry (1,2) in the n-th power of the matrix M=[1,-1;-1,3]. - Simone Severini, Feb 15 2006
Hankel transform of this sequence is [1,-2,0,0,0,0,0,0,0,0,0,...]. - Philippe Deléham, Nov 21 2007
A204089 convolved with A000225, e.g., a(4) = 164 = (1*31 + 1*15 + 4*7 + 14*3 + 48*1) = (31 + 15 + 28 + 42 + 48). - Gary W. Adamson, Dec 23 2008
Equals INVERT transform of A000225: (1, 3, 7, 15, 31, ...). - Gary W. Adamson, May 03 2009
For n>=1, a(n-1) is the number of generalized compositions of n when there are 2^i-1 different types of the part i, (i=1,2,...). - Milan Janjic, Sep 24 2010
Binomial transform of A078057. - R. J. Mathar, Mar 28 2011
Pisano period lengths:  1, 1, 8, 1, 24, 8, 6, 1, 24, 24, 120, 8, 168, 6, 24, 1, 8, 24, 360, 24, ... . - R. J. Mathar, Aug 10 2012
a(n) is the diagonal of array A228405. - Richard R. Forberg, Sep 02 2013
From Wolfdieter Lang, Oct 01 2013: (Start)
a(n) appears together with A106731, both interspersed with zeros, in the representation of nonnegative powers of the algebraic number rho(8) = 2*cos(Pi/8) = A179260 of degree 4, which is the length ratio of the smallest diagonal and the side in the regular octagon.
The minimal polynomial for rho(8) is C(8,x) = x^4 - 4*x^2 + 2, hence rho(8)^n = A(n+1)*1 + A(n)*rho(8) + B(n+1)*rho(8)^2 + B(n)*rho(8)^3, n >= 0, with A(2*k) = 0, k >= 0, A(1) = 1, A(2*k+1) = A106731(k-1), k >= 1, and  B(2*k) = 0, k >= 0, B(1) = 0, B(2*k+1) = a(k-1), k >= 1. See also the P. Steinbach reference given under A049310. (End)
The ratio a(n)/A006012(n) converges to 1+sqrt(2). - Karl V. Keller, Jr., May 16 2015
From Tom Copeland, Dec 04 2015: (Start)
An aerated version of this sequence is given by the o.g.f. = 1 / (1 - 4 x^2 + 2 x^4) =  1 / [x^4 a_4(1/x)] = 1 / determinant(I - x M) = exp[-log(1 -4 x + 2 x^4)], where M is the adjacency matrix for the simple Lie algebra B_4 given in A265185 with the characteristic polynomial a_4(x) = x^4 - 4 x^2 + 2 = 2 T_4(x/2) = A127672(4,x), where T denotes a Chebyshev polynomial of the first kind.
A133314 relates a(n) to the reciprocal of the e.g.f. 1 - 4 x + 4 x^2/2!. (End)
a(n) is the number of vertices of the Minkowski sum of n simplices with vertices e_(2*i+1), e_(2*i+2), e_(2*i+3), e_(2*i+4) for i=0,...,n-1, where e_i is a standard basis vector. - Alejandro H. Morales, Oct 03 2022

			a(3) = 48 = 3 * 4 + 4 + 1 + 1 = 3*a(2) + a(1) + a(0) + 1.
Example for the octagon rho(8) powers: rho(8)^4  = 2 + sqrt(2) = -2*1 + 4*rho(8)^2  = A(5)*1 + A(4)*rho(8) + B(5)*rho(8)^2 + B(4)*rho(8)^3, with a(5) = A106731(1) = -2, B(5) = a(1) = 4, A(4) = 0, B(4) = 0. - _Wolfdieter Lang_, Oct 01 2013

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
C. Bautista-Ramos and C. Guillen-Galvan, Fibonacci numbers of generalized Zykov sums, J. Integer Seq., 15 (2012), Article 12.7.8.
A. Bernini, F. Disanto, R. Pinzani, and S. Rinaldi, Permutations defining convex permutominoes, J. Int. Seq. 10 (2007) # 07.9.7.
A. Burstein, S. Kitaev, and T. Mansour, Independent sets in certain classes of (almost) regular graphs, arXiv:math/0310379 [math.CO], 2003.
Tomislav Doslic, Planar polycyclic graphs and their Tutte polynomials, Journal of Mathematical Chemistry, Volume 51, Issue 6, 2013, pp. 1599-1607. (See Cor. 3.7(e).)
Tomislav Doslic and I. Zubac, Counting maximal matchings in linear polymers, Ars Mathematica Contemporanea 11 (2016) 255-276.
G. Dresden and Y. Li, Periodic Weighted Sums of Binomial Coefficients, arXiv:2210.04322 [math.NT], 2022.
L. Escobar, P. Gallardo, J. González-Anaya, J. L. González, G. Montúfar, and A. H. Morales, Enumeration of max-pooling responses with generalized permutohedra, arXiv:2209.14978 [math.CO], 2022. (See Ex. 5.1)
S. Falcon, Iterated Binomial Transforms of the k-Fibonacci Sequence, British Journal of Mathematics & Computer Science, 4 (22): 2014.
Pamela Fleischmann, Jonas Höfer, Annika Huch, and Dirk Nowotka, alpha-beta-Factorization and the Binary Case of Simon's Congruence, arXiv:2306.14192 [math.CO], 2023.
A. S. Fraenkel and C. Kimberling, Generalized Wythoff arrays, shuffles and interspersions, Discr. Math. 126 (1-3) (1994) 137-149.
Sela Fried, Even-up words and their variants, arXiv:2505.14196 [math.CO], 2025. See p. 8.
Sela Fried, Toufik Mansour, and Mark Shattuck, Counting k-ary words by number of adjacency differences of a prescribed size, arXiv:2504.03013 [math.CO], 2025. See p. 7.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 440
László Németh and László Szalay, Sequences Involving Square Zig-Zag Shapes, J. Int. Seq., Vol. 24 (2021), Article 21.5.2.
J. Riordan, The distribution of crossings of chords joining pairs of 2n points on a circle, Math. Comp., 29 (1975), 215-222.
J. Riordan, The distribution of crossings of chords joining pairs of 2n points on a circle, Math. Comp., 29 (1975), 215-222. [Annotated scanned copy]
Michael Z. Spivey and Laura L. Steil, The k-Binomial Transforms and the Hankel Transform, J. Integ. Seqs. Vol. 9 (2006), #06.1.1.
Index entries for sequences related to poker
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (4,-2).

Row sums of A059474. - David W. Wilson, Aug 14 2006
Cf. A000129, A000225, A002307, A002315, A006012 (same recurrence), A007052, A075271, A078057, A106731, A179260, A204089, A228405.
Equals 2 * A003480, n>0.
Row sums of A140071.
Cf. A127672, A265185, A133314.

a007070 n = a007070_list !! n
a007070_list = 1 : 4 : (map (* 2) $ zipWith (-)
   (tail $ map (* 2) a007070_list) a007070_list)
-- Reinhard Zumkeller, Jan 16 2012

Z:=PolynomialRing(Integers()); N:=NumberField(x^2-8); S:=[ ((4+r)^(1+n)-(4-r)^(1+n))/((2^(1+n))*r): n in [0..20] ]; [ Integers()!S[j]: j in [1..#S] ]; // Vincenzo Librandi, Mar 27 2011

[n le 2 select 3*n-2 else 4*Self(n-1)-2*Self(n-2): n in [1..23]];  // Bruno Berselli, Mar 28 2011

A007070 :=proc(n) option remember; if n=0 then 1 elif n=1 then 4 else 4*procname(n-1)-2*procname(n-2); fi; end:
seq(A007070(n), n=0..30); # Wesley Ivan Hurt, Dec 06 2015

LinearRecurrence[{4,-2}, {1,4}, 30] (* Harvey P. Dale, Sep 16 2014 *)

a(n)=polcoeff(1/(1-4*x+2*x^2)+x*O(x^n),n)

a(n)=if(n<1,1,ceil((2+sqrt(2))*a(n-1)))

[lucas_number1(n,4,2) for n in range(1, 24)]# Zerinvary Lajos, Apr 22 2009

G.f.: 1/(1 - 4*x + 2*x^2).
Preceded by 0, this is the binomial transform of the Pell numbers A000129. Its e.g.f. is then exp(2*x)*sinh(sqrt(2)*x)/sqrt(2). - Paul Barry, May 09 2003
a(n) = ((2+sqrt(2))^(n+1) - (2-sqrt(2))^(n+1))/sqrt(8). - Al Hakanson (hawkuu(AT)gmail.com), Dec 27 2008, corrected Mar 28 2011
a(n) = (2 - sqrt(2))^n*(1/2 - sqrt(2)/2) + (2 + sqrt(2))^n*(1/2 + sqrt(2)/2). - Paul Barry, May 09 2003
a(n) = ceiling((2 + sqrt(2))*a(n-1)). - Benoit Cloitre, Aug 15 2003
a(n) = U(n, sqrt(2))*sqrt(2)^n. - Paul Barry, Nov 19 2003
a(n) = (1/4)*Sum_{r=1..7} sin(r*Pi/8)*sin(r*Pi/2)*(2*cos(r*Pi/8))^(2*n+3). - Herbert Kociemba, Jun 11 2004
a(n) = center term in M^n * [1 1 1], where M = the 3 X 3 matrix [1 1 1 / 1 2 1 / 1 1 1]. M^n * [1 1 1] = [A007052(n) a(n) A007052(n)]. E.g., a(3) = 48 since M^3 * [1 1 1] = [34 48 34], where 34 = A007052(3). - Gary W. Adamson, Dec 18 2004
This is the binomial mean transform of A002307. See Spivey and Steil (2006). - Michael Z. Spivey (mspivey(AT)ups.edu), Feb 26 2006
a(2n) = Sum_{r=0..n} 2^(2n-1-r)*(4*binomial(2n-1,2r) + 3*binomial(2n-1,2r+1)) a(2n-1) = Sum_{r=0..n} 2^(2n-2-r)*(4*binomial(2n-2,2r) + 3*binomial(2n-2,2r+1)). - Jeffrey Liese, Oct 12 2006
a(n) = 3*a(n - 1) + a(n - 2) + a(n - 3) + ... + a(0) + 1. - Gary W. Adamson, Feb 18 2011
G.f.: 1/(1 - 4*x + 2*x^2) = 1/( x*(1 + U(0)) ) - 1/x  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
G.f.: A(x) = G(0)/(1-2*x) where G(k) = 1 + 2*x/(1 - 2*x - x*(1-2*x)/(x + (1-2*x)/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Jan 04 2013
G.f.: G(0)/(2*x) - 1/x, where G(k) = 1 + 1/(1 - x*(2*k-1)/(x*(2*k+1) - (1-x)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013
a(n-1) = Sum_{k=0..n} binomial(2*n, n+k)*(k|8) where (k|8) is the Kronecker symbol. - Greg Dresden, Oct 11 2022
E.g.f.: exp(2*x)*(cosh(sqrt(2)*x) + sqrt(2)*sinh(sqrt(2)*x)). - Stefano Spezia, May 20 2024

A204091 The number of 1 X n Haunted Mirror Maze puzzles with a unique solution ending with a mirror.

Original entry on oeis.org

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

David Nacin, Jan 10 2012

The final slot can refer to a mirror of either type ('/' or '\'.)

A204092 counts the situation dropping the requirement that a board ends with a mirror. A204089 counts the situation where all mirrors have the same orientation. A204090 counts the boards with same orientation where the last square need not be a mirror.

			For n=2 we get the following ten boards:
('Z', '/')
('Z', '|')
('V', '/')
('V', '|')
('G', '/')
('G', '|')
('/', '/')
('/', '|')
('|', '/')
('|', '|')

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).

Cf. A204089-A204092.

Apart from signs and the initial term, same as A106709.

Join[{1}, LinearRecurrence[{5, -2}, {2, 10}, 40]]

Vec((1-x)*(1-2*x)/(1-5*x+2*x^2) + O(x^30)) \\ Colin Barker, Nov 26 2016

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=', ')

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))

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

A204090 The number of 1 X n Haunted Mirror Maze puzzles with a unique solution where mirror orientation is fixed.

Original entry on oeis.org

1, 2, 8, 34, 134, 498, 1786, 6274, 21778, 75074, 257762, 882946, 3020354, 10323714, 35270530, 120467458, 411394306, 1404773378, 4796567042, 16377245698, 55916897282, 190915194882, 651831179266, 2225502715906, 7598365282306, 25942489251842, 88573293551618

Offset: 0

David Nacin, Jan 10 2012

Since the uniqueness of a solution is unaffected by the orientation of the mirrors in this 1 X n case, we assume mirror orientation is fixed for this sequence.

Connected to A204089, which counts the 1 X n boards with unique solutions that end in a mirror. Dropping the mirror orientation restriction would give A204092. Dropping the orientation restriction and requiring a mirror in the last slot gives A204091.

			For n=3 we would get the following 34 boards with unique solutions:
('Z', 'Z', '/')
('Z', 'G', '/')
('Z', '/', 'Z')
('Z', '/', 'V')
('Z', '/', 'G')
('Z', '/', '/')
('V', 'V', '/')
('V', 'G', '/')
('V', '/', 'Z')
('V', '/', 'V')
('V', '/', 'G')
('V', '/', '/')
('G', 'Z', '/')
('G', 'V', '/')
('G', 'G', 'G')
('G', 'G', '/')
('G', '/', 'Z')
('G', '/', 'V')
('G', '/', 'G')
('G', '/', '/')
('/', 'Z', 'Z')
('/', 'Z', 'G')
('/', 'Z', '/')
('/', 'V', 'V')
('/', 'V', 'G')
('/', 'V', '/')
('/', 'G', 'Z')
('/', 'G', 'V')
('/', 'G', 'G')
('/', 'G', '/')
('/', '/', 'Z')
('/', '/', 'V')
('/', '/', 'G')
('/', '/', '/')

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Samples of these types of puzzles can be found at this and other sites.
Index entries for linear recurrences with constant coefficients, signature (7,-16,14,-4).

Cf. A204089, A204091, A204092.

LinearRecurrence[{7, -16, 14, -4}, {1, 2, 8, 34}, 40]

Vec((1-5*x+10*x^2-4*x^3) / ((1-x)*(1-2*x)*(1-4*x+2*x^2)) + O(x^30)) \\ Colin Barker, Nov 26 2016

def a(n, d={0:1,1:2,2:8,3:34}):
 if n in d:
  return d[n]
 d[n]=7*a(n-1) - 16*a(n-2) + 14*a(n-3) - 4*a(n-4)
 return d[n]

#Produces a(n) through enumeration and also displays boards:
def Hprint(n):
 print('The following generate boards with a unique solution')
 s=0
 for x in product(['Z','V','G','/'],repeat=n):
  #Taking care of the no mirror case
  if '/' not in x:
   if 'Z' not in x and 'V' not in x:
    s+=1
    print(x)
  else:
   #Splitting x up into a list pieces
   y=list(x)
   z=list()
   while y:
    if '/' in y:
     if y[0] != '/': #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

G.f.: (1 - 5*x + 10*x^2 - 4*x^3) / ((1 - x)*(1 - 2*x)*(1 - 4*x + 2*x^2)).
a(n) = A204089(n+1) - 2^(n+1) + 2.
a(n) = 7*a(n-1) - 16*a(n-2) + 14*a(n-3) - 4*a(n-4), a(0)=1, a(1)=2, a(2)=8, a(3)=34.
a(n) = 2 - 2^(1+n) + ((2+sqrt(2))^(1+n) - (2-sqrt(2))^(1+n))/(2*sqrt(2)). - Colin Barker, Nov 26 2016

A204092 The number of 1 by n Haunted Mirror Maze puzzles with a unique solution.

Original entry on oeis.org

1, 3, 17, 91, 449, 2123, 9841, 45211, 206881, 945003, 4313297, 19680571, 89784449, 409577483, 1868351281, 8522666971, 38876763361, 177338745003, 808940722577, 3690027171451, 16832256509249, 76781232397643, 350241657358321, 1597645838773531, 7287745912705441

Offset: 0

David Nacin, Jan 10 2012

Requiring the last slot contain a mirror results in A204091. Fixing orientation of mirrors results in A204090. Fixing orientation and requiring the last slot be a mirror results in A204089.

			The following 17 boards illustrate K(2):
('Z', '/')
('Z', '|')
('V', '/')
('V', '|')
('G', 'G')
('G', '/')
('G', '|')
('/', 'Z')
('/', 'V')
('/', 'G')
('/', '/')
('/', '|')
('|', 'Z')
('|', 'V')
('|', 'G')
('|', '/')
('|', '|')

Samples of these types of puzzles can be found at this site.
Index entries for linear recurrences with constant coefficients, signature (8,-19,16,-4).

Cf. A204089, A204090, A204091.

LinearRecurrence[{8, -19, 16, -4}, {1, 3, 17, 91}, 40]

def a(n, d={0:1,1:3,2:17,3:91}):
 if n in d:
  return d[n]
 d[n]=8*a(n-1) - 19*a(n-2) + 16*a(n-3) - 4*a(n-4)
 return d[n]

#Produces a(n) through enumeration and also displays boards:
def Kprint(n):
 print('The following generate boards with a unique solution')
 s=0
 for x in product(['Z','V','G','/','|'],repeat=n):
  #Taking care of the no mirror case
  if '/' not in x and '|' not in x:
   if 'Z' not in x and 'V' not in x:
    s+=1
    print(x)
  else:
   #Splitting x up into a list pieces
   y=list(x)
   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 ['/','|']: #Don't need to add blank pieces to z
      z.append(y[:m])
     y=y[m+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

a(n) = A204091(n+1)/2 - 2^(n+1) + 2.
a(n) = 8*a(n-1) - 19*a(n-2) + 16*a(n-3) - 4*a(n-4), a(0) = 1, a(1) = 3, a(2) = 17, a(3) = 91.
G.f.: (1-5x+12x^2-4x^3)/((1-x)(1-2x)(1-5x+2x^2)).

A206306 Riordan array (1, x/(1-3x+2x^2)).

Original entry on oeis.org

1, 0, 1, 0, 3, 1, 0, 7, 6, 1, 0, 15, 23, 9, 1, 0, 31, 72, 48, 12, 1, 0, 63, 201, 198, 82, 15, 1, 0, 127, 522, 699, 420, 125, 18, 1, 0, 255, 1291, 2223, 1795, 765, 177, 21, 1, 0, 511, 3084, 6562, 6768, 3840, 1260, 238, 24, 1

Offset: 0

Philippe Deléham, Feb 06 2012

The convolution triangle of the Mersenne numbers A000225. - Peter Luschny, Oct 09 2022

			Triangle begins:
  1;
  0,    1;
  0,    3,    1;
  0,    7,    6,     1;
  0,   15,   23,     9,     1;
  0,   31,   72,    48,    12,     1;
  0,   63,  201,   198,    82,    15,    1;
  0,  127,  522,   699,   420,   125,   18,    1;
  0,  255, 1291,  2223,  1795,   765,  177,   21,   1;
  0,  511, 3084,  6562,  6768,  3840, 1260,  238,  24,  1;
  0, 1023, 7181, 18324, 23276, 16758, 7266, 1932, 308, 27,  1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Columns: A000007, A000225 (Mersenne numbers), A045618, A055582.
Cf. A008585, A009545, A084938, A062725, A110441, A204089, A204091.
Cf. A045741, A078020, A244038.

function T(n,k) // T = A206306
  if k lt 0 or k gt n then return 0;
  elif k eq n then return 1;
  elif k eq 0 then return 0;
  else return 3*T(n-1, k) +T(n-1, k-1) -2*T(n-2, k);
  end if; return T;
end function;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Dec 20 2022

# Uses function PMatrix from A357368.
PMatrix(10, n -> 2^n - 1); # Peter Luschny, Oct 09 2022

T[n_, k_]:= T[n, k]= If[k<0 || k>n, 0, If[k==n, 1, If[k==0, 0, 3*T[n- 1, k] +T[n-1, k-1] -2*T[n-2, k]]]];
Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Dec 20 2022 *)

def T(n,k): # T = A206306
    if (k<0 or k>n): return 0
    elif (k==n): return 1
    elif (k==0): return 0
    else: return 3*T(n-1, k) +T(n-1, k-1) -2*T(n-2, k)
flatten([[T(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Dec 20 2022

Triangle T(n,k), read by rows, given by (0, 3, -2/3, 2/3, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938.
Diagonals sums are even-indexed Fibonacci numbers.
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A204089(n), A204091(n) for x = 0, 1, 2 respectively.
G.f.: (1-3*x+2*x^)/(1-(3+y)*x+2*x^2).
From Philippe Deléham, Nov 17 2013; corrected Feb 13 2020: (Start)
T(n, n) = 1.
T(n+1, n) = 3n = A008585(n).
T(n+2, n) = A062725(n).
T(n,k) = 3*T(n-1,k)+T(n-1,k-1)-2*T(n-2,k), T(0,0)=T(1,1)=T(2,2)=1, T(1,0)=T(2,0)=0, T(2,1)=3, T(n,k)=0 if k<0 or if k>n. (End)
From G. C. Greubel, Dec 20 2022: (Start)
Sum_{k=0..n} (-1)^k*T(n,k) = [n=1] - A009545(n).
Sum_{k=0..n} (-2)^k*T(n,k) = [n=1] + A078020(n+1).
T(2*n, n+1) = A045741(n+2), n >= 0.
T(2*n+1, n+1) = A244038(n). (End)

A217730 Expansion of (1+2x-x^3)/(1-4x^2+2*x^4).

Original entry on oeis.org

1, 2, 4, 7, 14, 24, 48, 82, 164, 280, 560, 956, 1912, 3264, 6528, 11144, 22288, 38048, 76096, 129904, 259808, 443520, 887040, 1514272, 3028544, 5170048, 10340096, 17651648, 35303296, 60266496, 120532992, 205762688, 411525376, 702517760, 1405035520, 2398545664, 4797091328, 8189147136, 16378294272, 27959497216

Offset: 0

Philippe Deléham, Mar 22 2013

In general, a(n,j,m) = Sum_{r=1..m} (2^n*(1-(-1)^r)*cos(Pi*r/(m+1))^n*cot(Pi*r/(2*(m+1)))*sin(j*Pi*r/(m+1)))/(m+1) gives the number of paths of length n starting at the j-th node on the path graph P_m. Here we have the case m=7 and j=3. - Herbert Kociemba, Sep 17 2020

Shaun V. Ault and Charles Kicey, Counting paths in corridors using circular Pascal arrays, Discrete Math. 332 (2014), 45--54. MR3227977. See Fig. 5. - _N. J. A. Sloane_, Aug 04 2014
Index entries for linear recurrences with constant coefficients, signature (0,4,0,-2).

Cf. A007070, A077957, A204089, A216232.

First differences are in A062113.

m:=40; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((1+2*x-x^3)/(1-4*x^2+2*x^4))); // Bruno Berselli, Mar 22 2013

CoefficientList[Series[(1 + 2 x - x^3)/(1 - 4 x^2 + 2 x^4), {x, 0, 40}], x] (* Bruno Berselli, Mar 22 2013 *)
a[n_,j_,m_]:=Sum[(2^(n+1)Cos[Pi r/(m+1)]^n Cot[Pi r/(2(m+1))] Sin[j Pi r/(m+1)])/(m+1),{r,1,m,2}]
Table[a[n,3,7],{n,0,40}]//Round (* Herbert Kociemba, Sep 17 2020 *)

makelist(coeff(taylor((1+2*x-x^3)/(1-4*x^2+2*x^4), x, 0, n), x, n), n, 0, 40); /* Bruno Berselli, Mar 22 2013 */

G.f.: (1+x)*(1+x-x^2)/(1-4*x^2+2*x^4).
a(n) = Sum_{k=0..n} A216232(n-k,k).
a(n) = 4*a(n-2) - 2*a(n-4) for n>=4, a(0)=1, a(1)=2, a(2)=4, a(3)=7.
a(2*n) = A007070(n), a(2*n-1) = a(2*n)/2 = A007070(n)/2.
a(n)*a(n+1)-a(n-1)*a(n+2) = (1-(-1)^n)*2^floor(n/2-1) for n>0. - Bruno Berselli, Mar 22 2013
a(n) = Sum_{r=1..7} (2^n*(1-(-1)^r)*cos(Pi*r/8)^n*cot(Pi*r/16)*sin(3*Pi*r/8))/8. - Herbert Kociemba, Sep 17 2020

cp's OEIS Frontend

A007070 a(n) = 4*a(n-1) - 2*a(n-2) with a(0) = 1, a(1) = 4.

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A204091 The number of 1 X n Haunted Mirror Maze puzzles with a unique solution ending with a mirror.

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A204090 The number of 1 X n Haunted Mirror Maze puzzles with a unique solution where mirror orientation is fixed.

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A204092 The number of 1 by n Haunted Mirror Maze puzzles with a unique solution.

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A206306 Riordan array (1, x/(1-3*x+2*x^2)).

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A217730 Expansion of (1+2*x-x^3)/(1-4*x^2+2*x^4).

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A007070 a(n) = 4a(n-1) - 2a(n-2) with a(0) = 1, a(1) = 4.

A206306 Riordan array (1, x/(1-3x+2x^2)).

A217730 Expansion of (1+2x-x^3)/(1-4x^2+2*x^4).