A138395 -id:A138395 - cp's OEIS frontend

A190958 a(n) = 2a(n-1) - 10a(n-2), with a(0) = 0, a(1) = 1.

0, 1, 2, -6, -32, -4, 312, 664, -1792, -10224, -2528, 97184, 219648, -532544, -3261568, -1197696, 30220288, 72417536, -157367808, -1038910976, -504143872, 9380822016, 23803082752, -46202054656, -330434936832, -198849327104, 2906650714112, 7801794699264

Offset: 0

Vladimir Joseph Stephan Orlovsky, May 24 2011

For the difference equation a(n) = c*a(n-1) - d*a(n-2), with a(0) = 0, a(1) = 1, the solution is a(n) = d^((n-1)/2) * ChebyshevU(n-1, c/(2*sqrt(d))) and has the alternate form a(n) = ( ((c + sqrt(c^2 - 4*d))/2)^n - ((c - sqrt(c^2 - 4*d))/2)^n )/sqrt(c^2 - 4*d). In the case c^2 = 4*d then the solution is a(n) = n*d^((n-1)/2). The generating function is x/(1 - c*x + d^2) and the exponential generating function takes the form (2/sqrt(c^2 - 4*d))*exp(c*x/2)*sinh(sqrt(c^2 - 4*d)*x/2) for c^2 > 4*d, (2/sqrt(4*d - c^2))*exp(c*x/2)*sin(sqrt(4*d - c^2)*x/2) for 4*d > c^2, and x*exp(sqrt(d)*x) if c^2 = 4*d. - G. C. Greubel, Jun 10 2022

Vincenzo Librandi, Table of n, a(n) for n = 0..190
Index entries for linear recurrences with constant coefficients, signature (2,-10).

Sequences of the form a(n) = c*a(n-1) - d*a(n-2), with a(0)=0, a(1)=1:
c/d...1.......2.......3.......4.......5.......6.......7.......8.......9......10
.A128834,A107920,A106852,A106853,A106854,A145934,A145976,A145978,A146078,A146080
.A000027,A009545,A088137,A088138,A045873,A088139,A089181,A090591,A127357,A190958
.A001906,A000225,A057083,A049072,A190959,A190960,A190961,A190962,A190963,A190964
.A001353,A007070,A003462,A001787,A099456,A190965,A168175,A190966,A190967,A190968
.A004254,A107839,A116415,A002450,A030191,A001047,A099450,A190969,A190970,A190971
.A001109,A154244,A138395,A084326,A003463,A030192,A081179,A006516,A027471,A106392
.A004187,A186446,A190972,A190973,A190974,A003464,A030240,A152268,A099459,A016127
.A001090,A190975,A190976,A099156,A190977,A190978,A023000,A057084,A154245,A154235
.A018913,A190979,A190980,A190981,A190982,A190983,A190984,A023001,A057085,A178869
A004189,A190869,A190985,A190986,A190987,A190988,A190989,A190990,A002452,A057086

I:=[0,1]; [n le 2 select I[n] else 2*Self(n-1)-10*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Sep 17 2011

Mathematica
```
LinearRecurrence[{2,-10}, {0,1}, 50]
```

a(n)=([0,1; -10,2]^n*[0;1])[1,1] \\ Charles R Greathouse IV, Apr 08 2016

[lucas_number1(n,2,10) for n in (0..50)] # G. C. Greubel, Jun 10 2022

G.f.: x / ( 1 - 2*x + 10*x^2 ). - R. J. Mathar, Jun 01 2011
E.g.f.: (1/3)*exp(x)*sin(3*x). - Franck Maminirina Ramaharo, Nov 13 2018
a(n) = 10^((n-1)/2) * ChebyshevU(n-1, 1/sqrt(10)). - G. C. Greubel, Jun 10 2022
a(n) = (1/3)*10^(n/2)*sin(n*arctan(3)) = Sum_{k=0..floor(n/2)} (-1)^k*3^(2*k)*binomial(n,2*k+1). - Gerry Martens, Oct 15 2022

A084120 a(n) = 6a(n-1) - 3a(n-2), a(0)=1, a(1)=3.

Original entry on oeis.org

1, 3, 15, 81, 441, 2403, 13095, 71361, 388881, 2119203, 11548575, 62933841, 342957321, 1868942403, 10184782455, 55501867521, 302456857761, 1648235544003, 8982042690735, 48947549512401, 266739169002201

Offset: 0

Paul Barry, May 13 2003

Binomial transform of A084059.

			G.f. = 1 + 3*x + 15*x^2 + 81*x^3 + 441*x^4 + 2403*x^5 + 13095*x^6 + ...

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,-3).

Cf. A084059, A138395, A147720.

[n le 2 select 3^(n-1) else 6*Self(n-1) -3*Self(n-2): n in [1..41]]; // G. C. Greubel, Oct 13 2022

LinearRecurrence[{6,-3},{1,3},30] (* Harvey P. Dale, Feb 25 2014 *)

{a(n) = if( n<0, 0, polsym(x^2 - 6*x + 3, n)[1+n] / 2)};

[lucas_number2(n,6,3)/2 for n in range(0,27)] # Zerinvary Lajos, Jul 08 2008

a(n) = ((3+sqrt(6))^n + (3-sqrt(6))^n)/2.
G.f.: (1-3*x)/(1-6*x+3*x^2).
E.g.f.: exp(3*x)*cosh(sqrt(6)*x).
a(n) = 3^n * Sum_{k=0..floor(n/2)} C(n, 2*k)*(2/3)^k. - Paul Barry, Sep 10 2005
Lim_{n -> oo} a(n)/a(n-1) = (3 + sqrt(6)) = 5.445489742... - Gary W. Adamson, Mar 19 2008
a(n) = Sum_{k=0..n} A147720(n,k)*3^k. - Philippe Deléham, Nov 15 2008
G.f.: G(0)/2, where G(k)= 1 + 1/(1 - x*(2*k-3)/(x*(2*k-1) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 27 2013
a(n) = A138395(n) - 3*A138395(n-1). - R. J. Mathar, May 11 2022

A090018 a(n) = 6a(n-1) + 3a(n-2) for n > 2, a(0)=1, a(1)=6.

Original entry on oeis.org

1, 6, 39, 252, 1629, 10530, 68067, 439992, 2844153, 18384894, 118841823, 768205620, 4965759189, 32099171994, 207492309531, 1341251373168, 8669985167601, 56043665125110, 362271946253463, 2341762672896108, 15137391876137037, 97849639275510546, 632510011281474387

Offset: 0

Paul Barry, Nov 19 2003

From Johannes W. Meijer, Aug 09 2010: (Start)

a(n) represents the number of n-move routes of a fairy chess piece starting in a given corner or side square on a 3 X 3 chessboard. This fairy chess piece behaves like a white queen on the eight side and corner squares but on the central square the queen explodes with fury and turns into a red queen, see A180032. The central square leads to A180028. (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (6,3).

Cf. A141041, A180028, A180032.

Sequences with g.f. of the form 1/(1 - 6*x - k*x^2): A106392 (k=-10), A027471 (k=-9), A006516 (k=-8), A081179 (k=-7), A030192 (k=-6), A003463 (k=-5), A084326 (k=-4), A138395 (k=-3), A154244 (k=-2), A001109 (k=-1), A000400 (k=0), A005668 (k=1), A135030 (k=2), this sequence (k=3), A135032 (k=4), A015551 (k=5), A057089 (k=6), A015552 (k=7), A189800 (k=8), A189801 (k=9), A190005 (k=10), A015553 (k=11).

[n le 2 select 6^(n-1) else 6*Self(n-1)+3*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 15 2011

a:= n-> (<<0|1>, <3|6>>^n. <<1,6>>)[1,1]:
seq(a(n), n=0..30);  # Alois P. Heinz, Jan 17 2011

Join[{a=1,b=6},Table[c=6*b+3*a;a=b;b=c,{n,100}]] (* Vladimir Joseph Stephan Orlovsky, Jan 16 2011 *)
LinearRecurrence[{6,3}, {1,6}, 41] (* G. C. Greubel, Oct 10 2022 *)

my(x='x+O('x^30)); Vec(1/(1-6*x-3*x^2)) \\ G. C. Greubel, Jan 24 2018

[lucas_number1(n,6,-3) for n in range(1, 31)] # Zerinvary Lajos, Apr 24 2009

a(n) = (3+2*sqrt(3))^n*(sqrt(3)/4+1/2) + (1/2-sqrt(3)/4)*(3-2*sqrt(3))^n.
a(n) = (-i*sqrt(3))^n * ChebyshevU(n, isqrt(3)), i^2=-1.
From Johannes W. Meijer, Aug 09 2010: (Start)
G.f.: 1/(1 - 6*x - 3*x^2).
Limit_{k->oo} a(n+k)/a(k) = A141041(n) + A090018(n-1)*sqrt(12) for n >= 1.
Limit_{n->oo} A141041(n)/A090018(n-1) = sqrt(12). (End)
a(n) = Sum_{k=0..n} A099089(n,k)*3^k. - Philippe Deléham, Nov 21 2011
E.g.f.: exp(3*x)*(2*cosh(2*sqrt(3)*x) + sqrt(3)*sinh(2*sqrt(3)*x))/2. - Stefano Spezia, Apr 23 2025

Typo in Mathematica program corrected by Vincenzo Librandi, Nov 15 2011

A123347 Number of words of length n over the alphabet {1,2,3,4,5} such that 1 is not followed by an odd letter.

Original entry on oeis.org

1, 5, 22, 98, 436, 1940, 8632, 38408, 170896, 760400, 3383392, 15054368, 66984256, 298045760, 1326151552, 5900697728, 26255094016, 116821771520, 519797274112, 2312832639488, 10290925106176, 45789365703680, 203739313027072, 906535983515648, 4033622560116736

Offset: 0

N. J. A. Sloane, Oct 10 2006

Appears to be Kekulé numbers for certain benzenoids (see the Cyvin-Gutman book for details).

			a(2) = 22 because all 25 words of length 2 are included except 11, 13 and 15.

S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (see p. 78).

Andrew Howroyd, Table of n, a(n) for n = 0..500
D. Birmajer, J. B. Gil, and M. D. Weiner, On the Enumeration of Restricted Words over a Finite Alphabet, J. Int. Seq. 19 (2016) # 16.1.3, example 17.
Index entries for linear recurrences with constant coefficients, signature (4, 2).

Cf. A138395.

I:=[1, 5]; [n le 2 select I[n] else 4*Self(n-1) + 2*Self(n-2): n in [1..30]]; // G. C. Greubel, Nov 29 2018

seq(coeff(series((1+x)/(1-4*x-2*x^2),x,n+1), x, n), n = 0 .. 25); # Muniru A Asiru, Nov 27 2018

LinearRecurrence[{4, 2}, {1, 5}, 30] (* Amiram Eldar, Nov 26 2018 *)

Vec((1 + x)/(1 - 4*x - 2*x^2) + O(x^30)) \\ Andrew Howroyd, Nov 25 2018

s=((1+x)/(1-4*x-2*x^2)).series(x, 50); s.coefficients(x, sparse=False) # G. C. Greubel, Nov 29 2018

From Klaus Brockhaus, Oct 03 2009: (Start)
Inverse binomial transform of A138395.
a(n) = ((2+sqrt(6))^(n+1) + (2-sqrt(6))^(n+1))/4.
a(n) = 4*a(n-1) + 2*a(n-2) for n > 1.
G.f.: (1 + x)/(1 - 4*x - 2*x^2).
(End)
a(n) = A090017(n+1)+A090017(n). - R. J. Mathar, Aug 04 2019

Edited and new name by Armend Shabani and Andrew Howroyd, Nov 25 2018

A180032 Eight white queens and one red queen on a 3 X 3 chessboard. G.f.: (1+x)/(1-5x-7x^2).

Original entry on oeis.org

1, 6, 37, 227, 1394, 8559, 52553, 322678, 1981261, 12165051, 74694082, 458625767, 2815987409, 17290317414, 106163498933, 651849716563, 4002393075346, 24574913392671, 150891318490777, 926480986202582, 5688644160448349

Offset: 0

Johannes W. Meijer, Aug 09 2010

The a(n) represent the number of n-move routes of a fairy chess piece starting in a given corner or side square (m = 1, 3, 7, 9; 2, 4, 6, 8) on a 3 X 3 chessboard. This fairy chess piece behaves like a white chess queen on the eight side and corner squares but on the central square the queen explodes with fury and turns into a red queen.
On a 3 X 3 chessboard there are 2^9 = 512 ways to explode with fury on the central square (we assume here that a red queen might behave like a white queen). The red queen is represented by the A[5] vector in the fifth row of the adjacency matrix A, see the Maple program. For the corner and side squares the 512 red queens lead to 17 red queen sequences, see the cross-references for the complete set.
The sequence above corresponds to 8 red queen vectors, i.e., A[5] vectors, with decimal values 239, 367, 431, 463, 487, 491, 493 and 494. The central square leads for these vectors to A152240.
This sequence belongs to a family of sequences with g.f. (1+x)/(1 - 5*x - k*x^2). The members of this family that are red queen sequences are A180030 (k=8), A180032 (k=7; this sequence), A000400 (k=6), A180033 (k=5), A126501 (k=4), A180035 (k=3), A180037 (k=2) A015449 (k=1) and A003948 (k=0). Other members of this family are A030221 (k=-1), A109114 (k=-3), A020989 (k=-4), A166060 (k=-6).
Inverse binomial transform of A054413.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Wikipedia, Alice in Wonderland (2010 film)
Index entries for linear recurrences with constant coefficients, signature (5, 7).

Cf. A180028 (Central square).

Cf. Red queen sequences corner and side squares [decimal value A[5]]: A090018 [511], A135030 [255], A180030 [495], A005668 [127], A180032 [239], A000400 [63], A180033 [47], A001109 [31], A126501 [15], A154244 [23], A180035 [7], A138395 [19], A180037 [3], A084326 [17], A015449 [1], A003463 [16], A003948 [0].

I:=[1,6]; [n le 2 select I[n] else 5*Self(n-1)+7*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 15 2011

with(LinearAlgebra): nmax:=20; m:=1; A[5]:= [1,1,1,1,0,1,1,1,0]: A:=Matrix([[0,1,1,1,1,0,1,0,1], [1,0,1,1,1,1,0,1,0], [1,1,0,0,1,1,1,0,1], [1,1,0,0,1,1,1,1,0], A[5], [0,1,1,1,1,0,0,1,1], [1,0,1,1,1,0,0,1,1], [0,1,0,1,1,1,1,0,1], [1,0,1,0,1,1,1,1,0]]): for n from 0 to nmax do B(n):=A^n: a(n):= add(B(n)[m,k],k=1..9): od: seq(a(n), n=0..nmax);

LinearRecurrence[{5,7},{1,6},40] (* Vincenzo Librandi, Nov 15 2011 *)
CoefficientList[Series[(1+x)/(1-5x-7x^2),{x,0,30}],x] (* Harvey P. Dale, Apr 04 2024 *)

G.f.: (1+x)/(1 - 5*x - 7*x^2).
a(n) = 5*a(n-1) + 7*a(n-2) with a(0) = 1 and a(1) = 6.
a(n) = ((7+9*A)*A^(-n-1) + (7+9*B)*B^(-n-1))/53 with A = (-5+sqrt(53))/14 and B = (-5-sqrt(53))/14.

A342133 Square array T(n,k), n>=0, k>=0, read by antidiagonals, where column k is the expansion of g.f. 1/(1 - 2kx + k*x^2).

Original entry on oeis.org

1, 1, 0, 1, 2, 0, 1, 4, 3, 0, 1, 6, 14, 4, 0, 1, 8, 33, 48, 5, 0, 1, 10, 60, 180, 164, 6, 0, 1, 12, 95, 448, 981, 560, 7, 0, 1, 14, 138, 900, 3344, 5346, 1912, 8, 0, 1, 16, 189, 1584, 8525, 24960, 29133, 6528, 9, 0, 1, 18, 248, 2548, 18180, 80750, 186304, 158760, 22288, 10, 0

Offset: 0

Seiichi Manyama, Mar 01 2021

			Square array begins:
  1, 1,   1,    1,     1,     1, ...
  0, 2,   4,    6,     8,    10, ...
  0, 3,  14,   33,    60,    95, ...
  0, 4,  48,  180,   448,   900, ...
  0, 5, 164,  981,  3344,  8525, ...
  0, 6, 560, 5346, 24960, 80750, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened
Index entries for sequences related to Chebyshev polynomials.

Columns 0..5 give A000007, A000027(n+1), A007070, A138395, A099156(n+1), A190987(n+1).
Rows 0..2 give A000012, A005843, A033991.
Main diagonal gives (-1) * A109520(n+1).
Cf. A342120, A342129, A342134.

T:= (n, k)-> (<<0|1>, <-k|2*k>>^(n+1))[1, 2]:
seq(seq(T(n, d-n), n=0..d), d=0..12);  # Alois P. Heinz, Mar 01 2021

T[n_, k_] := Sum[If[k == j == 0, 1, (2*k)^j] * (-2)^(j - n) * Binomial[j, n - j], {j, 0, n}]; Table[T[k, n - k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, Apr 27 2021 *)

T(n, k) = sum(j=0, n\2, (2*k)^(n-j)*(-2)^(-j)*binomial(n-j, j));

T(n, k) = sum(j=0, n, (2*k)^j*(-2)^(j-n)*binomial(j, n-j));

T(n, k) = round(sqrt(k)^n*polchebyshev(n, 2, sqrt(k)));

T(0,k) = 1, T(1,k) = 2*k and T(n,k) = k*(2*T(n-1,k) - T(n-2,k)) for n > 1.
T(n,k) = Sum_{j=0..floor(n/2)} (2*k)^(n-j) * (-1/2)^j * binomial(n-j,j) = Sum_{j=0..n} (2*k)^j * (-1/2)^(n-j) * binomial(j,n-j).
T(n,k) = sqrt(k)^n * U(n, sqrt(k)) where U(n, x) is a Chebyshev polynomial of the second kind.

A158869 Number of ways of filling a 2 X 3 X 2n hole with 1 X 2 X 2 bricks.

Original entry on oeis.org

1, 5, 27, 147, 801, 4365, 23787, 129627, 706401, 3849525, 20977947, 114319107, 622980801, 3394927485, 18500622507, 100818952587, 549411848001, 2994014230245, 16315849837467, 88913056334067

Offset: 0

Martin Griffiths, Mar 28 2009

Note that it is not possible to fill a 2 X 3 X (2*n-1) hole using 1 X 2 X 2 bricks.
a(n+1) of the Jacobsthal sequence A001045 gives the number of ways of filling a 2 X 2 X n hole with 1 X 2 X 2 bricks.
Will the pattern of rightmost digits (1,5,7,7) be continued? - Bill McEachen, May 20 2009
The answer to the question in a previous comment is: the linear recurrence proves that the pattern 1, 5, 7, 7 of the least significant digits will continue. - R. J. Mathar, Jun 20 2010
a(n) is the number of compositions of n when there are 5 types of 1 and 2 types of other natural numbers. - Milan Janjic, Aug 13 2010

Martin Griffiths, Filling cuboidal holes with bricks, Mathematical Spectrum (Applied Probability Trust) 42(2) (2010), 91-92.
Dimana Miroslavova Pramatarova, Investigating the Periodicity of Weighted Catalan Numbers and Generalizing Them to Higher Dimensions, MIT Res. Sci. Instit. (2025). See p. 12.
Index entries for linear recurrences with constant coefficients, signature (6,-3).

Simplify[Table[ 1/6 * ((3 + Sqrt[6])^(n + 1) + (3 - Sqrt[6])^(n + 1)), {n, 0, 19}]]
Table[3^n * Hypergeometric2F1[ -((n + 1)/2), -(n/2), 1/2, 2/3], {n, 0, 19}]
LinearRecurrence[{6,-3},{1,5},30] (* Harvey P. Dale, May 28 2015 *)

def A158869(n): return 3^n*lucas_number2(n+1, 2, 1/3)/2
[A158869(n) for n in (0..19)]  # Peter Luschny, May 06 2013

a(0)=1, a(1)=5 and a(n) = 6*a(n-1) - 3*a(n-2) for n > 1.
a(n) = (3^n) * 2F1[-((n + 1)/2), -(n/2); 1/2; 2/3], using Gauss' hypergeometric function.
From Martin Griffiths, Apr 02 2009: (Start)
G.f.: A(x) = (1-x)/(1-6x+3x^2).
a(n) = (1/6)*((3+sqrt(6))^(n+1) + (3-sqrt(6))^(n+1)). (End)
From R. J. Mathar, Mar 29 2009: (Start)
G.f.: -(-1+x)/(1-6*x+3*x^2).
a(n) = A138395(n+1) - A138395(n). (End)
G.f.: G(0)/(6*x) -1/(3*x), where G(k) = 1 + 1/(1 - x*(2*k-3)/(x*(2*k-1) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 13 2013

Edited by Charles R Greathouse IV, Mar 08 2011

A201972 Triangle T(n,k), read by rows, given by (2,1/2,-1/2,0,0,0,0,0,0,0,...) DELTA (2,-1/2,1/2,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 2, 2, 5, 8, 3, 12, 28, 20, 4, 29, 88, 94, 40, 5, 70, 262, 372, 244, 70, 6, 169, 752, 1333, 1184, 539, 112, 7, 408, 2104, 4472, 5016, 3144, 1064, 168, 8, 985, 5776, 14316, 19408, 15526, 7344, 1932, 240, 9

Offset: 0

Philippe Deléham, Dec 07 2011

Diagonal sums: A201967(n), row sums: A000302(n) (powers of 4).

			Triangle begins:
    1;
    2,   2;
    5,   8,    3;
   12,  28,   20,    4;
   29,  88,   94,   40,   5;
   70, 262,  372,  244,  70,   6;
  169, 752, 1333, 1184, 539, 112, 7;

G. C. Greubel, Rows n = 0..100 of triangle, flattened

Cf. A000027, A000129, A006645, A007290.

T:= proc(n, k) option remember;
      if k=0 and n=0 then 1
    elif k<0 or  k>n  then 0
    else 2*T(n-1,k) + 2*T(n-1,k-1) + T(n-2,k) - T(n-2,k-2)
      fi; end:
seq(seq(T(n, k), k=0..n), n=0..10); # G. C. Greubel, Feb 17 2020

With[{m = 8}, CoefficientList[CoefficientList[Series[1/(1-2*(y+1)*x+(y+1)*(y-1)*x^2), {x, 0 , m}, {y, 0, m}], x], y]] // Flatten (* Georg Fischer, Feb 17 2020 *)

T(n,k) = if(nMichel Marcus, Feb 17 2020

@CachedFunction
def T(n, k):
    if (k<0 or k>n): return 0
    elif (k==0 and n==0): return 1
    else: return 2*T(n-1,k) + 2*T(n-1,k-1) + T(n-2,k) - T(n-2,k-2)
[[T(n, k) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Feb 17 2020

G.f.: 1/(1-2*(y+1)*x+(y+1)*(y-1)*x^2).
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A000129(n+1), A000302(n), A138395(n), A057084(n) for x = -1, 0, 1, 2, 3, respectively.
Sum_{k=0..n} T(n,k)*x^(n-k) = A000027(n), A000302(n), A090018(n), A057091(n) for x = 0, 1, 2, 3, respectively.
T(n,k) = 2*T(n-1,k) + 2*T(n-1,k-1) + T(n-2,k) - T(n-2,k-2) with T(0,0) = 1, T(n,k) = 0 if k < 0 or if n < k.

a(40) corrected by Georg Fischer, Feb 17 2020

A109112 a(n) = 6a(n-1) - 3a(n-2), a(0)=2, a(1)=13.

Original entry on oeis.org

2, 13, 72, 393, 2142, 11673, 63612, 346653, 1889082, 10294533, 56099952, 305716113, 1665996822, 9078832593, 49475005092, 269613532773, 1469256181362, 8006696489853, 43632410395032, 237774372900633, 1295749006218702

Offset: 0

Emeric Deutsch, Jun 19 2005

Kekulé numbers for certain benzenoids.

S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (p. 302, P_{14}).

Index entries for linear recurrences with constant coefficients, signature (6,-3).

a[0]:=2:a[1]:=13: for n from 2 to 24 do a[n]:=6*a[n-1]-3*a[n-2] od: seq(a[n],n=0..24);

LinearRecurrence[{6,-3},{2,13},30] (* Harvey P. Dale, Dec 15 2014 *)

a(n) = (1/(2*sqrt(6)))*((2*sqrt(6) + 7)*(3 + sqrt(6))^n + (2*sqrt(6) - 7)*(3 - sqrt(6))^n).
G.f.: (2+z)/(1 - 6z + 3z^2).
a(n) = 2*A138395(n) + A138395(n-1). - R. J. Mathar, Jul 22 2022

cp's OEIS Frontend

A190958 a(n) = 2*a(n-1) - 10*a(n-2), with a(0) = 0, a(1) = 1.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

SageMath

Formula

A084120 a(n) = 6*a(n-1) - 3*a(n-2), a(0)=1, a(1)=3.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Sage

Formula

A090018 a(n) = 6*a(n-1) + 3*a(n-2) for n > 2, a(0)=1, a(1)=6.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Sage

Formula

Extensions

A123347 Number of words of length n over the alphabet {1,2,3,4,5} such that 1 is not followed by an odd letter.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Sage

Formula

Extensions

A180032 Eight white queens and one red queen on a 3 X 3 chessboard. G.f.: (1+x)/(1-5*x-7*x^2).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

Formula

A342133 Square array T(n,k), n>=0, k>=0, read by antidiagonals, where column k is the expansion of g.f. 1/(1 - 2*k*x + k*x^2).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

A190958 a(n) = 2a(n-1) - 10a(n-2), with a(0) = 0, a(1) = 1.

A084120 a(n) = 6a(n-1) - 3a(n-2), a(0)=1, a(1)=3.

A090018 a(n) = 6a(n-1) + 3a(n-2) for n > 2, a(0)=1, a(1)=6.

A180032 Eight white queens and one red queen on a 3 X 3 chessboard. G.f.: (1+x)/(1-5x-7x^2).

A342133 Square array T(n,k), n>=0, k>=0, read by antidiagonals, where column k is the expansion of g.f. 1/(1 - 2kx + k*x^2).

A109112 a(n) = 6a(n-1) - 3a(n-2), a(0)=2, a(1)=13.