A052899 -id:A052899 - cp's OEIS frontend

A087131 a(n) = 2^n*Lucas(n), where Lucas = A000032.

2, 2, 12, 32, 112, 352, 1152, 3712, 12032, 38912, 125952, 407552, 1318912, 4268032, 13811712, 44695552, 144637952, 468058112, 1514668032, 4901568512, 15861809152, 51329892352, 166107021312, 537533612032, 1739495309312

Offset: 0

Paul Barry, Aug 16 2003

Number of ways to tile an n-bracelet with two types of colored squares and four types of colored dominoes.
Inverse binomial transform of even Lucas numbers (A014448).
From L. Edson Jeffery, Apr 25 2011: (Start)
Let A be the unit-primitive matrix (see [Jeffery])
A=A_(10,4)=
(0 0 0 0 1)
(0 0 0 2 0)
(0 0 2 0 1)
(0 2 0 2 0)
(2 0 2 0 1).
Then a(n)=(Trace(A^n)-1)/2. Also a(n)=Trace((2*A_(5,1))^n), where A_(5,1)=[(0,1); (1,1)] is also a unit-primitive matrix. (End)
Also the number of connected dominating sets in the n-sun graph for n >= 3. - Eric W. Weisstein, May 02 2017
Also the number of total dominating sets in the n-sun graph for n >= 3. - Eric W. Weisstein, Apr 27 2018

Arthur T. Benjamin and Jennifer J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A., 2003, identity 237, p. 132.

L. Edson Jeffery, Unit-primitive matrices
Eric Weisstein's World of Mathematics, Connected Dominating Set.
Eric Weisstein's World of Mathematics, Sun Graph.
Eric Weisstein's World of Mathematics, Total Dominating Set.
Index entries for linear recurrences with constant coefficients, signature (2,4).

First differences of A006483 and A103435.

Cf. A000032, A014334, A014335, A084057, A269992.

[2] cat [2^n*Lucas(n): n in [1..30]]; // G. C. Greubel, Dec 18 2017

Table[Tr[MatrixPower[{{2, 2}, {2, 0}}, x]], {x, 1, 20}] (* Artur Jasinski, Jan 09 2007 *)
Join[{2}, Table[2^n LucasL[n], {n, 20}]] (* Eric W. Weisstein, May 02 2017 *)
Join[{2}, 2^# LucasL[#] & [Range[20]]] (* Eric W. Weisstein, May 02 2017 *)
LinearRecurrence[{2, 4}, {2, 12}, {0, 20}] (* Eric W. Weisstein, Apr 27 2018 *)
CoefficientList[Series[(2 (-1 + x))/(-1 + 2 x + 4 x^2), {x, 0, 20}], x] (* Eric W. Weisstein, Apr 27 2018 *)

for(n=0,30, print1(if(n==0, 2, 2^n*(fibonacci(n+1) + fibonacci(n-1))), ", ")) \\ G. C. Greubel, Dec 18 2017

first(n) = Vec(2*(1-x)/(1-2*x-4*x^2) + O(x^n)) \\ Iain Fox, Dec 19 2017

[lucas_number2(n,2,-4) for n in range(0, 25)] # Zerinvary Lajos, Apr 30 2009

a(n) = 2*A084057(n).
Recurrence: a(n) = 2a(n-1) + 4a(n-2), a(0)=2, a(1)=2.
G.f.: 2*(1-x)/(1-2*x-4*x^2).
a(n) = (1+sqrt(5))^n + (1-sqrt(5))^n.
For n>=2, a(n) = Trace of matrix [({2,2},{2,0})^n]. - Artur Jasinski, Jan 09 2007
a(n) = 2*[A063727(n)-A063727(n-1)]. - R. J. Mathar, Nov 16 2007
a(n) = (5*A052899(n)-1)/2. - L. Edson Jeffery, Apr 25 2011
a(n) = [x^n] ( 1 + x + sqrt(1 + 2*x + 5*x^2) )^n for n >= 1. - Peter Bala, Jun 23 2015
Sum_{n>=1} 1/a(n) = (1/2) * A269992. - Amiram Eldar, Nov 17 2020
From Amiram Eldar, Jan 15 2022: (Start)
a(n) == 2 (mod 10).
a(n) = 5 * A014334(n) + 2.
a(n) = 10 * A014335(n) + 2. (End)

Edited by Ralf Stephan, Feb 08 2005

A189318 Expansion of 5(1-2x)/(1-3x-2x^2+4*x^3).

Original entry on oeis.org

5, 5, 25, 65, 225, 705, 2305, 7425, 24065, 77825, 251905, 815105, 2637825, 8536065, 27623425, 89391105, 289275905, 936116225, 3029336065, 9803137025, 31723618305, 102659784705, 332214042625, 1075067224065, 3478990618625, 11258250133505

Offset: 0

L. Edson Jeffery, Apr 20 2011

(Start) Let A be the unit-primitive matrix (see [Jeffery])
A=A_(10,4)=
(0 0 0 0 1)
(0 0 0 2 0)
(0 0 2 0 1)
(0 2 0 2 0)
(2 0 2 0 1).
Then a(n)=Trace(A^n). (End)
Evidently one of a class of accelerator sequences for Catalan's constant based on traces of successive powers of a unit-primitive matrix A_(N,r) (0

L. E. Jeffery, Unit-primitive matrices.
Index entries for linear recurrences with constant coefficients, signature (3, 2, -4).

Cf. A052899.

A189315, A189316, A189317.

CoefficientList[Series[5(1-2x)/(1-3x-2x^2+4x^3),{x,0,30}],x] (* or *) LinearRecurrence[{3,2,-4},{5,5,25},30] (* Harvey P. Dale, Jun 02 2014 *)

Vec(5*(1-2*x)/(1-3*x-2*x^2+4*x^3)+O(x^99)) \\ Charles R Greathouse IV, Sep 25 2012

G.f.: 5*(1-2*x)/(1-3*x-2*x^2+4*x^3).
a(n)=3*a(n-1)+2*a(n-2)-4*a(n-3), n>3, a(0)=5, a(1)=5, a(2)=25, a(3)=65.
a(n)=Sum_{k=1..5} ((w_k)^4-3*(w_k)^2+1)^n, w_k=2*cos((2*k-1)*Pi/10).
a(n)=1+2*(1-Sqrt(5))^n+2*(1+Sqrt(5))^n.
a(n)=5*A052899(n).

A081057 E.g.f.: Sum_{n>=0} a(n)x^n/n! = {Sum_{n>=0} F(n+1)x^n/n!}^2, where F(n) is the n-th Fibonacci number.

Original entry on oeis.org

1, 2, 6, 18, 58, 186, 602, 1946, 6298, 20378, 65946, 213402, 690586, 2234778, 7231898, 23402906, 75733402, 245078426, 793090458, 2566494618, 8305351066, 26876680602, 86974765466, 281456253338, 910811568538, 2947448150426, 9538142575002, 30866077751706

Offset: 0

Paul D. Hanna, Mar 03 2003

nonn

a(n) ~ c*(sqrt(5)+1)^n, where c = (sqrt(5)+3)/10.
The inverse binomial transform is 1,1,3,5,... (1 followed by A056487). Partial sum of 1,1,4,12,..., i.e., 1 plus n-th partial sum of A087206. [R. J. Mathar, Oct 04 2010]
From R. J. Mathar, Oct 12 2010: (Start)
Apparently the row n=4 of an array which counts walks with k steps on an n X n board, starting at a corner, each step to one of the <= 4 adjacent squares:
1,2,4,8,16,32,64,128,256,512,1024,2048,4096,
1,2,6,16,48,128,384,1024,3072,8192,24576,65536,196608,
1,2,6,18,58,186,602,1946,6298,20378,65946,213402,690586,
1,2,6,18,60,198,684,2322,8100,27702,96876,331938,1161540,
1,2,6,18,60,200,698,2432,8658,30762,110374,395428,1422916,
1,2,6,18,60,200,700,2448,8800,31552,115104,418176,1537536,
1,2,6,18,60,200,700,2450,8818,31730,116182,425172,1573416,
1,2,6,18,60,200,700,2450,8820,31750,116400,426600,1583400,
1,2,6,18,60,200,700,2450,8820,31752,116422,426862,1585246,
1,2,6,18,60,200,700,2450,8820,31752,116424,426886,1585556,
1,2,6,18,60,200,700,2450,8820,31752,116424,426888,1585582,
(End)
Decomposing rook walks of length=n on a 4 X 4 board into combinations of independent vertical and horizontal walks in 4-wide corridors leads to an exponential convolution of the Fibonacci numbers, cf. A052899. [David Scambler, Oct 17 2010]

Index entries for linear recurrences with constant coefficients, signature (3,2,-4).

a(n) = A052899(n-1) + A052899(n). a(n) - 2*a(n-1) = A014334(n).
Cf. A000032, A000045, A000204.
Row sums of A109906.

G.f.: (1-x-2x^2)/(1-3x-2x^2+4x^3). - Michael Somos, Mar 04 2003
a(n) - 2*a(n-1) = A014334(n), n > 0. - Vladeta Jovovic, Mar 05 2003
From Vladeta Jovovic, Mar 05 2003: (Start)
a(n) = 2/5 + (3/10 - 1/10*5^(1/2))*(1 - 5^(1/2))^n + (3/10 + 1/10*5^(1/2))*(1 + 5^(1/2))^n.
Recurrence: a(n) = 3*a(n-1) + 2*a(n-2) - 4*a(n-3).
G.f.: (1+x)*(1-2*x)/(1-2*x-4*x^2)/(1-x). (End)
a(n) = Sum_{k=0..n} ( F(k+1) * F(n-k+1) * C(n,k) ), where F(k) = Fibonacci(k). - David Scambler, Oct 17 2010
a(n) = (2^n*Lucas(n+2)+2)/5. - Ira M. Gessel, Mar 06 2022

Corrected and extended by Vladeta Jovovic and Michael Somos, Mar 05 2003

A376837 a(n) is the number of paths to reach a position outside an 8 X 8 chessboard after n steps, starting in one of the corners, when performing a walk with unit steps on the square lattice.

Original entry on oeis.org

2, 2, 6, 12, 40, 100, 350, 982, 3542, 10738, 39556, 127272, 475332, 1602458, 6030830, 21056830, 79514918, 284645860, 1075801928, 3917238476, 14799350958, 54498514998, 205721183302, 763140403282, 2878050335900, 10726898070952, 40421307665420, 151112554663930, 569043610134622, 2131459901180670

Offset: 1

Ruediger Jehn, Oct 06 2024

a(n)/4^n is the probability that the 1-step rook falls off the chess board at step n. The average number of steps it takes this piece to fall off the board is Sum_{n>0} n*a(n)/4^n = A376606(8)/A376607(8) = 4374/901 or approximately 4.855 steps.

Because of the mirror symmetry of the problem to the board diagonal, all terms are even.

			a(3) = 6. Starting on square a1 there are 6 paths to leave the chess board: up-up-left, up-down-left, up-down-down, right-right-down, right-left-down and right-left-left.

Andrew Howroyd, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (5,9,-69,21,225,-171,-162,108,32,-16).

Cf. A376606, A376607, {A052899}+1 is the analog for the 4X4 chessboard.

LinearRecurrence[{5, 9, -69, 21, 225, -171, -162, 108, 32, -16}, {2, 2, 6, 12, 40, 100, 350, 982, 3542, 10738}, 30] (* Hugo Pfoertner, Oct 16 2024 *)

Vec(2*(1 - 4*x - 11*x^2 + 51*x^3 + 11*x^4 - 143*x^5 + 42*x^6 + 78*x^7 - 12*x^8 - 8*x^9)/((1 - 2*x)*(1 - 3*x^2 + x^3)*(1 - 3*x + x^3)*(1 - 12*x^2 - 8*x^3)) + O(x^30)) \\ Andrew Howroyd, Oct 16 2024

a(n) == 0 (mod 2).

G.f.: 2*x*(1 - 4*x - 11*x^2 + 51*x^3 + 11*x^4 - 143*x^5 + 42*x^6 + 78*x^7 - 12*x^8 - 8*x^9)/((1 - 2*x)*(1 - 3*x^2 + x^3)*(1 - 3*x + x^3)*(1 - 12*x^2 - 8*x^3)). - Andrew Howroyd, Oct 16 2024

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cp's OEIS Frontend

A087131 a(n) = 2^n*Lucas(n), where Lucas = A000032.

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A189318 Expansion of 5*(1-2*x)/(1-3*x-2*x^2+4*x^3).

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A081057 E.g.f.: Sum_{n>=0} a(n)*x^n/n! = {Sum_{n>=0} F(n+1)*x^n/n!}^2, where F(n) is the n-th Fibonacci number.

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A376837 a(n) is the number of paths to reach a position outside an 8 X 8 chessboard after n steps, starting in one of the corners, when performing a walk with unit steps on the square lattice.

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Formula

A189318 Expansion of 5(1-2x)/(1-3x-2x^2+4*x^3).

A081057 E.g.f.: Sum_{n>=0} a(n)x^n/n! = {Sum_{n>=0} F(n+1)x^n/n!}^2, where F(n) is the n-th Fibonacci number.