A015523 -id:A015523 - cp's OEIS frontend

A072271 A partial product representation of f(n) = A015523(n) and L(n) = A072263(n).

3, 1, 24, 19, 431, 14, 7589, 311, 5559, 241, 2345179, 286, 41223001, 4229, 70051, 95471, 12736968311, 5309, 223887209309, 88321, 21607111, 1306469, 69176042380099, 94846, 2821250547551, 22964761, 160204320879, 27289081, 375703599163598591, 119641

Offset: 1

Miklos Kristof, Jul 09 2002

nonn

For even n, f(n) = Product_{d|n} a(d); for odd n, f(n) = Product_{d|n} a(2d).
For odd prime p, a(p) = L(p)/3, where L(n) = 5*f(n-1) + f(n+1).
a(1)=3, a(2)=1.
a(2p) = f(p) for odd primes p.
a(2^(k+1)) = L(2^k).
a(3*2^k) = L(2^k) - 5^k.
For odd n, L(n) = Product_{d|n} a(d).
For k > 0 and odd n, L(n*2^k) = Product_{d|n} a(d*2^(k+1)).

			f(12) = a(1)*a(2)*a(3)*a(4)*a(6)*a(12) = 3*1*24*19*14*286 = 5477472 for even n;
f(7) = a(2)*a(14) = 1*4229 = 4229 for odd n.
L(6) = a(4)*a(12) = 19*286 = 5434 = 5*f(5) + f(7) = 5*241 + 4229 for even n;
L(15) = a(1)*a(3)*a(5)*a(15) = 3*24*431*70051 = 2173822632 for odd n.

Cf. A072270, A072280, A072183, A127259, A127607.

a(n) = (h-3)^g(n) * K(n, h^2/5) for n > 2 where h = (3+sqrt(29))/2, Phi(n, x) = n-th cyclotomic polynomial and g(n) is the order of Phi(n, x).

More terms and entry revised by Sean A. Irvine, Sep 19 2024

A052918 a(0) = 1, a(1) = 5, a(n+1) = 5*a(n) + a(n-1).

Original entry on oeis.org

1, 5, 26, 135, 701, 3640, 18901, 98145, 509626, 2646275, 13741001, 71351280, 370497401, 1923838285, 9989688826, 51872282415, 269351100901, 1398627786920, 7262490035501, 37711077964425, 195817879857626

Offset: 0

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

A087130(n)^2 - 29*a(n-1)^2 = 4*(-1)^n, n >= 1. - Gary W. Adamson, Jul 01 2003, corrected Oct 07 2008, corrected by Jianing Song, Feb 01 2019
a(p-1) == 29^((p-1)/2) (mod p), for odd primes p. - Gary W. Adamson, Feb 22 2009 [See A087475 for more info about this congruence. - Jason Yuen, Apr 05 2025]
For more information about this type of recurrence, follow the Khovanova link and see A054413, A086902 and A178765. - Johannes W. Meijer, Jun 12 2010
Binomial transform of A015523. - Johannes W. Meijer, Aug 01 2010
For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 5's along the main diagonal and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
a(n) equals the number of words of length n on alphabet {0,1,...,5} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
From Michael A. Allen, Feb 15 2023: (Start)
Also called the 5-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence.
a(n) is the number of tilings of an n-board (a board with dimensions n X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are 5 kinds of squares available. (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
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 8.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 901
Milan Janjić, Hessenberg Matrices and Integer Sequences, J. Int. Seq. 13 (2010) # 10.7.8, section 3.
Milan Janjić, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Tanya Khovanova, Recursive Sequences
Kai Wang, On k-Fibonacci Sequences And Infinite Series List of Results and Examples, 2020.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (5,1).

Row 5 of A073133, A172236, and A352361.

Cf. A087130, A099365 (squares), A100237, A175184 (Pisano periods), A201005 (prime subsequence).

a:=[1,5];; for n in [3..30] do a[n]:=5*a[n-1]+a[n-2]; od; a; # G. C. Greubel, Oct 16 2019

I:=[1, 5]; [n le 2 select I[n] else 5*Self(n-1)+Self(n-2): n in [1..30]]; // Vincenzo Librandi, Feb 23 2013

R:=PowerSeriesRing(Integers(), 22); Coefficients(R!( 1/(1 - 5*x - x^2) )); // Marius A. Burtea, Oct 16 2019

spec := [S,{S=Sequence(Union(Z,Z,Z,Z,Z,Prod(Z,Z)))},unlabeled]: seq(combstruct[count](spec,size=n), n=0..30);
a[0]:=1: a[1]:=5: for n from 2 to 26 do a[n]:=5*a[n-1]+a[n-2] od: seq(a[n], n=0..30); # Zerinvary Lajos, Jul 26 2006
with(combinat):a:=n->fibonacci(n,5):seq(a(n),n=1..30); # Zerinvary Lajos, Dec 07 2008

LinearRecurrence[{5, 1}, {1, 5}, 30] (* Vincenzo Librandi, Feb 23 2013 *)
Table[Fibonacci[n+1, 5], {n,0,30}] (* Vladimir Reshetnikov, May 08 2016 *)

Vec(1/(1-5*x-x^2)+O(x^30)) \\ Charles R Greathouse IV, Nov 20 2011

[lucas_number1(n,5,-1) for n in range(1, 22)] # Zerinvary Lajos, Apr 24 2009

G.f.: 1/(1 - 5*x - x^2).
a(3n) = A041047(5n), a(3n+1) = A041047(5n+3), a(3n+2) = 2*A041047(5n+4). - Henry Bottomley, May 10 2000
a(n) = Sum_{alpha=RootOf(-1+5*z+z^2)} (1/29)*(5+2*alpha)*alpha^(-1-n).
a(n-1) = (((5 + sqrt(29))/2)^n - ((5 - sqrt(29))/2)^n)/sqrt(29). - Gary W. Adamson, Jul 01 2003
a(n) = U(n, 5*i/2)*(-i)^n with i^2 = -1 and Chebyshev's U(n, x/2) = S(n, x) polynomials. See triangle A049310.
Let M = {{0, 1}, {1, 5}}, then a(n) is the lower-right term of M^n. - Roger L. Bagula, May 29 2005
a(n) = F(n, 5), the n-th Fibonacci polynomial evaluated at x = 5. - T. D. Noe, Jan 19 2006
a(n) = denominator of n-th convergent to [1, 4, 5, 5, 5, ...], for n > 0. Continued fraction [1, 4, 5, 5, 5, ...] = 0.807417596..., the inradius of a right triangle with legs 2 and 5. n-th convergent = A100237(n)/A052918(n), the first few being: 1/1, 4/5, 21/26, 109/135, 566/701, ... - Gary W. Adamson, Dec 21 2007
From Johannes W. Meijer, Jun 12 2010: (Start)
a(2n+1) = 5*A097781(n), a(2n) = A097835(n).
Limit_{k->oo} a(n+k)/a(k) = (A087130(n) + a(n-1)*sqrt(29))/2.
Limit_{n->oo} A087130(n)/a(n-1) = sqrt(29). (End)
From L. Edson Jeffery, Jan 07 2012: (Start)
Define the 2 X 2 matrix A = {{1, 1}, {5, 4}}. Then:
a(n) is the upper-left term of (1/5)*(A^(n+2) - A^(n+1));
a(n) is the upper-right term of A^(n+1);
a(n) is the lower-left term of (1/5)*A^(n+1);
a(n) is the lower-right term of (Sum_{k=0..n} A^k). (End)
Sum_{n>=0} (-1)^n/(a(n)*a(n+1)) = (sqrt(29) - 5)/2. - Vladimir Shevelev, Feb 23 2013
G.f.: x/(1 - 5*x - x^2) = Sum_{n >= 0} x^(n+1) *( Product_{k = 1..n} (m*k + 5 - m + x)/(1 + m*k*x) ) for arbitrary m (a telescoping series). - Peter Bala, May 08 2024

A057088 Scaled Chebyshev U-polynomials evaluated at i*sqrt(5)/2. Generalized Fibonacci sequence.

Original entry on oeis.org

1, 5, 30, 175, 1025, 6000, 35125, 205625, 1203750, 7046875, 41253125, 241500000, 1413765625, 8276328125, 48450468750, 283633984375, 1660422265625, 9720281250000, 56903517578125, 333118994140625, 1950112558593750, 11416157763671875, 66831351611328125, 391237546875000000

Offset: 0

Wolfdieter Lang, Aug 11 2000

a(n) gives the length of the word obtained after n steps with the substitution rule 0->11111, 1->111110, starting from 0. The number of 1's and 0's of this word is 5*a(n-1) and 5*a(n-2), resp.
a(n) / a(n-1) converges to (5 + (3 * sqrt(5))) / 2 as n approaches infinity. (5 + (3 * sqrt(5))) / 2 can also be written as phi^2 + (2 * phi), phi^3 + phi, phi + sqrt(5) + 2, (3 * phi) + 1, (3 * phi^2) - 2, phi^4 - 1 and (5 + (3 * (L(n) / F(n)))) / 2, where L(n) is the n-th Lucas number and F(n) is the n-th Fibonacci number as n approaches infinity. - Ross La Haye, Aug 18 2003, on another version
Pisano period lengths: 1, 3, 3, 6, 1, 3, 24, 12, 9, 3, 10, 6, 56, 24, 3, 24,288, 9, 18, 6, ... - R. J. Mathar, Aug 10 2012

Indranil Ghosh, Table of n, a(n) for n = 0..1300
Martin Burtscher, Igor Szczyrba, Rafał Szczyrba, Analytic Representations of the n-anacci Constants and Generalizations Thereof, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.
A. F. Horadam, Special properties of the sequence W_n(a,b; p,q), Fib. Quart., 5.5 (1967), 424-434. Case n->n+1, a=0,b=1; p=5, q=5.
Tanya Khovanova, Recursive Sequences
W. Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38 (2000) 408-419. Eqs.(39) and (45),rhs, m=5.
Eric Weisstein's World of Mathematics, Horadam Sequence
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (5,5)

Cf. A001076, A006190, A007482, A015520, A015521, A015523, A015524, A015525, A015528, A015529, A015530, A015531, A015532, A015533, A015534, A015535, A015536, A015537, A015443, A015447, A030195, A053404, A057087, A083858, A085939, A090017, A091914, A099012, A180222, A180226.

I:=[1, 5]; [n le 2 select I[n] else 5*Self(n-1) + 5*Self(n-2): n in [0..30]]; // G. C. Greubel, Jan 16 2018

a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=5*a[n-1]+5*a[n-2]od: seq(a[n], n=1..33); # Zerinvary Lajos, Dec 14 2008

LinearRecurrence[{5,5}, {1,5}, 30] (* G. C. Greubel, Jan 16 2018 *)

x='x+O('x^30); Vec(1/(1 - 5*x - 5*x^2)) \\ G. C. Greubel, Jan 16 2018

[lucas_number1(n,5,-5) for n in range(1, 22)] # Zerinvary Lajos, Apr 24 2009

a(n) = 5*(a(n-1) + a(n-2)), a(-1)=0, a(0)=1.
a(n) = S(n, i*sqrt(5))*(-i*sqrt(5))^n with S(n, x) := U(n, x/2), Chebyshev's polynomials of the 2nd kind, A049310.
G.f.: 1/(1 - 5*x - 5*x^2).
a(n) = (1/3)*Sum_{k=0..n} binomial(n, k)*Fibonacci(k)*3^k. - Benoit Cloitre, Oct 25 2003
a(n) = ((5 + 3*sqrt(5))/2)^n(1/2 + sqrt(5)/6) + (1/2 - sqrt(5)/6)((5 - 3*sqrt(5))/2)^n. - Paul Barry, Sep 22 2004
(a(n)) appears to be given by the floretion - 0.75'i - 0.5'j + 'k - 0.75i' + 0.5j' + 0.5k' + 1.75'ii' - 1.25'jj' + 1.75'kk' - 'ij' - 0.5'ji' - 0.75'jk' - 0.75'kj' - 1.25e ("jes"). - Creighton Dement, Nov 28 2004
a(n) = Sum_{k=0..n} 4^k*A063967(n,k). - Philippe Deléham, Nov 03 2006
G.f.: G(0)/(2-5*x), where G(k)= 1 + 1/(1 - x*(9*k-5)/(x*(9*k+4) - 2/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 17 2013
From Ehren Metcalfe, Nov 18 2017: (Start)
With F(n) = A000045(n), L(n) = A000032(n), beta = (1-sqrt(5))/2:
a(2*n-1) = 5^n*F(4*n)/3 = (5^(n-1/2)*L(4*n) - 2*5^(n-1/2)*beta^(4*n))/3.
a(2*n) = 5^n*L(4*n+2)/3 = (5^(n+1/2)*F(4*n+2) + 2*5^n*beta^(4*n+2))/3.
a(n) = round 5^((n+1)/2)*F(2*(n+1))/3.
a(n) = round 5^(n/2)*L(2*(n+1))/3. (End)

A179596 Eight white kings and one red king on a 3 X 3 chessboard. G.f.: (1 + x)/(1 - 2x - 11x^2 - 6*x^3).

Original entry on oeis.org

1, 3, 17, 73, 351, 1607, 7513, 34809, 161903, 751783, 3493353, 16227737, 75393055, 350251335, 1627192697, 7559508409, 35119644495, 163157037671, 757987215241, 3521419711833, 16359641017343, 76002822156295, 353090213774361

Offset: 0

Johannes W. Meijer, Jul 28 2010; edited Jun 21 2013

The a(n) represent the number of n-move routes of a fairy chess piece starting in a given corner square (m = 1, 3, 7 or 9) on a 3 X 3 chessboard. This fairy chess piece behaves like a king on the eight side and corner squares but on the center square the king goes crazy and turns into a red king.
On a 3 X 3 chessboard there are 2^9 = 512 ways to go crazy on the center square (off the center the piece behaves like a normal king). The red king is represented by the A[5] vector in the fifth row of the adjacency matrix A, see the Maple program and A180140. For the corner squares the 512 red kings lead to 47 different red king sequences, see the overview of the red king sequences.
The sequence above corresponds to four A[5] vectors with decimal [binary] values 367 [101 101 111], 463 [111 001 111], 487 [111 100 111] and 493 [111 101 101]. These vectors lead for the side squares to A126473 and for the central square to A179597.
This sequence belongs to a family of sequences with g.f. (1+x)/(1 - 2*x - (k+8)*x^2 - 2*k*x^3). Red king sequences that are members of this family are A083424 (k=0), A179604 (k=1), A179600 (k=2), A179596 (k=3; this sequence) and A086346 (k=4). Other members of this family are A015528 (k=5) and A179608 (k=-4).

Gary Chartrand, Introductory Graph Theory, pp. 217-221, 1984.

Vincenzo Librandi, Table of n, a(n) for n = 0..100
Charles Krauthammer, Did Chess Make Him Crazy?, Time, April 26, 2005.
Johannes W. Meijer, The red king sequences.
Index entries for linear recurrences with constant coefficients, signature (2,11,6).

Cf. A180140 (berserker sequences).

Cf. Red king sequences corner squares [decimal value A[5]]: A086346 [495], A015525 [239], A179596 [367], A179600 [335], A015524 [95], A083858 [31], A179604 [327], A015523 [27], A179610 [85], A083424 [325], A015521 [11], A007482 [2], A014335 [16].

nmax:=22; m:=1; A[1]:= [0,1,0,1,1,0,0,0,0]: A[2]:= [1,0,1,1,1,1,0,0,0]: A[3]:= [0,1,0,0,1,1,0,0,0]: A[4]:=[1,1,0,0,1,0,1,1,0]: A[5]:= [1,0,1,1,0,1,1,1,1]: A[6]:= [0,1,1,0,1,0,0,1,1]: A[7]:= [0,0,0,1,1,0,0,1,0]: A[8]:= [0,0,0,1,1,1,1,0,1]: A[9]:= [0,0,0,0,1,1,0,1,0]: A:=Matrix([A[1],A[2],A[3],A[4],A[5], A[6],A[7],A[8],A[9]]): 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[{2,11,6},{1,3,17},30] (* Harvey P. Dale, May 18 2011 *)

Vec((1+x)/(1-2*x-11*x^2-6*x^3)+O(x^99)) \\ Charles R Greathouse IV, Jul 16 2011

G.f.: (1+x)/(1 - 2*x - 11*x^2 - 6*x^3).
a(n) = 2*a(n-1) + 11*a(n-2) + 6*a(n-3) with a(0)=1, a(1)=3 and a(2)=17.
a(n) = (-1)^(-n)*2^(n+1)/9 + ((49+17*sqrt(7))*A^(-n) + (49-17*sqrt(7))*B^(-n))/126 with A = (-2+sqrt(7))/3 and B = (-2-sqrt(7))/3.
Lim_{k->infinity} a(n+k)/a(k) = (-1)^(n+1)*A000244(n)/(A015530(n)*sqrt(7) - A108851(n)).

A015442 a(n) = a(n-1) + 7*a(n-2), a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 1, 8, 15, 71, 176, 673, 1905, 6616, 19951, 66263, 205920, 669761, 2111201, 6799528, 21577935, 69174631, 220220176, 704442593, 2245983825, 7177081976, 22898968751, 73138542583, 233431323840, 745401121921, 2379420388801

Offset: 0

Olivier Gérard

One obtains A015523 through a binomial transform, and A197189 by shifting one place left (starting 1,1,8 with offset 0) followed by a binomial transform. - R. J. Mathar, Oct 11 2011
The compositions of n in which each positive integer is colored by one of p different colors are called p-colored compositions of n. For n>=2, 8*a(n-1) equals the number of 8-colored compositions of n, with all parts >=2, such that no adjacent parts have the same color. - Milan Janjic, Nov 26 2011
a(n+1) is the number of compositions (ordered partitions) of n into parts 1 and 2, where there are 7 sorts of part 2. - Joerg Arndt, Jan 16 2024
Pisano period lengths: 1, 3, 8, 6, 4, 24, 1, 6, 24, 12, 60, 24, 12, 3, 8, 6, 288, 24, 120, 12, ... - R. J. Mathar, Aug 10 2012

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Joerg Arndt, Matters Computational (The Fxtbook), section 14.8 "Strings with no two consecutive nonzero digits", pp.317-318
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Index entries for linear recurrences with constant coefficients, signature (1,7).

Cf. A015440, A015441, A049310.

I:=[0, 1]; [n le 2 select I[n] else Self(n-1) + 7*Self(n-2): n in [1..40]]; // Vincenzo Librandi, Oct 17 2012

LinearRecurrence[{1, 7}, {0, 1}, 30] (* Vincenzo Librandi, Oct 17 2012 *)
nxt[{a_,b_}]:={b,7a+b}; NestList[nxt,{0,1},30][[All,1]] (* Harvey P. Dale, Feb 25 2022 *)

concat(0,Vec(1/(1-x-7*x^2)+O(x^99))) \\ Charles R Greathouse IV, Mar 12 2014

[lucas_number1(n,1,-7) for n in range(0, 27)] # Zerinvary Lajos, Apr 22 2009

O.g.f.: x/(1-x-7x^2). - R. J. Mathar, May 06 2008
a(n) = ( ((1+sqrt(29))/2)^(n+1) - ((1-sqrt(29))/2)^(n+1) )/sqrt(29).
a(n) = 8*a(n-2) + 7*a(n-3) with characteristic polynomial x^3 - 8*x - 7. - Roger L. Bagula, May 30 2007
a(n+1) = Sum_{k=0..n} A109466(n,k)*(-7)^(n-k). - Philippe Deléham, Oct 26 2008
a(n) = (Sum_{1<=k<=n, k odd} C(n,k)*29^((k-1)/2))/2^(n-1). - Vladimir Shevelev, Feb 05 2014
a(n) = sqrt(-7)^(n-1)*S(n-1, 1/sqrt(-7)), with the Chebyshev polynomial S(n, x), and S(-1, x) = 1 (see A049310). - Wolfdieter Lang, Nov 26 2023

A083858 Expansion of x/(1 - 3x - 6x^2).

Original entry on oeis.org

0, 1, 3, 15, 63, 279, 1215, 5319, 23247, 101655, 444447, 1943271, 8496495, 37149111, 162426303, 710173575, 3105078543, 13576277079, 59359302495, 259535569959, 1134762524847, 4961500994295, 21693078131967, 94848240361671, 414703189876815, 1813199011800471

Offset: 0

Paul Barry, May 06 2003

Binomial transform of A015443. A row of array A083857.

Pisano period lengths: 1, 1, 1, 1, 12, 1, 8, 1, 1, 12, 110, 1, 168, 8, 12, 2, 16, 1, 360, 12, ... - R. J. Mathar, Aug 10 2012

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,6).

Cf. A015523, A015524.

I:=[0,1]; [n le 2 select I[n] else 3*Self(n-1) + 6*Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 16 2018

a[n_]:=(MatrixPower[{{1,2},{1,-4}},n].{{1},{1}})[[2,1]]; Table[Abs[a[n]],{n,-1,40}] (* Vladimir Joseph Stephan Orlovsky, Feb 19 2010 *)
LinearRecurrence[{3,6}, {0,1}, 30] (* G. C. Greubel, Jan 16 2018 *)

x='x+O('x^30); concat([0], Vec(x/(1-3*x-6*x^2))) \\ G. C. Greubel, Jan 16 2018

[lucas_number1(n,3,-6) for n in range(0, 24)] # Zerinvary Lajos, Apr 22 2009

a(n) = 3*a(n-1) + 6*a(n-2), a(0)=0, a(1)=1.
a(n) = (3*sqrt(33)/2 + 21/2)^(n/2)/sqrt(33) - (21/2 - 3*sqrt(33)/2)^(n/2)*(-1)^n/sqrt(33).
G.f.: Q(0)*x/2, where Q(k) = 1 + 1/(1 - x*(6*k+3 + 6*x )/( x*(6*k+6 + 6*x ) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Sep 21 2013
a(n) = B(n, k + 2^(n-1)) - B(n,k) where B(n,k) is formed by the family of recursions b(n) = 3*(b(n-1) + b(n-2))/2, with b(0) = 1 and b(1) = k, as explained further in A249861. - Richard R. Forberg, Nov 04 2014
a(n) = Sum_{k=0..n-1} 3^k * 2^(n-1-k) * binomial(k,n-1-k). - Seiichi Manyama, Aug 31 2025

A057089 Scaled Chebyshev U-polynomials evaluated at i*sqrt(6)/2. Generalized Fibonacci sequence.

Original entry on oeis.org

1, 6, 42, 288, 1980, 13608, 93528, 642816, 4418064, 30365280, 208700064, 1434392064, 9858552768, 67757668992, 465697330560, 3200729997312, 21998563967232, 151195763787264, 1039165966526976, 7142170381885440

Offset: 0

Wolfdieter Lang, Aug 11 2000

a(n) gives the length of the word obtained after n steps with the substitution rule 0->1^6, 1->(1^6)0, starting from 0. The number of 1's and 0's of this word is 6*a(n-1) and 6*a(n-2), resp.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Martin Burtscher, Igor Szczyrba, and Rafał Szczyrba, Analytic Representations of the n-anacci Constants and Generalizations Thereof, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.
A. F. Horadam, Special properties of the sequence W_n(a,b; p,q), Fib. Quart., 5.5 (1967), 424-434. Case n->n+1, a=0,b=1; p=6, q=6.
Tanya Khovanova, Recursive Sequences
Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38 (2000) 408-419. Eqs.(39) and (45),rhs, m=6.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (6,6).

Cf. A001076, A006190, A007482, A015520, A015521, A015523, A015524, A015525, A015528, A015529, A015530, A015531, A015532, A015533, A015534, A015535, A015536, A015537, A015440, A015441, A015443, A015444, A015445, A015447, A015548, A030195, A053404, A057087, A057088, A083858, A085939, A090017, A091914, A099012, A135030, A135032, A180222, A180226, A180250.

I:=[1,6]; [n le 2 select I[n] else 6*Self(n-1)+6*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 14 2011

Join[{a=0,b=1},Table[c=6*b+6*a;a=b;b=c,{n,100}]] (* Vladimir Joseph Stephan Orlovsky, Jan 16 2011 *)
LinearRecurrence[{6,6},{1,6},40] (* Harvey P. Dale, Nov 05 2011 *)

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

[lucas_number1(n,6,-6) for n in range(1, 21)] # Zerinvary Lajos, Apr 24 2009

a(n) = 6*a(n-1) + 6*a(n-2); a(0)=1, a(1)=6.
a(n) = S(n, i*sqrt(6))*(-i*sqrt(6))^n with S(n, x) := U(n, x/2), Chebyshev's polynomials of the 2nd kind, A049310.
G.f.: 1/(1-6*x-6*x^2).
a(n) = Sum_{k=0..n} 5^k*A063967(n,k). - Philippe Deléham, Nov 03 2006

A015537 Expansion of x/(1 - 5x - 4x^2).

Original entry on oeis.org

0, 1, 5, 29, 165, 941, 5365, 30589, 174405, 994381, 5669525, 32325149, 184303845, 1050819821, 5991314485, 34159851709, 194764516485, 1110461989261, 6331368012245, 36098688018269, 205818912140325, 1173489312774701, 6690722212434805

Offset: 0

Olivier Gérard

First differences give A122690(n) = {1, 4, 24, 136, 776, 4424, 25224, ...}. Partial sums of a(n) are {0, 1, 6, 35, 200, ...} = (A123270(n) - 1)/8. - Alexander Adamchuk, Nov 03 2006
For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 5's along the main diagonal, and 2's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 19 2011
Pisano period lengths:  1, 1, 8, 1, 4, 8, 48, 1, 24, 4, 40, 8, 42, 48, 8, 2, 72, 24, 360, 4, ... - R. J. Mathar, Aug 10 2012

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Lucyna Trojnar-Spelina and Iwona Włoch, On Generalized Pell and Pell-Lucas Numbers, Iranian Journal of Science and Technology, Transactions A: Science (2019), 1-7.
Index entries for linear recurrences with constant coefficients, signature (5,4).

Cf. A001076, A006190, A007482, A015520, A015521, A015523, A015524, A015525, A015528, A015529, A015530, A015531, A015532, A015533, A015534, A015535, A015536, A015443, A015447, A030195, A053404, A057087, A083858, A085939, A090017, A091914, A099012, A122690, A123270, A180222, A180226.

a:=[0,1];; for n in [3..30] do a[n]:=5*a[n-1]+4*a[n-2]; od; a; # G. C. Greubel, Dec 26 2019

[n le 2 select n-1 else 5*Self(n-1)+4*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 12 2012

seq( simplify((2/I)^(n-1)*ChebyshevU(n-1, 5*I/4)), n=0..20); # G. C. Greubel, Dec 26 2019

LinearRecurrence[{5,4}, {0,1}, 30] (* Vincenzo Librandi, Nov 12 2012 *)
Table[2^(n-1)*Fibonacci[n, 5/2], {n, 0, 30}] (* G. C. Greubel, Dec 26 2019 *)

x='x+O('x^30); concat([0], Vec(x/(1-5*x-4*x^2))) \\ G. C. Greubel, Jan 01 2018

[lucas_number1(n,5,-4) for n in range(0, 22)] # Zerinvary Lajos, Apr 24 2009

a(n) = 5*a(n-1) + 4*a(n-2).
a(n) = Sum_{k=0..floor((n-1)/2)} C(n-k-1, k)*4^k*5^(n-2*k-1). - Paul Barry, Apr 23 2005
a(n) = Sum_{k=0..(n-1)} A122690(k). - Alexander Adamchuk, Nov 03 2006
a(n) = 2^(n-1)*Fibonacci(n, 5/2) = (2/i)^(n-1)*ChebyshevU(n-1, 5*i/4). - G. C. Greubel, Dec 26 2019

A180140 Eight rooks and one berserker on a 3 X 3 chessboard. G.f.: (1+x+x^2)/(1-3x-5x^2).

Original entry on oeis.org

1, 4, 18, 74, 312, 1306, 5478, 22964, 96282, 403666, 1692408, 7095554, 29748702, 124723876, 522915138, 2192364794, 9191670072, 38536834186, 161568852918, 677390729684, 2840016453642, 11907003009346, 49921091296248

Offset: 0

Johannes W. Meijer, Aug 13 2010, Jun 15 2013

a(n) gives the number of n-move routes of a fairy chess piece starting in a given side square (m = 2, 4, 6,  8) on a 3 X 3 chessboard. This fairy chess piece behaves like a rook on the four side and four corner (m = 1, 3, 7, 9) squares but on the center square (m = 5) it goes berserk and turns into a berserker. For this sequence, the berserker can move to three of the side squares and three of the corners from the center.
The berserker is one of the Lewis chessmen which were discovered in 1831 on the Isle of Lewis. They are carved from walrus ivory in Scandinavian style of the 12th century. The pawns look like decorated tombstones. The pieces have all human representations with facial expressions varying from gloom to anger. Some of the rooks show men biting their shield in the manner of berserkers. According to Hooper and Whyld none looks happy.
Let A be the adjacency matrix of the graph G, where V(G) = {v1, v2, v3, v4, v5, v6, v7, v8, v9}. Then the (m, k) entry of A^n is the number of different vm-vk walks of length n in G, see the Chartrand reference. In the adjacency matrix A, see the Maple program, the A[1], A[3], A[7] and A[9] vectors represent the rook moves on the corner squares, the A[2], A[4], A[6] and A[8] vectors represent the rook moves on the side squares and the A[5] vector represents the moves of the berserker. On a 3 X 3 chessboard there are 2^9 = 512 ways a berserker could move from the center square (off the center the berserker behaves like a rook) so there are 512 different berserkers.
For the side squares the 512 berserker vectors lead to 42 different sequences, see the overview of berserker sequences. There are 16 berserker vectors that lead to the sequence given above. Their decimal [binary] values are: 111 [001 101 111] , 207 [011 001 111], 231 [011 100 111], 237 [011 101 101], 303 [100 101 111], 363 [101 101 011], 366 [101 101 110], 399 [110 001 111], 423 [110 100 111], 429 [110 101 101], 459 [111 001 011], 462 [111 001 110], 483 [111 100 011], 486 [111 100 110], 489 [111 101 001] and 492 [111 101 100]. These berserker vectors lead for the corner squares to sequence 4*A179606 (with leading term 1 added) and for the central square to sequence 6*A179606 (with leading term 1 added).
This sequence belongs to a family of sequences with GF(x)=(1+x-k*x^2)/(1-3*x+(k-4)*x^2), see A180142.

Gary Chartrand, Introductory Graph Theory, pp. 217-221, 1984.
David Hooper and Kenneth Whyld, The Oxford Companion to Chess, pp. 131, 225, 1992.

Indranil Ghosh, Table of n, a(n) for n = 0..1603
Dougie MacLean, The Lewis Chessmen: "Marching Mystery"
Johannes W. Meijer, The berserker sequences.
Wikipedia, Lewis Chessmen
Index entries for linear recurrences with constant coefficients, signature (3,5)

Cf. A180141 (corner squares) and A180147 (central square).

Cf. Berserker sequences side squares: 4*A007482 (with leading 1 added), A180144, A003500 (n>=1 and a(0)=1), A180142, A000302, A180140 (this sequence), 2*A001077 (n>=1 and a(0)=1), A180146, 4*A154964 (n>=1 and a(0)=1), 4*A123347 (with leading 1 added).

nmax:=22; m:=2; A[1]:=[0, 1, 1, 1, 0, 0, 1, 0, 0]: A[2]:=[1, 0, 1, 0, 1, 0, 0, 1, 0]: A[3]:= [1, 1, 0, 0, 0, 1, 0, 0, 1]: A[4]:= [1, 0, 0, 0, 1, 1, 1, 0, 0]: A[5]:=[0, 0, 1, 1, 0, 1, 1, 1, 1]: A[6]:=[0, 0, 1, 1, 1, 0, 0, 0, 1]: A[7]:=[1, 0, 0, 1, 0, 0, 0, 1, 1]: A[8]:=[0, 1, 0, 0, 1, 0, 1, 0, 1]: A[9]:=[0, 0, 1, 0, 0, 1, 1, 1, 0]: A:= Matrix([A[1], A[2], A[3], A[4], A[5], A[6], A[7], A[8], A[9]]): 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);

CoefficientList[Series[(1+x+x^2)/(1-3*x-5*x^2), {x, 0, 22}],x] (* or *) LinearRecurrence[{3,5,0},{1,4,18},23] (* Indranil Ghosh, Mar 05 2017 *)

print(Vec((1 + x + x^2)/(1- 3*x - 5*x^2) + O(x^23))); \\ Indranil Ghosh, Mar 05 2017

G.f.: (1+x+x^2)/(1-3*x-5*x^2).
a(n) = 3*a(n-1) + 5*a(n-2) for n>=3  with a(0)=1, a(1)=4 and a(2)=18.
a(n) = ((22+54*A)*A^(-n-1) + (22+54*B)*B^(-n-1))/145  with A=(-3+sqrt(29))/10 and B=(-3-sqrt(29))/10 for n>=1 with a(0)=1.
5*a(n) = 2*( A015523(n) + 3*A015523(n+1)), n>0 - R. J. Mathar, May 11 2013

A179606 Eight white kings and one red king on a 3 X 3 chessboard. G.f.: (1 + x)/(1 - 3x - 5x^2).

Original entry on oeis.org

1, 4, 17, 71, 298, 1249, 5237, 21956, 92053, 385939, 1618082, 6783941, 28442233, 119246404, 499950377, 2096083151, 8788001338, 36844419769, 154473265997, 647641896836, 2715292020493, 11384085545659, 47728716739442

Offset: 0

Johannes W. Meijer, Jul 28 2010

a(n) represents the number of n-move routes of a fairy chess piece starting in the central square (m = 5) on a 3 X 3 chessboard. This fairy chess piece behaves like a king on the eight side and corner squares but on the central square the king goes crazy and turns into a red king, see A179596.
The sequence above corresponds to 24 red king vectors, i.e., A[5] vectors, with decimal values 27, 30, 51, 54, 57, 60, 90, 114, 120, 147, 150, 153, 156, 177, 180, 210, 216, 240, 282, 306, 312, 402, 408 and 432. These vectors lead for the corner squares to A015523 and for the side squares to A152187.
This sequence belongs to a family of sequences with g.f. (1 + (k-4)*x)/(1 - 3*x - k*x^2). Red king sequences that are members of this family are A007483 (k= 2), A015521 (k=4), A179606 (k=5; this sequence), A154964 (k=6), A179603 (k=7) and A179599 (k=8). We observe that there is no red king sequence for k=3. Other members of this family are A006190 (k=1), A133494 (k=0) and A168616 (k=-2).
Inverse binomial transform of A052918.
The sequence b(n+1) = 6*a(n), n >= 0 with b(0)=1, is a berserker sequence, see A180147. The b(n) sequence corresponds to 16 A[5] vectors with decimal values between 111 and 492. These vectors lead for the corner squares to sequence c(n+1)=4*A179606(n), n >= 0 with c(0)=1, and for the side squares to A180140. - Johannes W. Meijer, Aug 14 2010
Equals the INVERT transform of A063782: (1, 3, 10, 32, 104, ...). Example: a(3) = 71 = (1, 1, 4, 7) dot (32, 10, 3, 1) = (32 + 10 + 12 + 17). - Gary W. Adamson, Aug 14 2010

Indranil Ghosh, Table of n, a(n) for n = 0..1603
Index entries for linear recurrences with constant coefficients, signature (3,5).

Cf. A179597 (central square).

Cf. A072263, A072264, A152187, A197189.

with(LinearAlgebra): nmax:=22; m:=5; A[1]:= [0,1,0,1,1,0,0,0,0]: A[2]:= [1,0,1,1,1,1,0,0,0]: A[3]:= [0,1,0,0,1,1,0,0,0]: A[4]:= [1,1,0,0,1,0,1,1,0]: A[5]:= [0,0,0,1,1,1,0,0,1]: A[6]:= [0,1,1,0,1,0,0,1,1]: A[7]:= [0,0,0,1,1,0,0,1,0]: A[8]:= [0,0,0,1,1,1,1,0,1]: A[9]:= [0,0,0,0,1,1,0,1,0]: A:=Matrix([A[1],A[2],A[3],A[4],A[5],A[6],A[7],A[8],A[9]]): 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);

CoefficientList[Series[(1+x)/(1-3*x-5*x^2), {x, 0, 22}],x] (* or *) LinearRecurrence[{3,5,0},{1,4},23] (* Indranil Ghosh, Mar 05 2017 *)

print(Vec((1 + x)/(1- 3*x - 5*x^2) + O(x^23))); \\ Indranil Ghosh, Mar 05 2017

G.f.: (1+x)/(1 - 3*x - 5*x^2).
a(n) = A015523(n) + A015523(n+1).
a(n) = 3*a(n-1) + 5*a(n-2) with a(0) = 1 and a(1) = 4.
a(n) = ((29 + 7*sqrt(29))*A^(-n-1) + (29-7*sqrt(29))*B^(-n-1))/290 with A = (-3+sqrt(29))/10 and B = (-3-sqrt(29))/10
Limit_{k->oo} a(n+k)/a(k) = (-1)^(n+1)*A000351(n)*A130196(n)/(A015523(n)*sqrt(29) - A072263(n)) for n >= 1.

cp's OEIS Frontend

A072271 A partial product representation of f(n) = A015523(n) and L(n) = A072263(n).

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A052918 a(0) = 1, a(1) = 5, a(n+1) = 5*a(n) + a(n-1).

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A057088 Scaled Chebyshev U-polynomials evaluated at i*sqrt(5)/2. Generalized Fibonacci sequence.

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A179596 Eight white kings and one red king on a 3 X 3 chessboard. G.f.: (1 + x)/(1 - 2*x - 11*x^2 - 6*x^3).

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A015442 a(n) = a(n-1) + 7*a(n-2), a(0)=0, a(1)=1.

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A083858 Expansion of x/(1 - 3*x - 6*x^2).

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A057089 Scaled Chebyshev U-polynomials evaluated at i*sqrt(6)/2. Generalized Fibonacci sequence.

A179596 Eight white kings and one red king on a 3 X 3 chessboard. G.f.: (1 + x)/(1 - 2x - 11x^2 - 6*x^3).

A083858 Expansion of x/(1 - 3x - 6x^2).

A015537 Expansion of x/(1 - 5x - 4x^2).

A180140 Eight rooks and one berserker on a 3 X 3 chessboard. G.f.: (1+x+x^2)/(1-3x-5x^2).

A179606 Eight white kings and one red king on a 3 X 3 chessboard. G.f.: (1 + x)/(1 - 3x - 5x^2).