A055099 -id:A055099 - cp's OEIS frontend

A256960 a(0)=1, a(1)=4; thereafter a(n) = a(n-2)+2*A055099(n-1)+2^(n-1).

1, 4, 11, 36, 119, 408, 1419, 4988, 17631, 62528, 222163, 790180, 2812135, 10011304, 35647259, 126942540, 452078447, 1610033040, 5734081251, 20421960308, 72733344375, 259042555640, 922591559467, 3285854197276, 11702734525951, 41679889602784, 148445093121011, 528694969090116

Offset: 0

N. J. A. Sloane, Apr 14 2015

Colin Barker, Table of n, a(n) for n = 0..1000
J. Goldwasser et al., The density of ones in Pascal's rhombus, Discrete Math., 204 (1999), 231-236.
Paul K. Stockmeyer, The Pascal Rhombus and the Stealth Configuration, arXiv:1504.04404 [math.CO], 2015.
Index entries for linear recurrences with constant coefficients, signature (4,1,-8,-4).

Cf. A055099.

Vec(-(2*x+1)*(2*x^2+2*x-1)/((x+1)*(2*x-1)*(2*x^2+3*x-1)) + O(x^100)) \\ Colin Barker, Jun 06 2015

G.f.: -(2*x+1)*(2*x^2+2*x-1) / ((x+1)*(2*x-1)*(2*x^2+3*x-1)). - Colin Barker, Jun 05 2015

A001333 Pell-Lucas numbers: numerators of continued fraction convergents to sqrt(2).

Original entry on oeis.org

1, 1, 3, 7, 17, 41, 99, 239, 577, 1393, 3363, 8119, 19601, 47321, 114243, 275807, 665857, 1607521, 3880899, 9369319, 22619537, 54608393, 131836323, 318281039, 768398401, 1855077841, 4478554083, 10812186007, 26102926097, 63018038201, 152139002499, 367296043199

Offset: 0

N. J. A. Sloane and R. K. Guy

Number of n-step non-selfintersecting paths starting at (0,0) with steps of types (1,0), (-1,0) or (0,1) [Stanley].
Number of n steps one-sided prudent walks with east, west and north steps. - Shanzhen Gao, Apr 26 2011
Number of ternary strings of length n-1 with subwords (0,2) and (2,0) not allowed. - Olivier Gérard, Aug 28 2012
Number of symmetric 2n X 2 or (2n-1) X 2 crossword puzzle grids: all white squares are edge connected; at least 1 white square on every edge of grid; 180-degree rotational symmetry. - Erich Friedman
a(n+1) is the number of ways to put molecules on a 2 X n ladder lattice so that the molecules do not touch each other.
In other words, a(n+1) is the number of independent vertex sets and vertex covers in the n-ladder graph P_2 X P_n. - Eric W. Weisstein, Apr 04 2017
Number of (n-1) X 2 binary arrays with a path of adjacent 1's from top row to bottom row, see A359576. - R. H. Hardin, Mar 16 2002
a(2*n+1) with b(2*n+1) := A000129(2*n+1), n >= 0, give all (positive integer) solutions to Pell equation a^2 - 2*b^2 = -1.
a(2*n) with b(2*n) := A000129(2*n), n >= 1, give all (positive integer) solutions to Pell equation a^2 - 2*b^2 = +1 (see Emerson reference).
Bisection: a(2*n) = T(n,3) = A001541(n), n >= 0 and a(2*n+1) = S(2*n,2*sqrt(2)) = A002315(n), n >= 0, with T(n,x), resp. S(n,x), Chebyshev's polynomials of the first, resp. second kind. See A053120, resp. A049310.
Binomial transform of A077957. - Paul Barry, Feb 25 2003
For n > 0, the number of (s(0), s(1), ..., s(n)) such that 0 < s(i) < 4 and |s(i) - s(i-1)| <= 1 for i = 1,2,...,n, s(0) = 2, s(n) = 2. - Herbert Kociemba, Jun 02 2004
For n > 1, a(n) corresponds to the longer side of a near right-angled isosceles triangle, one of the equal sides being A000129(n). - Lekraj Beedassy, Aug 06 2004
Exponents of terms in the series F(x,1), where F is determined by the equation F(x,y) = xy + F(x^2*y,x). - Jonathan Sondow, Dec 18 2004
Number of n-words from the alphabet A={0,1,2} which two neighbors differ by at most 1. - Fung Cheok Yin (cheokyin_restart(AT)yahoo.com.hk), Aug 30 2006
Consider the mapping f(a/b) = (a + 2b)/(a + b). Taking a = b = 1 to start with and carrying out this mapping repeatedly on each new (reduced) rational number gives the following sequence 1/1, 3/2, 7/5, 17/12, 41/29, ... converging to 2^(1/2). Sequence contains the numerators. - Amarnath Murthy, Mar 22 2003 [Amended by Paul E. Black (paul.black(AT)nist.gov), Dec 18 2006]
Odd-indexed prime numerators are prime RMS numbers (A140480) and also NSW primes (A088165). - Ctibor O. Zizka, Aug 13 2008
The intermediate convergents to 2^(1/2) begin with 4/3, 10/7, 24/17, 58/41; essentially, numerators=A052542 and denominators here. - Clark Kimberling, Aug 26 2008
Equals right border of triangle A143966. Starting (1, 3, 7, ...) equals INVERT transform of (1, 2, 2, 2, ...) and row sums of triangle A143966. - Gary W. Adamson, Sep 06 2008
Inverse binomial transform of A006012; Hankel transform is := [1, 2, 0, 0, 0, 0, 0, 0, 0, ...]. - Philippe Deléham, Dec 04 2008
From Charlie Marion, Jan 07 2009: (Start)
In general, denominators, a(k,n) and numerators, b(k,n), of continued fraction convergents to sqrt((k+1)/k) may be found as follows:
let a(k,0) = 1, a(k,1) = 2k; for n>0, a(k,2n) = 2*a(k,2n-1) + a(k,2n-2) and a(k,2n+1) = (2k)*a(k,2n) + a(k,2n-1);
let b(k,0) = 1, b(k,1) = 2k+1; for n>0, b(k,2n) = 2*b(k,2n-1) + b(k,2n-2) and b(k,2n+1) = (2k)*b(k,2n) + b(k,2n-1).
For example, the convergents to sqrt(2/1) start 1/1, 3/2, 7/5, 17/12, 41/29.
In general, if a(k,n) and b(k,n) are the denominators and numerators, respectively, of continued fraction convergents to sqrt((k+1)/k) as defined above, then
k*a(k,2n)^2 - a(k,2n-1)*a(k,2n+1) = k = k*a(k,2n-2)*a(k,2n) - a(k,2n-1)^2 and
b(k,2n-1)*b(k,2n+1) - k*b(k,2n)^2 = k+1 = b(k,2n-1)^2 - k*b(k,2n-2)*b(k,2n);
for example, if k=1 and n=3, then b(1,n)=a(n+1) and
1*a(1,6)^2 - a(1,5)*a(1,7) = 1*169^2 - 70*408 = 1;
1*a(1,4)*a(1,6) - a(1,5)^2 = 1*29*169 - 70^2 = 1;
b(1,5)*b(1,7) - 1*b(1,6)^2 = 99*577 - 1*239^2 = 2;
b(1,5)^2 - 1*b(1,4)*b(1,6) = 99^2 - 1*41*239 = 2.
Cf. A000129, A142238-A142239, A153313-A153318.
(End)
This sequence occurs in the lower bound of the order of the set of equivalent resistances of n equal resistors combined in series and in parallel (A048211). - Sameen Ahmed Khan, Jun 28 2010
Let M = a triangle with the Fibonacci series in each column, but the leftmost column is shifted upwards one row. A001333 = lim_{n->infinity} M^n, the left-shifted vector considered as a sequence. - Gary W. Adamson, Jul 27 2010
a(n) is the number of compositions of n when there are 1 type of 1 and 2 types of other natural numbers. - Milan Janjic, Aug 13 2010
Equals the INVERTi transform of A055099. - Gary W. Adamson, Aug 14 2010
From L. Edson Jeffery, Apr 04 2011: (Start)
Let U be the unit-primitive matrix (see [Jeffery])
U = U_(8,2) = (0 0 1 0)
              (0 1 0 1)
              (1 0 2 0)
              (0 2 0 1).
Then a(n) = (1/4)*Trace(U^n). (See also A084130, A006012.)
(End)
For n >= 1, row sums of triangle
  m/k.|..0.....1.....2.....3.....4.....5.....6.....7
  ==================================================
  .0..|..1
  .1..|..1.....2
  .2..|..1.....2.....4
  .3..|..1.....4.....4.....8
  .4..|..1.....4....12.....8....16
  .5..|..1.....6....12....32....16....32
  .6..|..1.....6....24....32....80....32....64
  .7..|..1.....8....24....80....80...192....64...128
which is the triangle for numbers 2^k*C(m,k) with duplicated diagonals. - Vladimir Shevelev, Apr 12 2012
a(n) is also the number of ways to place k non-attacking wazirs on a 2 X n board, summed over all k >= 0 (a wazir is a leaper [0,1]). - Vaclav Kotesovec, May 08 2012
The sequences a(n) and b(n) := A000129(n) are entries of powers of the special case of the Brahmagupta Matrix - for details see Suryanarayan's paper. Further, as Suryanarayan remark, if we set A = 2*(a(n) + b(n))*b(n), B = a(n)*(a(n) + 2*b(n)), C = a(n)^2 + 2*a(n)*b(n) + 2*b(n)^2 we obtain integral solutions of the Pythagorean relation A^2 + B^2 = C^2, where A and B are consecutive integers. - Roman Witula, Jul 28 2012
Pisano period lengths: 1, 1, 8, 4, 12, 8, 6, 4, 24, 12, 24, 8, 28, 6, 24, 8, 16, 24, 40, 12, .... - R. J. Mathar, Aug 10 2012
This sequence and A000129 give the diagonal numbers described by Theon of Smyrna. - Sture Sjöstedt, Oct 20 2012
a(n) is the top left entry of the n-th power of any of the following six 3 X 3 binary matrices: [1, 1, 1; 1, 1, 1; 1, 0, 0] or [1, 1, 1; 1, 1, 0; 1, 1, 0] or [1, 1, 1; 1, 0, 1; 1, 1, 0] or [1, 1, 1; 1, 1, 0; 1, 0, 1] or [1, 1, 1; 1, 0, 1; 1, 0, 1] or [1, 1, 1; 1, 0, 0; 1, 1, 1]. - R. J. Mathar, Feb 03 2014
If p is prime, a(p) == 1 (mod p) (compare with similar comment for A000032). - Creighton Dement, Oct 11 2005, modified by Davide Colazingari, Jun 26 2016
a(n) = A000129(n) + A000129(n-1), where A000129(n) is the n-th Pell Number; e.g., a(6) = 99 = A000129(6) + A000129(5) = 70 + 29. Hence the sequence of fractions has the form 1 + A000129(n-1)/A000129(n), and the ratio A000129(n-1)/A000129(n)converges to sqrt(2) - 1. - Gregory L. Simay, Nov 30 2018
For n > 0, a(n+1) is the length of tau^n(1) where tau is the morphism: 1 -> 101, 0 -> 1. See Song and Wu. - Michel Marcus, Jul 21 2020
For n > 0, a(n) is the number of nonisomorphic quasitrivial semigroups with n elements, see Devillet, Marichal, Teheux. A292932 is the number of labeled quasitrivial semigroups. - Peter Jipsen, Mar 28 2021
a(n) is the permanent of the n X n tridiagonal matrix defined in A332602. - Stefano Spezia, Apr 12 2022
From Greg Dresden, May 08 2023: (Start)
For n >= 2, 4*a(n) is the number of ways to tile this T-shaped figure of length n-1 with two colors of squares and one color of domino; shown here is the figure of length 5 (corresponding to n=6), and it has 4*a(6) = 396 different tilings.
   _
  || _  
  |||_|||
  |_|
(End)
12*a(n) = number of walks of length n in the cyclic Kautz digraph CK(3,4). - Miquel A. Fiol, Feb 15 2024

			Convergents are 1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378, 8119/5741, 19601/13860, 47321/33461, 114243/80782, ... = A001333/A000129.
The 15 3 X 2 crossword grids, with white squares represented by an o:
  ooo ooo ooo ooo ooo ooo ooo oo. o.o .oo o.. .o. ..o oo. .oo
  ooo oo. o.o .oo o.. .o. ..o ooo ooo ooo ooo ooo ooo .oo oo.
G.f. = 1 + x + 3*x^2 + 7*x^3 + 17*x^4 + 41*x^5 + 99*x^6 + 239*x^7 + 577*x^8 + ...

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Index entries for "core" sequences.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (2,1).

For denominators see A000129.
See A040000 for the continued fraction expansion of sqrt(2).
See also A078057 which is the same sequence without the initial 1.
Cf. also A002203, A152113.
Row sums of unsigned Chebyshev T-triangle A053120. a(n)= A054458(n, 0) (first column of convolution triangle).
Row sums of A140750, A160756, A135837.
Equals A034182(n-1) + 2 and A084128(n)/2^n. First differences of A052937. Partial sums of A052542. Pairwise sums of A048624. Bisection of A002965.
The following sequences (and others) belong to the same family: A001333, A000129, A026150, A002605, A046717, A015518, A084057, A063727, A002533, A002532, A083098, A083099, A083100, A015519.
Second row of the array in A135597.
Cf. A048211, A153588, A174283, A174284, A174285, A174286, A176499, A176500, A176501, A176502.
Cf. A055099.
Cf. A028859, A001906 / A088305, A033303, A000225, A095263, A003945, A006356, A002478, A214260, A001911 and A000217 for other restricted ternary words.
Cf. Triangle A106513 (alternating row sums).
Equals A293004 + 1.
Cf. A033539, A332602, A086395 (subseq. of primes).

a001333 n = a001333_list !! n
a001333_list = 1 : 1 : zipWith (+)
                       a001333_list (map (* 2) $ tail a001333_list)
-- Reinhard Zumkeller, Jul 08 2012

[n le 2 select 1 else 2*Self(n-1)+Self(n-2): n in [1..35]]; // Vincenzo Librandi, Nov 10 2018

A001333 := proc(n) option remember; if n=0 then 1 elif n=1 then 1 else 2*procname(n-1)+procname(n-2) fi end;
Digits := 50; A001333 := n-> round((1/2)*(1+sqrt(2))^n);
with(numtheory): cf := cfrac (sqrt(2),1000): [seq(nthnumer(cf,i), i=0..50)];
a:= n-> (M-> M[2, 1]+M[2, 2])(<<2|1>, <1|0>>^n):
seq(a(n), n=0..33);  # Alois P. Heinz, Aug 01 2008
A001333List := proc(m) local A, P, n; A := [1,1]; P := [1,1];
for n from 1 to m - 2 do P := ListTools:-PartialSums([op(A), P[-2]]);
A := [op(A), P[-1]] od; A end: A001333List(32); # Peter Luschny, Mar 26 2022

Insert[Table[Numerator[FromContinuedFraction[ContinuedFraction[Sqrt[2], n]]], {n, 1, 40}], 1, 1] (* Stefan Steinerberger, Apr 08 2006 *)
Table[((1 - Sqrt[2])^n + (1 + Sqrt[2])^n)/2, {n, 0, 29}] // Simplify (* Robert G. Wilson v, May 02 2006 *)
a[0] = 1; a[1] = 1; a[n_] := a[n] = 2a[n - 1] + a[n - 2]; Table[a@n, {n, 0, 29}] (* Robert G. Wilson v, May 02 2006 *)
Table[ MatrixPower[{{1, 2}, {1, 1}}, n][[1, 1]], {n, 0, 30}] (* Robert G. Wilson v, May 02 2006 *)
a=c=0;t={b=1}; Do[c=a+b+c; AppendTo[t,c]; a=b;b=c,{n,40}]; t (* Vladimir Joseph Stephan Orlovsky, Mar 23 2009 *)
LinearRecurrence[{2, 1}, {1, 1}, 40] (* Vladimir Joseph Stephan Orlovsky, Mar 23 2009 *)
Join[{1}, Numerator[Convergents[Sqrt[2], 30]]] (* Harvey P. Dale, Aug 22 2011 *)
Table[(-I)^n ChebyshevT[n, I], {n, 10}] (* Eric W. Weisstein, Apr 04 2017 *)
CoefficientList[Series[(-1 + x)/(-1 + 2 x + x^2), {x, 0, 20}], x] (* Eric W. Weisstein, Sep 21 2017 *)
Table[Sqrt[(ChebyshevT[n, 3] + (-1)^n)/2], {n, 0, 20}] (* Eric W. Weisstein, Apr 17 2018 *)

{a(n) = if( n<0, (-1)^n, 1) * contfracpnqn( vector( abs(n), i, 1 + (i>1))) [1, 1]}; /* Michael Somos, Sep 02 2012 */

{a(n) = polchebyshev(n, 1, I) / I^n}; /* Michael Somos, Sep 02 2012 */

a(n) = real((1 + quadgen(8))^n); \\ Michel Marcus, Mar 16 2021

{ for (n=0, 4000, a=contfracpnqn(vector(n, i, 1+(i>1)))[1, 1]; if (a > 10^(10^3 - 6), break); write("b001333.txt", n, " ", a); ); } \\ Harry J. Smith, Jun 12 2009

from functools import cache
@cache
def a(n): return 1 if n < 2 else 2*a(n-1) + a(n-2)
print([a(n) for n in range(32)]) # Michael S. Branicky, Nov 13 2022

from sage.combinat.sloane_functions import recur_gen2
it = recur_gen2(1,1,2,1)
[next(it) for i in range(30)] ## Zerinvary Lajos, Jun 24 2008

[lucas_number2(n,2,-1)/2 for n in range(0, 30)] # Zerinvary Lajos, Apr 30 2009

a(n) = A055642(A125058(n)). - Reinhard Zumkeller, Feb 02 2007
a(n) = 2a(n-1) + a(n-2);
a(n) = ((1-sqrt(2))^n + (1+sqrt(2))^n)/2.
a(n)+a(n+1) = 2 A000129(n+1). 2*a(n) = A002203(n).
G.f.: (1 - x) / (1 - 2*x - x^2) = 1 / (1 - x / (1 - 2*x / (1 + x))). - Simon Plouffe in his 1992 dissertation.
A000129(2n) = 2*A000129(n)*a(n). - John McNamara, Oct 30 2002
a(n) = (-i)^n * T(n, i), with T(n, x) Chebyshev's polynomials of the first kind A053120 and i^2 = -1.
a(n) = a(n-1) + A052542(n-1), n>1. a(n)/A052542(n) converges to sqrt(1/2). - Mario Catalani (mario.catalani(AT)unito.it), Apr 29 2003
E.g.f.: exp(x)cosh(x*sqrt(2)). - Paul Barry, May 08 2003
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2k)2^k. - Paul Barry, May 13 2003
For n > 0, a(n)^2 - (1 + (-1)^(n))/2 = Sum_{k=0..n-1} ((2k+1)*A001653(n-1-k)); e.g., 17^2 - 1 = 288 = 1*169 + 3*29 + 5*5 + 7*1; 7^2 = 49 = 1*29 + 3*5 + 5*1. - Charlie Marion, Jul 18 2003
a(n+2) = A078343(n+1) + A048654(n). - Creighton Dement, Jan 19 2005
a(n) = A000129(n) + A000129(n-1) = A001109(n)/A000129(n) = sqrt(A001110(n)/A000129(n)^2) = ceiling(sqrt(A001108(n))). - Henry Bottomley, Apr 18 2000
Also the first differences of A000129 (the Pell numbers) because A052937(n) = A000129(n+1) + 1. - Graeme McRae, Aug 03 2006
a(n) = Sum_{k=0..n} A122542(n,k). - Philippe Deléham, Oct 08 2006
For another recurrence see A000129.
a(n) = Sum_{k=0..n} A098158(n,k)*2^(n-k). - Philippe Deléham, Dec 26 2007
a(n) = upper left and lower right terms of [1,1; 2,1]^n. - Gary W. Adamson, Mar 12 2008
If p[1]=1, and p[i]=2, (i>1), and if A is Hessenberg matrix of order n defined by: A[i,j]=p[j-i+1], (i<=j), A[i,j]=-1, (i=j+1), and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det A. - Milan Janjic, Apr 29 2010
For n>=2, a(n)=F_n(2)+F_(n+1)(2), where F_n(x) is Fibonacci polynomial (cf. A049310): F_n(x) = Sum_{i=0..floor((n-1)/2)} binomial(n-i-1,i)x^(n-2*i-1). - Vladimir Shevelev, Apr 13 2012
a(-n) = (-1)^n * a(n). - Michael Somos, Sep 02 2012
Dirichlet g.f.: (PolyLog(s,1-sqrt(2)) + PolyLog(s,1+sqrt(2)))/2. - Ilya Gutkovskiy, Jun 26 2016
a(n) = A000129(n) - A000129(n-1), where A000129(n) is the n-th Pell Number. Hence the continued fraction is of the form 1-(A000129(n-1)/A000129(n)). - Gregory L. Simay, Nov 09 2018
a(n) = (A000129(n+3) + A000129(n-3))/10, n>=3. - Paul Curtz, Jun 16 2021
a(n) = (A000129(n+6) - A000129(n-6))/140, n>=6. - Paul Curtz, Jun 20 2021
a(n) = round((1/2)*sqrt(Product_{k=1..n} 4*(1 + sin(k*Pi/n)^2))), for n>=1. - Greg Dresden, Dec 28 2021
a(n)^2 + a(n+1)^2 = A075870(n+1) = 2*(b(n)^2 + b(n+1)^2) for all n in Z where b(n) := A000129(n). - Michael Somos, Apr 02 2022
a(n) = 2*A048739(n-2)+1. - R. J. Mathar, Feb 01 2024
Sum_{n>=1} 1/a(n) = 1.5766479516393275911191017828913332473... - R. J. Mathar, Feb 05 2024
From Peter Bala, Jul 06 2025: (Start)
G.f.: Sum_{n >= 1} (-1)^(n+1) * x^(n-1) * Product_{k = 1..n} (1 - k*x)/(1 - 3*x + k*x^2).
The following series telescope:
Sum_{n >= 1} (-1)^(n+1)/(a(2*n) + 1/a(2*n)) = 1/4, since 1/(a(2*n) + 1/a(2*n)) = 1/A077445(n) + 1/A077445(n+1).
Sum_{n >= 1} (-1)^(n+1)/(a(2*n+1) - 1/a(2*n+1)) = 1/8, since. 1/(a(2*n+1) - 1/a(2*n+1)) = 1/(4*Pell(2*n)) + 1/(4*Pell(2*n+2)), where Pell(n) = A000129(n).
Sum_{n >= 1} (-1)^(n+1)/(a(2*n+1) + 9/a(2*n+1)) = 1/10, since 1/(a(2*n+1) + 9/a(2*n+1)) = b(n) + b(n+1), where b(n) = A001109(n)/(2*Pell(2*n-1)*Pell(2*n+1)).
Sum_{n >= 1} (-1)^(n+1)/(a(n)*a(n+1)) = 1 - sqrt(2)/2 = A268682, since (-1)^(n+1)/(a(n)*a(n+1)) = Pell(n)/a(n) - Pell(n+1)/a(n+1). (End)

Chebyshev comments from Wolfdieter Lang, Jan 10 2003

A126473 Number of strings over a 5 symbol alphabet with adjacent symbols differing by three or less.

Original entry on oeis.org

1, 5, 23, 107, 497, 2309, 10727, 49835, 231521, 1075589, 4996919, 23214443, 107848529, 501037445, 2327695367, 10813893803, 50238661313, 233396326661, 1084301290583, 5037394142315, 23402480441009, 108722104190981, 505095858086951, 2346549744920747

Offset: 0

R. H. Hardin, Dec 27 2006

[Empirical] a(base,n) = a(base-1,n) + 7^(n-1) for base >= 3n-2; a(base,n) = a(base-1,n) + 7^(n-1)-2 when base = 3n-3.
From Johannes W. Meijer, Aug 01 2010: (Start)
The a(n) represent the number of n-move routes of a fairy chess piece starting in a given side square (m = 2, 4, 6 or 8) 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.
For the side squares the 512 red kings lead to 47 different red king sequences, see the cross-references for some examples.
The sequence above corresponds to four A[5] vectors with the decimal [binary] values 367 [1,0,1,1,0,1,1,1,1], 463 [1,1,1,0,0,1,1,1,1], 487 [1,1,1,1,0,0,1,1,1] and 493 [1,1,1,1,0,1,1,0,1]. These vectors lead for the corner squares to A179596 and for the central square to A179597.
This sequence belongs to a family of sequences with g.f. (1+x)/(1-4*x-k*x^2). Red king sequences that are members of this family are A003947 (k=0), A015448 (k=1), A123347 (k=2), A126473 (k=3; this sequence) and A086347 (k=4). Other members of this family are A000351 (k=5), A001834 (k=-1), A111567 (k=-2), A048473 (k=-3) and A053220 (k=-4)
Inverse binomial transform of A154244. (End)
Equals the INVERT transform of A055099: (1, 4, 14, 50, 178, ...). - Gary W. Adamson, Aug 14 2010
Number of one-sided n-step walks taking steps from {E, W, N, NE, NW}. - Shanzhen Gao, May 10 2011
For n>=1, a(n) equals the numbers of words of length n-1 on alphabet {0,1,2,3,4} containing no subwords 00 and 11. - Milan Janjic, Jan 31 2015

Shanzhen Gao and Keh-Hsun Chen, Tackling Sequences From Prudent Self-Avoiding Walks, FCS'14, The 2014 International Conference on Foundations of Computer Science.
S. Gao and H. Niederhausen, Sequences Arising From Prudent Self-Avoiding Walks, 2010.
Index entries for linear recurrences with constant coefficients, signature (4,3).

Cf. 5 symbol differing by two or less A126392, one or less A057960.
Cf. Red king sequences side squares [numerical value A[5]]: A086347 [495], A179598 [239], A126473 [367], A123347 [335], A179602 [95], A154964 [31], A015448 [327], A152187 [27], A003947 [325], A108981 [11], A007483 [2]. - Johannes W. Meijer, Aug 01 2010
Cf. A055099.

with(LinearAlgebra): nmax:=19; m:=2; A[5]:= [1,0,1,1,0,1,1,1,1]: A:=Matrix([[0,1,0,1,1,0,0,0,0],[1,0,1,1,1,1,0,0,0],[0,1,0,0,1,1,0,0,0],[1,1,0,0,1,0,1,1,0],A[5],[0,1,1,0,1,0,0,1,1],[0,0,0,1,1,0,0,1,0],[0,0,0,1,1,1,1,0,1],[0,0,0,0,1,1,0,1,0]]): 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); # Johannes W. Meijer, Aug 01 2010
# second Maple program:
a:= n-> (M-> M[1,2]+M[2,2])(<<0|1>, <3|4>>^n):
seq(a(n), n=0..24);  # Alois P. Heinz, Jun 28 2021

LinearRecurrence[{4, 3}, {1, 5}, 24] (* Jean-François Alcover, Dec 10 2024 *)

a(n)=([0,1; 3,4]^n*[1;5])[1,1] \\ Charles R Greathouse IV, May 10 2016

From Johannes W. Meijer, Aug 01 2010: (Start)
G.f.: (1+x)/(1-4*x-3*x^2).
a(n) = 4*a(n-1) + 3*a(n-2) with a(0) = 1 and a(1) = 5.
a(n) = ((1+3/sqrt(7))/2)*(A)^(-n) + ((1-3/sqrt(7))/2)*(B)^(-n) with A = (-2 + sqrt(7))/3 and B = (-2-sqrt(7))/3.
Lim_{k->oo} a(n+k)/a(k) = (-1)^(n+1)*A000244(n)/(A015530(n)*sqrt(7)-A108851(n))
(End)
a(n) = A015330(n)+A015330(n+1). - R. J. Mathar, May 09 2023

Edited by Johannes W. Meijer, Aug 10 2010

A287804 Number of quinary sequences of length n such that no two consecutive terms have distance 1.

Original entry on oeis.org

1, 5, 17, 59, 205, 713, 2481, 8635, 30057, 104629, 364225, 1267923, 4413861, 15365465, 53490097, 186209299, 648230545, 2256616133, 7855718641, 27347281995, 95201200637, 331413874569, 1153716087665, 4016309864843, 13981555011321, 48672509644725

Offset: 0

David Nacin, Jun 01 2017

			For n=2 the a(2)=17=25-8 sequences contain every combination except these eight: 01,10,12,21,23,32,34,43.

Index entries for linear recurrences with constant coefficients, signature (5,-5,-1).

Cf. A040000, A003945, A083318, A078057, A003946, A126358, A003946, A055099, A003947, A015448, A126473. A287804-A287819.

LinearRecurrence[{5, -5, -1}, {1, 5, 17}, 50]

def a(n):
    if n in [0,1,2]:
        return [1,5,17][n]
    return 5*a(n-1)-5*a(n-2)-a(n-3)

a(n) = 5*a(n-1) - 5a(n-2) - a(n-3), a(0)=1, a(1)=5, a(2)=17.

G.f.: (1 - 3*x^2)/(1 - 5*x + 5*x^2 + x^3).

A287819 Number of nonary sequences of length n such that no two consecutive terms have distance 4.

Original entry on oeis.org

1, 9, 71, 561, 4433, 35031, 276827, 2187585, 17287073, 136608591, 1079529611, 8530826457, 67413620993, 532726379847, 4209793089371, 33267280400913, 262889866978817, 2077449112980255, 16416740845208075, 129730917736941417, 1025179795159015841

Offset: 0

David Nacin, Jun 02 2017

			For n=2 the a(2) = 81 - 10 = 71 sequences contain every combination except these ten: 04,40,15,51,26,62,37,73,48,84.

Index entries for linear recurrences with constant coefficients, signature (8,1,-14).

Cf. A040000, A003945, A083318, A078057, A003946, A126358, A003946, A055099, A003947, A015448, A126473. A287804-A287819.

LinearRecurrence[{8, 1, -14}, {1, 9, 71, 561}, 40]

def a(n):
    if n in [0, 1, 2, 3]:
        return [1, 9, 71, 561][n]
    return 8*a(n-1)+a(n-2)-14*a(n-3)

For n>2, a(n) = 8*a(n-1) + a(n-2) - 14*a(n-3), a(0)=1, a(1)=9, a(2)=71, a(3)=561.

G.f.: (1 + x - 2 x^2 - 2 x^3)/(1 - 8 x - x^2 + 14 x^3).

A107839 a(n) = 5a(n-1) - 2a(n-2); a(0)=1, a(1)=5.

Original entry on oeis.org

1, 5, 23, 105, 479, 2185, 9967, 45465, 207391, 946025, 4315343, 19684665, 89792639, 409593865, 1868384047, 8522732505, 38876894431, 177339007145, 808941246863, 3690028220025, 16832258606399, 76781236591945, 350241665746927, 1597645855550745, 7287745946259871

Offset: 0

Emeric Deutsch, Jun 12 2005

Kekulé numbers for certain benzenoids.
This is the number of spanning, connected subgraphs of the "ladder graph" of n squares (ladder graph = the vertices and edges of the tiling of a 1 X n rectangle by unit squares). - David Pasino (davepasino(AT)yahoo.com), Sep 18 2007
a(n) equals the number of words of length n over {0,1,2,3,4} avoiding 01 and 02. - Milan Janjic, Dec 17 2015

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

Alois P. Heinz, Table of n, a(n) for n = 0..1000
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).
A. M. Hinz, S. Klavžar, U. Milutinović, and C. Petr, The Tower of Hanoi - Myths and Maths, Birkhäuser 2013. See page 117. Book's website
Index entries for linear recurrences with constant coefficients, signature (5,-2).

Cf. A020698, A055099 (inverse binomial transform).

I:=[1,5]; [n le 2 select I[n] else 5*Self(n-1)-2*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Dec 17 2015

a:= n-> (<<0|1>, <-2|5>>^n)[2$2]:
seq(a(n), n=0..25);  # Alois P. Heinz, Nov 21 2020

a[n_] := (MatrixPower[{{1, 2}, {1, 4}}, n].{{1}, {1}})[[2, 1]]; Table[a[n], {n, 0, 40}] (* Vladimir Joseph Stephan Orlovsky, Feb 19 2010 *)
Table[(((5 + Sqrt[17])/2)^n - ((5 - Sqrt[17])/2)^n)/Sqrt[17], {n, 20}] // Expand (* Eric W. Weisstein, Nov 03 2024 *)
LinearRecurrence[{5, -2}, {1, 5}, 20] (* Eric W. Weisstein, Nov 03 2024 *)
CoefficientList[Series[1/(1 - 5 x + 2 x^2), {x, 0, 20}], x] (* Eric W. Weisstein, Nov 03 2024 *)

Vec(1/(1-5*x+2*x^2) + O(x^100)) \\ Altug Alkan, Dec 17 2015

[lucas_number1(n,5,2) for n in range(27)] # Zerinvary Lajos, Jun 25 2008

a(n) = A020698(n)-2*A020698(n-1) (n>=1).
a(n) = [M^(n+1)]1,2, where M is the 3 X 3 matrix defined as follows: M = [2,1,2; 1,1,1; 2,1,2]. - _Simone Severini, Jun 12 2006
a(n) = (((5 + s)/2)^(n+1) - ((5 - s)/2)^(n+1))/s with s = 17^(1/2). - David Pasino (davepasino(AT)yahoo.com), Jan 09 2009
G.f.: 1/(1 - 5*x + 2*x^2). - R. J. Mathar, Apr 07 2009
E.g.f.: exp(5*x/2)*(17*cosh(sqrt(17)*x/2) + 5*sqrt(17)*sinh(sqrt(17)*x/2))/17. - Stefano Spezia, Jun 17 2025

A287825 Number of sequences over the alphabet {0,1,...,9} such that no two consecutive terms have distance 1.

Original entry on oeis.org

1, 10, 82, 674, 5540, 45538, 374316, 3076828, 25291120, 207889674, 1708825732, 14046322404, 115458919774, 949057110644, 7801124426174, 64124215108032, 527092600834054, 4332631742719370, 35613662169258228, 292739611493034596, 2406281042646218328

Offset: 0

David Nacin, Jun 02 2017

Index entries for linear recurrences with constant coefficients, signature (9, -4, -21, 9, 5).

Cf. A040000, A003945, A083318, A078057, A003946, A126358, A003946, A055099, A003947, A015448, A126473. A287804-A287819. A287825-A287831.

LinearRecurrence[{9, -4, -21, 9, 5}, {1, 10, 82, 674, 5540, 45538}, 40]

def a(n):
    if n in [0, 1, 2, 3, 4, 5]:
        return [1, 10, 82, 674, 5540, 45538][n]
    return 9*a(n-1) - 4*a(n-2) - 21*a(n-3) + 9*a(n-4) + 5*a(n-5)

For n>5, a(n) = 9*a(n-1) - 4*a(n-2) - 21*a(n-3) + 9*a(n-4) + 5*a(n-5), a(0)=1, a(1)=10, a(2)=82, a(3)=674, a(4)=5540, a(5)=45538.

G.f.: (-1 - x + 4*x^2 + 3*x^3 - 3*x^4 - x^5)/(-1 + 9*x - 4*x^2 - 21*x^3 + 9*x^4 + 5*x^5).

A104934 Expansion of (1-x)/(1 - 3x - 2x^2).

Original entry on oeis.org

1, 2, 8, 28, 100, 356, 1268, 4516, 16084, 57284, 204020, 726628, 2587924, 9217028, 32826932, 116914852, 416398420, 1483024964, 5281871732, 18811665124, 66998738836, 238619546756, 849856117940, 3026807447332, 10780134577876, 38394018628292, 136742325040628, 487015012378468, 1734529687216660

Offset: 0

Creighton Dement, Mar 29 2005

A floretion-generated sequence relating A007482, A007483, A007484. Inverse is A046717. Inverse of Fibonacci(3n+1), A033887. Binomial transform is A052984. Inverse binomial transform is A006131. Note: the conjectured relation 2*a(n) = A007482(n) + A007483(n-1) is a result of the FAMP identity dia[I] + dia[J] + dia[K] = jes + fam
Floretion Algebra Multiplication Program, FAMP Code: 1dia[I]tesseq[A*B] with A = - .25'i + .25'j + .25'k - .25i' + .25j' + .25k' - .25'ii' + .25'jj' + .25'kk' + .25'ij' + .25'ik' + .25'ji' + .25'jk' + .25'ki' + .25'kj' - .25e and B = + 'i + i' + 'ji' + 'ki' + e
a(n) is also the number of ways to build a (2 x 2 x n)-tower using (2 X 1 X 1)-bricks (see Exercise 3.15 in Aigner's book). - Vania Mascioni (vmascioni(AT)bsu.edu), Mar 09 2009
a(n) is the number of compositions of n when there are 2 types of 1 and 4 types of other natural numbers. - Milan Janjic, Aug 13 2010
Pisano period lengths: 1, 1, 4, 1, 24, 4, 48, 1, 12, 24, 30, 4, 12, 48, 24, 1,272, 12, 18, 24, ... - R. J. Mathar, Aug 10 2012

M. Aigner, A Course in Enumeration, Springer, 2007, p.103.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
I. M. Gessel and Ji Li, Compositions and Fibonacci identities, J. Int. Seq. 16 (2013) 13.4.5
Index entries for linear recurrences with constant coefficients, signature (3,2)

Cf. A006131, A007482 (partial sums), A007483, A007484, A033887, A046717, A052984, A055099, A104935, A122542.

# Following the Pari implementation.
function a(n)
   F = BigInt[0 1; 2 3]
   Fn = F^n * [1; 2]
   Fn[1, 1]
end # Peter Luschny, Jan 06 2019

m:=35; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(1 - x)/(1 - 3*x - 2*x^2)); // Vincenzo Librandi, Jul 13 2018

a := proc(n) option remember; `if`(n < 2, [1, 2][n+1], (3*a(n-1) + 2*a(n-2))) end:
seq(a(n), n=0..28); # Peter Luschny, Jan 06 2019

LinearRecurrence[{3, 2}, {1, 2}, 40] (* Vincenzo Librandi, Jul 13 2018 *)
CoefficientList[Series[(1-x)/(1-3x-2x^2),{x,0,40}],x] (* Harvey P. Dale, May 02 2019 *)

a(n)=([0,1; 2,3]^n*[1;2])[1,1] \\ Charles R Greathouse IV, Jun 20 2015

[(i*sqrt(2))^(n-1)*(i*sqrt(2)*chebyshev_U(n, -3*i/(2*sqrt(2))) - chebyshev_U(n-1, -3*i/(2*sqrt(2))) ) for n in (0..40)] # G. C. Greubel, Jun 27 2021

Define A007483(-1) = 1. Then 2*a(n) = A007482(n) + A007483(n-1) (conjecture);
  a(n+2) = 4*A007484(n) (thus 8*A007484(n) = A007482(n+2) + A007483(n+1));
  a(n+1) = 2*A055099(n);
  a(n+2) - a(n+1) - a(n) = A007484(n+1) - A007484(n).
a(0)=1, a(1)=2, a(n) = 3*a(n-1) + 2*a(n-2) for n > 1. - Philippe Deléham, Sep 19 2006
a(n) = Sum_{k=0..n} 2^k*A122542(n,k). - Philippe Deléham, Oct 08 2006
a(n) = ((17+sqrt(17))/34)*((3+sqrt(17))/2)^n + ((17-sqrt(17))/34)*((3-sqrt(17))/2)^n. - Richard Choulet, Nov 19 2008
a(n) = 2*a(n-1) + 4*Sum_{k=0..n-2} a(k) for n > 0. - Vania Mascioni (vmascioni(AT)bsu.edu), Mar 09 2009
G.f.: (1-x)/(1-3*x-2*x^2). - M. F. Hasler, Jul 12 2018
a(n) = (i*sqrt(2))^(n-1)*( i*sqrt(2)*ChebyshevU(n, -3*i/(2*sqrt(2))) - ChebyshevU(n-1, -3*i/(2*sqrt(2))) ). - G. C. Greubel, Jun 27 2021
E.g.f.: exp(3*x/2)*(sqrt(17)*cosh(sqrt(17)*x/2) + sinh(sqrt(17)*x/2))/sqrt(17). - Stefano Spezia, May 24 2024

A254602 Numbers of n-length words on alphabet {0..7} with no subwords ii, where i is from {0..2}.

Original entry on oeis.org

1, 8, 61, 467, 3574, 27353, 209341, 1602152, 12261769, 93843143, 718210846, 5496691637, 42067895689, 321958728008, 2464050574501, 18858147661547, 144327286503334, 1104581743831073, 8453708639334181, 64698869194494632, 495160627558133329, 3789618738879406463

Offset: 0

Milan Janjic, Feb 02 2015

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (7,5).

Cf. A015562, A055099, A126473, A126501, A126528.

[n le 1 select 8^n else 7*Self(n)+5*Self(n-1): n in [0..25]];

A254602:=n->(2^(-1-n) * ((7-sqrt(69))^n * (-9+sqrt(69)) + (7+sqrt(69))^n * (9+sqrt(69)))) / sqrt(69): seq(simplify(A254602(n)), n=0..30); # Wesley Ivan Hurt, Sep 08 2016

RecurrenceTable[{a[0] == 1, a[1] == 8, a[n] == 7 a[n - 1] + 5 a[n - 2]}, a[n], {n, 0, 25}]
LinearRecurrence[{7,5},{1,8},30] (* Harvey P. Dale, Jun 23 2017 *)

Vec((1+x)/(1-7*x-5*x^2) + O(x^30)) \\ Colin Barker, Sep 08 2016

G.f.: (1 + x)/(1 - 7*x - 5*x^2).
a(n) = 7*a(n-1) + 5*a(n-2) with n>1, a(0) = 1, a(1) = 8.
a(n) = (2^(-1-n) * ((7-sqrt(69))^n * (-9+sqrt(69)) + (7+sqrt(69))^n * (9+sqrt(69)))) / sqrt(69). - Colin Barker, Sep 08 2016

A026581 Expansion of (1 + 2x) / (1 - x - 4x^2).

Original entry on oeis.org

1, 3, 7, 19, 47, 123, 311, 803, 2047, 5259, 13447, 34483, 88271, 226203, 579287, 1484099, 3801247, 9737643, 24942631, 63893203, 163663727, 419236539, 1073891447, 2750837603, 7046403391, 18049753803, 46235367367, 118434382579, 303375852047, 777113382363

Offset: 0

Clark Kimberling

T(n,0) + T(n,1) + ... + T(n,2n), T given by A026568.
Row sums of Riordan array ((1+2x)/(1+x),x(1+2x)/(1+x)). Binomial transform is A055099. - Paul Barry, Jun 26 2008
Equals row sums of triangle A153341. - Gary W. Adamson, Dec 24 2008
Also, the number of walks of length n starting at vertex 0 in the graph with 4 vertices and edges {{0,1}, {0,2}, {0,3}, {1,2}, {2,3}}. - Sean A. Irvine, Jun 02 2025

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,4).

Cf. A006131, A026568, A026583, A026597, A026599, A052923, A055099.

Cf. A153341. - Gary W. Adamson, Dec 24 2008

a:=[1,3];; for n in [3..30] do a[n]:=a[n-1]+4*a[n-2]; od; a; # G. C. Greubel, Aug 03 2019

I:=[1,3]; [n le 2 select I[n] else Self(n-1) +4*Self(n-2): n in [1..30]]; // G. C. Greubel, Aug 03 2019

CoefficientList[Series[(1+2x)/(1-x-4x^2),{x,0,30}],x] (* or *) LinearRecurrence[{1,4},{1,3},30] (* Harvey P. Dale, Aug 04 2015 *)

Vec((1+2*x)/(1-x-4*x^2) + O(x^30)) \\ Colin Barker, Dec 22 2016

((1+2*x)/(1-x-4*x^2)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Aug 03 2019

G.f.: (1 + 2*x) / (1 - x - 4*x^2).
a(n) = a(n-1) + 4*a(n-2), n>1.
a(n) = 2*A006131(n-1) + A006131(n), n>0.
a(n) = (2^(-1-n)*((1-sqrt(17))^n*(-5+sqrt(17)) + (1+sqrt(17))^n*(5+sqrt(17))))/sqrt(17). - Colin Barker, Dec 22 2016

Edited by Ralf Stephan, Jul 20 2013

cp's OEIS Frontend

A256960 a(0)=1, a(1)=4; thereafter a(n) = a(n-2)+2*A055099(n-1)+2^(n-1).

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A001333 Pell-Lucas numbers: numerators of continued fraction convergents to sqrt(2).

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A126473 Number of strings over a 5 symbol alphabet with adjacent symbols differing by three or less.

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A287804 Number of quinary sequences of length n such that no two consecutive terms have distance 1.

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A287819 Number of nonary sequences of length n such that no two consecutive terms have distance 4.

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A107839 a(n) = 5*a(n-1) - 2*a(n-2); a(0)=1, a(1)=5.

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A107839 a(n) = 5a(n-1) - 2a(n-2); a(0)=1, a(1)=5.

A104934 Expansion of (1-x)/(1 - 3x - 2x^2).

A026581 Expansion of (1 + 2x) / (1 - x - 4x^2).