A095263 -id:A095263 - cp's OEIS frontend

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

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

A291000 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - S - S^2 - S^3.

Original entry on oeis.org

1, 3, 9, 26, 74, 210, 596, 1692, 4804, 13640, 38728, 109960, 312208, 886448, 2516880, 7146144, 20289952, 57608992, 163568448, 464417728, 1318615104, 3743926400, 10630080640, 30181847168, 85694918912, 243312448256, 690833811712, 1961475291648, 5569190816256

Offset: 0

Clark Kimberling, Aug 22 2017

Suppose s = (c(0), c(1), c(2),...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x).  Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).
In the following guide to p-INVERT sequences using s = (1,1,1,1,1,...) = A000012, in some cases t(1,1,1,1,1,...) is a shifted version of the cited sequence:
p(S)                             t(1,1,1,1,1,...)
1 - S                                A000079
1 - S^2                              A000079
1 - S^3                              A024495
1 - S^4                              A000749
1 - S^5                              A139761
1 - S^6                              A290993
1 - S^7                              A290994
1 - S^8                              A290995
1 - S - S^2                          A001906
1 - S - S^3                          A116703
1 - S - S^4                          A290996
1 - S^3 - S^6                        A290997
1 - S^2 - S^3                        A095263
1 - S^3 - S^4                        A290998
1 - 2 S^2                            A052542
1 - 3 S^2                            A002605
1 - 4 S^2                            A015518
1 - 5 S^2                            A163305
1 - 6 S^2                            A290999
1 - 7 S^2                            A291008
1 - 8 S^2                            A291001
(1 - S)^2                            A045623
(1 - S)^3                            A058396
(1 - S)^4                            A062109
(1 - S)^5                            A169792
(1 - S)^6                            A169793
(1 - S^2)^2                          A024007
1 - 2 S - 2 S^2                      A052530
1 - 3 S - 2 S^2                      A060801
(1 - S)(1 - 2 S)                     A053581
(1 - 2 S)(1 - 3 S)                   A291002
(1 - S)(1 - 2 S)(1 - 3 S)(1 - 4 S)   A291003
(1 - 2 S)^2                          A120926
(1 - 3 S)^2                          A291004
1 + S - S^2                          A000045  (Fibonacci numbers starting with -1)
1 - S - S^2 - S^3                    A291000
1 - S - S^2 - S^3 - S^4              A291006
1 - S - S^2 - S^3 - S^4 - S^5        A291007
1 - S^2 - S^4                        A290990
(1 - S)(1 - 3 S)                     A291009
(1 - S)(1 - 2 S)(1 - 3 S)            A291010
(1 - S)^2 (1 - 2 S)                  A291011
(1 - S^2)(1 - 2 S)                   A291012
(1 - S^2)^3                          A291013
(1 - S^3)^2                          A291014
1 - S - S^2 + S^3                    A045891
1 - 2 S - S^2 + S^3                  A291015
1 - 3 S + S^2                        A136775
1 - 4 S + S^2                        A291016
1 - 5 S + S^2                        A291017
1 - 6 S + S^2                        A291018
1 - S - S^2 - S^3 + S^4              A291019
1 - S - S^2 - S^3 - S^4 + S^5        A291020
1 - S - S^2 - S^3 + S^4 + S^5        A291021
1 - S - 2 S^2 + 2 S^3                A175658
1 - 3 S^2 + 2 S^3                    A291023
(1 - 2 S^2)^2                        A291024
(1 - S^3)^3                          A291143
(1 - S - S^2)^2                      A209917

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4, -4, 2)

Cf. A000012, A289780.

z = 60; s = x/(1 - x); p = 1 - s - s^2 - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291000 *)

G.f.: (-1 + x - x^2)/(-1 + 4 x - 4 x^2 + 2 x^3).

a(n) = 4*a(n-1) - 4*a(n-2) + 2*a(n-3) for n >= 4.

A052530 a(n) = 4*a(n-1) - a(n-2), with a(0) = 0, a(1) = 2.

Original entry on oeis.org

0, 2, 8, 30, 112, 418, 1560, 5822, 21728, 81090, 302632, 1129438, 4215120, 15731042, 58709048, 219105150, 817711552, 3051741058, 11389252680, 42505269662, 158631825968, 592022034210, 2209456310872, 8245803209278, 30773756526240

Offset: 0

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

a(n-1) and a(n+1) are the solutions for c if b = a(n) in (b^2 + c^2)/(b*c + 1) = 4 and there are no other pairs of solutions apart from consecutive pairs of terms in this sequence. Cf. A061167. - Henry Bottomley, Apr 18 2001
a(n)^2 for n >= 1 gives solutions to A007913(3*x+4) = A007913(x). - Benoit Cloitre, Apr 07 2002
For all terms k of the sequence, 3*k^2 + 4 is a perfect square. Limit_{n->oo} a(n)/a(n-1) = 2 + sqrt(3). - Gregory V. Richardson, Oct 06 2002
a(n) = the number of compositions of the integer 2*n into even parts, where each part 2*i comes in 2*i colors. (Dedrickson, Theorem 3.2.6) An example is given below. Cf. A052529, A095263. - Peter Bala, Sep 17 2013
Except for an initial 1, this is the p-INVERT of (1, 1, 1, 1, 1, ...) for p(S) = 1 - 2*S - 2*S^2; see A291000.  - Clark Kimberling, Aug 24 2017
a(n+1) is the number of spanning trees of the graph P_n, where P_n is a 2 X n grid with two additional vertices, u and v, where u is adjacent to (1,1) and (2,1), and v is adjacent to (1,n) and (2,n). - Kevin Long, May 04 2018
a(n) is also the output of Tesler's formula for the number of perfect matchings of an m X n Mobius band where m is even and n is odd, specialized to m=2. (The twist is on the length-n side.) - Sarah-Marie Belcastro, Feb 15 2022
In general, values of x and y which satisfy (x^2 + y^2)/(x*y + 1) = k^2 are any two adjacent terms of a second-order recurrence with initial terms 0 and k and signature (k^2,-1). This can also be expressed as a first-order recurrence a(n+1) = (k^2*a(n) + sqrt((k^4-4)*a(n)^2 + 4*k^2))/2, n > 1. - Gary Detlefs, Feb 27 2024

			Colored compositions. a(2) = 8: There are two compositions of 4 into even parts, namely 4 and 2 + 2. Using primes to indicate the coloring of parts, the 8 colored compositions are 4, 4', 4'', 4''', 2 + 2, 2 + 2', 2' + 2 and 2' + 2'. - _Peter Bala_, Sep 17 2013

T. D. Noe, Table of n, a(n) for n = 0..200
K. Andersen, L. Carbone, and D. Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Daniel Birmajer, Juan B. Gil, and Michael D. Weiner, (an + b)-color compositions, arXiv:1707.07798 [math.CO], 2017.
Niccolò Castronuovo, On the number of fixed points of the map gamma, arXiv:2102.02739 [math.NT], 2021. Mentions this sequence.
C. R. Dedrickson III, Compositions, Bijections, and Enumerations Thesis, 2012, Jack N. Averitt College of Graduate Studies, Georgia Southern University.
J.-P. Ehrmann et al., Problem POLYA002, Integer pairs (x,y) for which (x^2+y^2)/(1+pxy) is an integer.
F. Goebel and A. A. Jagers, On a conjecture of Tutte concerning minimal tree numbers, J. Combin. Theory Ser. B 26 (1979), no. 3, 346-348. MR0535948 (80m:05064). [From _N. J. A. Sloane_, Feb 20 2012]
A. F. Horadam, Basic Properties of a Certain Generalized Sequence of Numbers, Fibonacci Quarterly, Vol. 3, No. 3, 1965, pp. 161-176.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 460
N. J. A. Sloane, Transforms
G. Tesler, Matchings in graphs on non-orientable surfaces, Journal of Combinatorial Theory B, 78(2000), 198-231.
Index entries for linear recurrences with constant coefficients, signature (4,-1).

Cf. A001353, A007913, A003699, A217233.

Cf. A052529, A095263.

a052530 n = a052530_list !! n
a052530_list =
   0 : 2 : zipWith (-) (map (* 4) $ tail a052530_list) a052530_list
-- Reinhard Zumkeller, Sep 29 2011

I:=[0,2]; [n le 2 select I[n] else 4*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Feb 25 2019

spec := [S,{S=Sequence(Prod(Union(Z,Z),Sequence(Z),Sequence(Z)))},unlabeled]: seq(combstruct[count](spec, size=n), n=0..20);
s := sqrt(3): a := n -> ((2-s)^n-(s+2)^n)/(s*(s-2)*(s+2)):
seq(simplify(a(n)), n=0..24); # Peter Luschny, Apr 28 2020

p=1; c=2; a[0]=0; a[1]=c; a[n_]:=a[n]=p*c^2*a[n-1]-a[n-2]; Table[a[n], {n, 0, 20}]
NestList[2 # + Sqrt[4 + 3 #^2]&, 0, 200] (* Zak Seidov, Mar 31 2011 *)
LinearRecurrence[{4, -1}, {0, 2}, 25] (* T. D. Noe, Jan 09 2012 *)
CoefficientList[Series[2x/(1-4x+x^2),{x,0,30}],x] (* Harvey P. Dale, May 31 2023 *)

{ polya002(p,c,m) = local(v,w,j,a); w=0; print1(w,", "); v=c; print1(v,", "); j=1; while(j<=m,a=p*c^2*v-w; print1(a,", "); w=v; v=a; j++) };
polya002(1,2,25)

my(x='x+O('x^30)); concat([0], Vec(2*x/(1-4*x+x^2))) \\ G. C. Greubel, Feb 25 2019

first(n) = n = max(n, 2); my(res = vector(n)); res[1] = 0; res[2] = 2; for(i = 3, n, res[i] = 4 * res[i-1] - res[i-2]); res \\ David A. Corneth, Apr 28 2020

(2*x/(1-4*x+x^2)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Feb 25 2019

G.f.: 2*x/(1 - 4*x + x^2).
Invert transform of even numbers: a(n) = 2*Sum_{k=1..n} k*a(n-k). - Vladeta Jovovic, Apr 27 2001
From Gregory V. Richardson, Oct 06 2002: (Start)
a(n) = Sum_{alpha} -(1/3)*(-1 + 2*alpha)*alpha^(-1 - n), alpha = root of (1 - 4*Z + Z^2).
a(n) = (((2+sqrt(3))^(n+1) - (2-sqrt(3))^(n+1)) - ((2+sqrt(3))^n - (2-sqrt(3))^n) + ((2+sqrt(3))^(n-1) - (2-sqrt(3))^(n-1)))/(3*sqrt(3)). (End)
a(n) = A071954(n) - 2. - N. J. A. Sloane, Feb 20 2005
a(n) = (2*sinh(2n*arcsinh(1/sqrt(2))))/sqrt(3). - Herbert Kociemba, Apr 24 2008
a(n) = 2*A001353(n). - R. J. Mathar, Oct 26 2009
a(n) = ((3 - 2*sqrt(3))/3)*(2 - sqrt(3))^(n - 1) + ((3 + 2*sqrt(3))/3)*(2 + sqrt(3))^(n - 1). - Vincenzo Librandi, Nov 20 2010
a(n) = floor((2 + sqrt(3))^n/sqrt(3)). - Zak Seidov, Mar 31 2011
a(n) = ((2 + sqrt(3))^n - (2 - sqrt(3))^n)/sqrt(3). (See Horadam for construction.) - Johannes Boot, Jan 08 2012
a(n) = A217233(n) + A217233(n-1) with A217233(-1) = -1. - Bruno Berselli, Oct 01 2012
a(n) = A001835(n+1) - A001835(n). - Kevin Long, May 04 2018
E.g.f.: (exp((2 + sqrt(3))*x) - exp((2 - sqrt(3))*x))/sqrt(3). - Franck Maminirina Ramaharo, Nov 12 2018
a(n+1) = 2*a(n) + sqrt(3*a(n)^2 + 4), n > 1. - Gary Detlefs, Feb 27 2024

More terms from James Sellers, Jun 06 2000
Edited by N. J. A. Sloane, Nov 11 2006
a(0) changed to 0 and entry revised accordingly by Max Alekseyev, Nov 15 2007
Signs in definition corrected by John W. Layman, Nov 20 2007

A052529 Expansion of (1-x)^3/(1 - 4x + 3x^2 - x^3).

Original entry on oeis.org

1, 1, 4, 13, 41, 129, 406, 1278, 4023, 12664, 39865, 125491, 395033, 1243524, 3914488, 12322413, 38789712, 122106097, 384377665, 1209982081, 3808901426, 11990037126, 37743426307, 118812495276, 374009739309, 1177344897715

Offset: 0

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

a(n+1) is the number of distinct matrix products in (A+B+C+D)^n where commutator [A,B] = [A,C] = [B,C] = 0 but D does not commute with A, B, or C. - Paul D. Hanna and Max Alekseyev, Feb 01 2006
Starting (1, 4, 13, ...) = INVERT transform of the triangular series, (1, 3, 6, 10, ...). Example: a(5) = 129 = termwise products of (1, 1, 4, 13, 41) and (15, 10, 6, 3, 1) = (15 + 10 + 24 + 39 + 41). - Gary W. Adamson, Apr 10 2009
a(n) is the number of generalized compositions of n when there are i^2/2+i/2 different types of i, (i=1,2,...). - Milan Janjic, Sep 24 2010
Dedrickson (Section 4.2) gives a bijection between colored compositions of n, where each part k has one of binomial(k+1,2) colors, and 0,1,2,3 strings of length n-1 avoiding 10, 20 and 21. Cf. A095263. For a refinement of this sequence counting binomial(k+1,2)-colored compositions by the number of parts see A127893. - Peter Bala, Sep 17 2013

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 80.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
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 14.
C. R. Dedrickson III, Compositions, Bijections, and Enumerations Thesis, Jack N. Averitt College of Graduate Studies, Georgia Southern University, 2012.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 459
Milan Janjić, Pascal Matrices and Restricted Words, J. Int. Seq., Vol. 21 (2018), Article 18.5.2.
Index entries for linear recurrences with constant coefficients, signature (4,-3,1).

Cf. A001906, A052530, A055991, A095263.
Trisection of A000930.
First differences of A052544.
Row sums of triangle A127893.

I:=[1, 1, 4, 13, 41, 129]; [n le 6 select I[n] else 4*Self(n-1) -3*Self(n-2)+Self(n-3): n in [1..40]]; // Vincenzo Librandi, Jun 22 2012

spec := [S,{S=Sequence(Prod(Z,Sequence(Z),Sequence(Z),Sequence(Z)))},unlabeled]: seq(combstruct[count](spec,size=n), n=0..20);
f:= gfun:-rectoproc({a(n+4)-4*a(n+3)+3*a(n+2)-a(n+1), a(0) = 1, a(1) = 1, a(2) = 4, a(3) = 13},a(n),`remember`):
seq(f(n),n=0..40); # Robert Israel, Dec 19 2014

CoefficientList[Series[(-1+x)^3/(-1+4*x-3*x^2+x^3),{x,0,40}],x] (* Vincenzo Librandi, Jun 22 2012 *)
LinearRecurrence[{4,-3,1},{1,1,4,13},30] (* Harvey P. Dale, Oct 04 2015 *)

my(x='x+O('x^30)); Vec((1-x)^3/(1-4*x+3*x^2-x^3)) \\ G. C. Greubel, May 12 2019

((1-x)^3/(1-4*x+3*x^2-x^3)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, May 12 2019

a(n) = Sum_{a=0..n} (Sum_{b=0..n} (Sum_{c=0..n} C(n-b-c,a)*C(n-a-c,b)*C(n-a-b,c))).
G.f.: (1 - x)^3/(1 - 4*x + 3*x^2 - x^3).
a(n) = 4*a(n-1) - 3*a(n-2) + a(n-3) for n>=4.
a(n) = Sum_{alpha = RootOf(-1+4*x-3*x^2+x^3)} (1/31)*(6 - 5*alpha - 3*alpha^2) * alpha^(-1-n).
For n>0, a(n) = Sum_{k=0..n-1} Sum_{i=0..k} Sum_{j=0..i} a(j). - Benoit Cloitre, Jan 26 2003
a(n) = Sum_{k=0..n} binomial(n+2*k-1, n-k). - Vladeta Jovovic, Mar 23 2003
If p[i]=i(i+1)/2 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, May 02 2010
Recurrence equation: a(n) = Sum_{k = 1..n} 1/2*k*(k+1)*a(n-k) with a(0) = 1. - Peter Bala, Sep 19 2013
a(n) = Sum_{i=0..n} (n-i)*A052544(i) = A052544(n) - A052544(n-1) for n>=1. - Areebah Mahdia, Jul 07 2020

A034943 Binomial transform of Padovan sequence A000931.

Original entry on oeis.org

1, 1, 1, 2, 5, 12, 28, 65, 151, 351, 816, 1897, 4410, 10252, 23833, 55405, 128801, 299426, 696081, 1618192, 3761840, 8745217, 20330163, 47261895, 109870576, 255418101, 593775046, 1380359512, 3208946545

Offset: 0

N. J. A. Sloane

Trisection of the Padovan sequence: a(n) = A000931(3n). - Paul Barry, Jul 06 2004
a(n+1) gives diagonal sums of Riordan array (1/(1-x),x/(1-x)^3). - Paul Barry, Oct 11 2005
a(n+2) is the sum, over all Boolean n-strings, of the product of the lengths of the runs of 1. For example, the Boolean 7-string (0,1,1,0,1,1,1) has two runs of 1s. Their lengths, 2 and 3, contribute a product of 6 to a(9). The 8 Boolean 3-strings contribute to a(5) as follows: 000 (empty product), 001, 010, 100, 101 all contribute 1, 011 and 110 contribute 2, 111 contributes 3. - David Callan, Nov 29 2007
[a(n), a(n+1), a(n+2)], n > 0, = [0,1,0; 0,0,1; 1,-2,3]^n * [1,1,1]. - Gary W. Adamson, Mar 27 2008
Without the initial 1 and 1: 1, 2, 5, 12, 28, this is also the transform of 1 by the T_{1,0} transformation; see Choulet link. - Richard Choulet, Apr 11 2009
Without the first 1: transform of 1 by T_{0,0} transformation (see Choulet link). - Richard Choulet, Apr 11 2009
Starting (1, 2, 5, 12, ...) = INVERT transform of (1, 1, 2, 3, 4, 5, ...) and row sums of triangle A159974. - Gary W. Adamson, Apr 28 2009
a(n+1) is also the number of 321-avoiding separable permutations. (A permutation is separable if it avoids both 2413 and 3142.) - Vince Vatter, Sep 21 2009
a(n+1) is an eigensequence of the sequence array for (1,1,2,3,4,5,...). - Paul Barry, Nov 03 2010
Equals the INVERTi transform of A055588: (1, 2, 4, 9, 22, 56, ...) - Gary W. Adamson, Apr 01 2011
The Ca3 sums, see A180662, of triangle A194005 equal the terms of this sequence without a(0) and a(1). - Johannes W. Meijer, Aug 16 2011
Without the initial 1, a(n) = row sums of A182097(n)*A007318(n,k); i.e., a Triangular array T(n,k) multiplying the binomial (Pascal's) triangle by the Padovan sequence where a(0) = 1, a(1) = 0 and a(2) = 1. - Bob Selcoe, Jun 28 2013
a(n+1) is the top left entry of the n-th power of any of the 3 X 3 matrices [1, 1, 1; 0, 1, 1; 1, 0, 1] or [1, 1, 0; 1, 1, 1; 1, 0, 1] or [1, 1, 1; 1, 1, 0; 0, 1, 1] or [1, 0, 1; 1, 1, 0; 1, 1, 1]. - R. J. Mathar, Feb 03 2014
a(n) is the top left entry of the n-th power of the 3 X 3 matrix [1, 0, 1; 1, 1, 1; 0, 1, 1] or of the 3 X 3 matrix [1, 1, 0; 0, 1, 1; 1, 1, 1]. - R. J. Mathar, Feb 03 2014
Number of sequences (e(1), ..., e(n-1)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i) != e(j) < e(k) and e(i) <= e(k). [Martinez and Savage, 2.8] - Eric M. Schmidt, Jul 17 2017
a(n+1) is the number of words of length n over the alphabet {0,1,2} that do not contain the substrings 01 or 12 and do not start with a 2 and do not end with a 0. - Yiseth K. Rodríguez C., Sep 11 2020

			G.f. = 1 + x + x^2 + 2*x^3 + 5*x^4 + 12*x^5 + 28*x^6 + 65*x^7 + 151*x^8 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Miklos Bona and Rebecca Smith, Pattern avoidance in permutations and their squares, arXiv:1901.00026 [math.CO], 2018. See H(z), Ex. 4.1.
Richard Choulet, Curtz like Transformation
Michael Dairyko, Samantha Tyner, Lara Pudwell, and Casey Wynn, Non-contiguous pattern avoidance in binary trees, Electron. J. Combin. 19 (2012), no. 3, Paper 22, 21 pp. MR2967227. - From _N. J. A. Sloane_, Feb 01 2013
Stoyan Dimitrov, Sorting by shuffling methods and a queue, arXiv:2103.04332 [math.CO], 2021.
Phan Thuan Do, Thi Thu Huong Tran, and Vincent Vajnovszki, Exhaustive generation for permutations avoiding a (colored) regular sets of patterns, arXiv:1809.00742 [cs.DM], 2018.
Brian Hopkins and Hua Wang, Restricted Color n-color Compositions, arXiv:2003.05291 [math.CO], 2020.
Jia Huang and Erkko Lehtonen, Associative-commutative spectra for some varieties of groupoids, arXiv:2401.15786 [math.CO], 2024. See p. 18.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 904
H. Magnusson and H. Ulfarsson, Algorithms for discovering and proving theorems about permutation patterns, arXiv preprint arXiv:1211.7110 [math.CO], 2012.
Megan A. Martinez and Carla D. Savage, Patterns in Inversion Sequences II: Inversion Sequences Avoiding Triples of Relations, arXiv:1609.08106 [math.CO], 2016
Vincent Vatter, Finding regular insertion encodings for permutation classes, arXiv:0911.2683 [math.CO], 2009.
Chunyan Yan and Zhicong Lin, Inversion sequences avoiding pairs of patterns, arXiv:1912.03674 [math.CO], 2019.
Index entries for linear recurrences with constant coefficients, signature (3,-2,1).

First differences of A052921.
Cf. A000931, A007318, A055588, A095263, A097550.
Cf. A137531, A159974, A182097, A185963, A194005.

[n le 3 select 1 else 3*Self(n-1)-2*Self(n-2)+Self(n-3): n in [1..40]]; // Vincenzo Librandi, Feb 14 2012

A034943 := proc(n): add(binomial(n+k-1, 3*k), k=0..floor(n/2)) end: seq(A034943(n), n=0..28); # Johannes W. Meijer, Aug 16 2011

LinearRecurrence[{3,-2,1},{1,1,1},30] (* Harvey P. Dale, Aug 11 2017 *)

{a(n) = if( n<1, n = 0-n; polcoeff( (1 - x + x^2) / (1 - 2*x + 3*x^2 - x^3) + x * O(x^n), n), n = n-1; polcoeff( (1 - x + x^2) / (1 - 3*x + 2*x^2 - x^3) + x * O(x^n), n))} /* Michael Somos, Mar 31 2012 */

@CachedFunction
def a(n): # a = A034943
    if (n<3): return 1
    else: return 3*a(n-1) - 2*a(n-2) + a(n-3)
[a(n) for n in range(51)] # G. C. Greubel, Apr 22 2023

a(n) = 3*a(n-1) - 2*a(n-2) + a(n-3).
a(n) = Sum_{k=0..floor(n/2)} binomial(n+k-1, 3*k). - Paul Barry, Jul 06 2004
G.f.: (1 - 2*x)/(1 - 3*x + 2*x^2 - x^3). - Paul Barry, Jul 06 2005
G.f.: 1 + x / (1 - x / (1 - x / (1 - x / (1 + x / (1 - x))))). - Michael Somos, Mar 31 2012
a(-1 - n) = A185963(n). - Michael Somos, Mar 31 2012
a(n) = A095263(n) - 2*A095263(n-1). - G. C. Greubel, Apr 22 2023

Edited by Charles R Greathouse IV, Apr 20 2010

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