A002532 -id:A002532 - cp's OEIS frontend

A000129 Pell numbers: a(0) = 0, a(1) = 1; for n > 1, a(n) = 2*a(n-1) + a(n-2).

0, 1, 2, 5, 12, 29, 70, 169, 408, 985, 2378, 5741, 13860, 33461, 80782, 195025, 470832, 1136689, 2744210, 6625109, 15994428, 38613965, 93222358, 225058681, 543339720, 1311738121, 3166815962, 7645370045, 18457556052, 44560482149, 107578520350, 259717522849

Offset: 0

N. J. A. Sloane

Sometimes also called lambda numbers.
Also denominators of continued fraction convergents to sqrt(2): 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.
Number of lattice paths from (0,0) to the line x=n-1 consisting of U=(1,1), D=(1,-1) and H=(2,0) steps (i.e., left factors of Grand Schroeder paths); for example, a(3)=5, counting the paths H, UD, UU, DU and DD. - Emeric Deutsch, Oct 27 2002
a(2*n) with b(2*n) := A001333(2*n), n >= 1, give all (positive integer) solutions to Pell equation b^2 - 2*a^2 = +1 (see Emerson reference). a(2*n+1) with b(2*n+1) := A001333(2*n+1), n >= 0, give all (positive integer) solutions to Pell equation b^2 - 2*a^2 = -1.
Bisection: a(2*n+1) = T(2*n+1, sqrt(2))/sqrt(2) = A001653(n), n >= 0 and a(2*n) = 2*S(n-1,6) = 2*A001109(n), n >= 0, with T(n,x), resp. S(n,x), Chebyshev's polynomials of the first, resp. second kind. S(-1,x)=0. See A053120, resp. A049310. - Wolfdieter Lang, Jan 10 2003
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 denominators. - Amarnath Murthy, Mar 22 2003
This is also the Horadam sequence (0,1,1,2). Limit_{n->oo} a(n)/a(n-1) = sqrt(2) + 1 = A014176. - Ross La Haye, Aug 18 2003
Number of 132-avoiding two-stack sortable permutations.
From Herbert Kociemba, Jun 02 2004: (Start)
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) = 3.
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) = 1, s(n) = 2. (End)
Counts walks of length n from a vertex of a triangle to another vertex to which a loop has been added. - Mario Catalani (mario.catalani(AT)unito.it), Jul 23 2004
Apart from initial terms, Pisot sequence P(2,5). See A008776 for definition of Pisot sequences. - David W. Wilson
Sums of antidiagonals of A038207 [Pascal's triangle squared]. - Ross La Haye, Oct 28 2004
The Pell primality test is "If N is an odd prime, then P(N)-Kronecker(2,N) is divisible by N". "Most" composite numbers fail this test, so it makes a useful pseudoprimality test. The odd composite numbers which are Pell pseudoprimes (i.e., that pass the above test) are in A099011. - Jack Brennen, Nov 13 2004
a(n) = sum of n-th row of triangle in A008288 = A094706(n) + A000079(n). - Reinhard Zumkeller, Dec 03 2004
Pell trapezoids (cf. A084158); for n > 0, A001109(n) = (a(n-1) + a(n+1))*a(n)/2; e.g., 1189 = (12+70)*29/2. - Charlie Marion, Apr 01 2006
(0!a(1), 1!a(2), 2!a(3), 3!a(4), ...) and (1,-2,-2,0,0,0,...) form a reciprocal pair under the list partition transform and associated operations described in A133314. - Tom Copeland, Oct 29 2007
Let C = (sqrt(2)+1) = 2.414213562..., then for n > 1, C^n = a(n)*(1/C) + a(n+1). Example: C^3 = 14.0710678... = 5*(0.414213562...) + 12. Let X = the 2 X 2 matrix [0, 1; 1, 2]; then X^n * [1, 0] = [a(n-1), a(n); a(n), a(n+1)]. a(n) = numerator of n-th convergent to (sqrt(2)-1) = 0.414213562... = [2, 2, 2, ...], the convergents being [1/2, 2/5, 5/12, ...]. - Gary W. Adamson, Dec 21 2007
A = sqrt(2) = 2/2 + 2/5 + 2/(5*29) + 2/(29*169) + 2/(169*985) + ...; B = ((5/2) - sqrt(2)) = 2/2 + 2/(2*12) + 2/(12*70) + 2/(70*408) + 2/(408*2378) + ...; A+B = 5/2. C = 1/2 = 2/(1*5) + 2/(2*12) + 2/(5*29) + 2/(12*70) + 2/(29*169) + ... - Gary W. Adamson, Mar 16 2008
From Clark Kimberling, Aug 27 2008: (Start)
Related convergents (numerator/denominator):
  lower principal convergents: A002315/A001653
  upper principal convergents: A001541/A001542
  intermediate convergents: A052542/A001333
  lower intermediate convergents: A005319/A001541
  upper intermediate convergents: A075870/A002315
  principal and intermediate convergents: A143607/A002965
  lower principal and intermediate convergents: A143608/A079496
  upper principal and intermediate convergents: A143609/A084068. (End)
Equals row sums of triangle A143808 starting with offset 1. - Gary W. Adamson, Sep 01 2008
Binomial transform of the sequence:= 0,1,0,2,0,4,0,8,0,16,..., powers of 2 alternating with zeros. - Philippe Deléham, Oct 28 2008
a(n) is also the sum of the n-th row of the triangle formed by starting with the top two rows of Pascal's triangle and then each next row has a 1 at both ends and the interior values are the sum of the three numbers in the triangle above that position. - Patrick Costello (pat.costello(AT)eku.edu), Dec 07 2008
Starting with offset 1 = eigensequence of triangle A135387 (an infinite lower triangular matrix with (2,2,2,...) in the main diagonal and (1,1,1,...) in the subdiagonal). - Gary W. Adamson, Dec 29 2008
Starting with offset 1 = row sums of triangle A153345. - Gary W. Adamson, Dec 24 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 a(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. A001333, A142238, A142239, A153313, A153314, A153315, A153316, A153317, A153318.
(End)
Starting with offset 1 = row sums of triangle A155002, equivalent to the statement that the Fibonacci sequence convolved with the Pell sequence prefaced with a "1": (1, 1, 2, 5, 12, 29, ...) = (1, 2, 5, 12, 29, ...). - Gary W. Adamson, Jan 18 2009
It appears that P(p) == 8^((p-1)/2) (mod p), p = prime; analogous to [Schroeder, p. 90]: Fp == 5^((p-1)/2) (mod p). Example: Given P(11) = 5741, == 8^5 (mod 11). Given P(17) = 11336689, == 8^8 (mod 17) since 17 divides (8^8 - P(17)). - Gary W. Adamson, Feb 21 2009
Equals eigensequence of triangle A154325. - Gary W. Adamson, Feb 12 2009
Another combinatorial interpretation of a(n-1) arises from a simple tiling scenario. Namely, a(n-1) gives the number of ways of tiling a 1 X n rectangle with indistinguishable 1 X 2 rectangles and 1 X 1 squares that come in two varieties, say, A and B. For example, with C representing the 1 X 2 rectangle, we obtain a(4)=12 from AAA, AAB, ABA, BAA, ABB, BAB, BBA, BBB, AC, BC, CA and CB. - Martin Griffiths, Apr 25 2009
a(n+1) = 2*a(n) + a(n-1), a(1)=1, a(2)=2 was used by Theon from Smyrna. - Sture Sjöstedt, May 29 2009
The n-th Pell number counts the perfect matchings of the edge-labeled graph C_2 x P_(n-1), or equivalently, the number of domino tilings of a 2 X (n-1) cylindrical grid. - Sarah-Marie Belcastro, Jul 04 2009
As a fraction: 1/79 = 0.0126582278481... or 1/9799 = 0.000102051229...(1/119 and 1/10199 for sequence in reverse). - Mark Dols, May 18 2010
Limit_{n->oo} (a(n)/a(n-1) - a(n-1)/a(n)) tends to 2.0. Example: a(7)/a(6) - a(6)/a(7) = 169/70 - 70/169 = 2.0000845... - Gary W. Adamson, Jul 16 2010
Numbers k such that 2*k^2 +- 1 is a square. - Vincenzo Librandi, Jul 18 2010
Starting (1, 2, 5, ...) = INVERTi transform of A006190: (1, 3, 10, 33, 109, ...). - Gary W. Adamson, Aug 06 2010
[u,v] = [a(n), a(n-1)] generates all Pythagorean triples [u^2-v^2, 2uv, u^2+v^2] whose legs differ by 1. - James R. Buddenhagen, Aug 14 2010
An elephant sequence, see A175654. For the corner squares six A[5] vectors, with decimal values between 21 and 336, lead to this sequence (without the leading 0). For the central square these vectors lead to the companion sequence A078057. - Johannes W. Meijer, Aug 15 2010
Let the 2 X 2 square matrix A=[2, 1; 1, 0] then a(n) = the (1,1) element of A^(n-1). - Carmine Suriano, Jan 14 2011
Define a t-circle to be a first-quadrant circle tangent to the x- and y-axes. Such a circle has coordinates equal to its radius. Let C(0) be the t-circle with radius 1. Then for n > 0, define C(n) to be the next larger t-circle which is tangent to C(n - 1). C(n) has radius A001333(2n) + a(2n)*sqrt(2) and each of the coordinates of its point of intersection with C(n + 1) is a(2n + 1) + (A001333(2n + 1)*sqrt(2))/2. See similar Comments for A001109 and A001653, Sep 14 2005. - Charlie Marion, Jan 18 2012
A001333 and A000129 give the diagonal numbers described by Theon from Smyrna. - Sture Sjöstedt, Oct 20 2012
Pell numbers could also be called "silver Fibonacci numbers", since, for n >= 1, F(n+1) = ceiling(phi*F(n)), if n is even and F(n+1) = floor(phi*F(n)), if n is odd, where phi is the golden ratio, while a(n+1) = ceiling(delta*a(n)), if n is even and a(n+1) = floor(delta*a(n)), if n is odd, where delta = delta_S = 1+sqrt(2) is the silver ratio. - Vladimir Shevelev, Feb 22 2013
a(n) is the number of compositions (ordered partitions) of n-1 into two sorts of 1's and one sort of 2's. Example: the a(3)=5 compositions of 3-1=2 are 1+1, 1+1', 1'+1, 1'+1', and 2. - Bob Selcoe, Jun 21 2013
Between every two consecutive squares of a 1 X n array there is a flap that can be folded over one of the two squares. Two flaps can be lowered over the same square in 2 ways, depending on which one is on top. The n-th Pell number counts the ways n-1 flaps can be lowered. For example, a sideway representation for the case n = 3 squares and 2 flaps is \\., .//, \./, ./., .\., where . is an empty square. - Jean M. Morales, Sep 18 2013
Define a(-n) to be a(n) for n odd and -a(n) for n even. Then a(n) = A005319(k)*(a(n-2k+1) - a(n-2k)) + a(n-4k) = A075870(k)*(a(n-2k+2) - a(n-2k+1)) - a(n-4k+2). - Charlie Marion, Nov 26 2013
An alternative formulation of the combinatorial tiling interpretation listed above: Except for n=0, a(n-1) is the number of ways of partial tiling a 1 X n board with 1 X 1 squares and 1 X 2 dominoes. - Matthew Lehman, Dec 25 2013
Define a(-n) to be a(n) for n odd and -a(n) for n even. Then a(n) = A077444(k)*a(n-2k+1) + a(n-4k+2). This formula generalizes the formula used to define this sequence. - Charlie Marion, Jan 30 2014
a(n-1) is the top left entry of the n-th power of any of the 3 X 3 matrices [0, 1, 1; 1, 1, 1; 0, 1, 1], [0, 1, 1; 0, 1, 1; 1, 1, 1], [0, 1, 0; 1, 1, 1; 1, 1, 1] or [0, 0, 1; 1, 1, 1; 1, 1, 1]. - R. J. Mathar, Feb 03 2014
a(n+1) counts closed walks on K2 containing two loops on the other vertex. Equivalently the (1,1) entry of A^(n+1) where the adjacency matrix of digraph is A=(0,1;1,2). - David Neil McGrath, Oct 28 2014
For n >= 1, a(n) equals the number of ternary words of length n-1 avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
This is a divisibility sequence (i.e., if n|m then a(n)|a(m)). - Tom Edgar, Jan 28 2015
A strong divisibility sequence, that is, gcd(a(n), a(m)) = a(gcd(n, m)) for all positive integers n and m. - Michael Somos, Jan 03 2017
a(n) is the number of compositions (ordered partitions) of n-1 into two kinds of parts, n and n', when the order of the 1 does not matter, or equivalently, when the order of the 1' does not matter. Example: When the order of the 1 does not matter, the a(3)=5 compositions of 3-1=2 are 1+1, 1+1'=1+1, 1'+1', 2 and 2'. (Contrast with entry from Bob Selcoe dated Jun 21 2013.) - Gregory L. Simay, Sep 07 2017
Number of weak orderings R on {1,...,n} that are weakly single-peaked w.r.t. the total ordering 1 < ... < n and for which {1,...,n} has exactly one minimal element for the weak ordering R. - J. Devillet, Sep 28 2017
Also the number of matchings in the (n-1)-centipede graph. - Eric W. Weisstein, Sep 30 2017
Let A(r,n) be the total number of ordered arrangements of an n+r tiling of r red squares and white tiles of total length n, where the individual tile lengths can range from 1 to n. A(r,0) corresponds to a tiling of r red squares only, and so A(r,0)=1. Let A_1(r,n) = Sum_{j=0..n} A(r,j) and let A_s(r,n) = Sum_{j=0..n} A_(s-1)(r,j). Then A_0(1,n) + A_2(3,n-4) + A_4(5,n-8) + ... + A_(2j) (2j+1, n-4j) = a(n) without the initial 0. - Gregory L. Simay, May 25 2018
(1, 2, 5, 12, 29, ...) is the fourth INVERT transform of (1, -2, 5, -12, 29, ...), as shown in A073133. - Gary W. Adamson, Jul 17 2019
Number of 2-compositions of n restricted to odd parts (and allowed zeros); see Hopkins & Ouvry reference. - Brian Hopkins, Aug 17 2020
Also called the 2-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence. - Michael A. Allen, Jan 23 2023
Named by Lucas (1878) after the English mathematician John Pell (1611-1685). - Amiram Eldar, Oct 02 2023
a(n) is the number of compositions of n when there are F(i) parts of size i, with i,n >= 1, F(n) the Fibonacci numbers, A000045(n) (see example below). - Enrique Navarrete, Dec 15 2023

			G.f. = x + 2*x^2 + 5*x^3 + 12*x^4 + 29*x^5 + 70*x^6 + 169*x^7 + 408*x^8 + 985*x^9 + ...
From _Enrique Navarrete_, Dec 15 2023: (Start)
From the comment on compositions with Fibonacci number of parts, F(n), there are F(1)=1 type of 1, F(2)=1 type of 2, F(3)=2 types of 3, F(4)=3 types of 4, F(5)=5 types of 5 and F(6)=8 types of 6.
The following table gives the number of compositions of n=6 with Fibonacci number of parts:
Composition, number of such compositions, number of compositions of this type:
6,           1,     8;
5+1,         2,    10;
4+2,         2,     6;
3+3,         1,     4;
4+1+1,       3,     9;
3+2+1,       6,    12;
2+2+2,       1,     1;
3+1+1+1,     4,     8;
2+2+1+1,     6,     6;
2+1+1+1+1,   5,     5;
1+1+1+1+1+1, 1,     1;
for a total of a(6)=70 compositions of n=6. (End).

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

Partial sums of A001333.
2nd row of A172236.
a(n) = A054456(n-1, 0), n>=1 (first column of triangle).
Cf. A002203, A096669, A096670, A097075, A097076, A051927, A005409.
Cf. A175181 (Pisano periods), A214028 (Entry points), A214027 (number of zeros in a fundamental period).
A077985 is a signed version.
INVERT transform of Fibonacci numbers (A000045).
Cf. A038207.
The following sequences (and others) belong to the same family: A001333, A000129, A026150, A002605, A046717, A015518, A084057, A063727, A002533, A002532, A083098, A083099, A083100, A015519.
Cf. A034867, A131980, A133156, A143808, A135387, A153346, A001622, A006497, A014176 (growth power), A098316, A154325, A021083, A243399, A008555.
Cf. A048739.
Cf. A073133.
Cf. A041085.
Cf. A157103, A352361.
Sequences with g.f. 1/(1-k*x-x^2) or x/(1-k*x-x^2): A000045 (k=1), this sequence (k=2), A006190 (k=3), A001076 (k=4), A052918 (k=5), A005668 (k=6), A054413 (k=7), A041025 (k=8), A099371 (k=9), A041041 (k=10), A049666 (k=11), A041061 (k=12), A140455 (k=13), A041085 (k=14), A154597 (k=15), A041113 (k=16), A178765 (k=17), A041145 (k=18), A243399 (k=19), A041181 (k=20).

a := [0,1];; for n in [3..10^3] do a[n] := 2 * a[n-1] + a[n-2]; od; A000129 := a; # Muniru A Asiru, Oct 16 2017

a000129 n = a000129_list !! n
a000129_list = 0 : 1 : zipWith (+) a000129_list (map (2 *) $ tail a000129_list)
-- Reinhard Zumkeller, Jan 05 2012, Feb 05 2011

[0] cat [n le 2 select n else 2*Self(n-1) + Self(n-2): n in [1..35]]; // Vincenzo Librandi, Aug 08 2015

A000129 := proc(n) option remember; if n <=1 then n; else 2*procname(n-1)+procname(n-2); fi; end;
a:= n-> (<<2|1>, <1|0>>^n)[1, 2]: seq(a(n), n=0..40); # Alois P. Heinz, Aug 01 2008
A000129 := n -> `if`(n<2, n, 2^(n-1)*hypergeom([1-n/2, (1-n)/2], [1-n], -1)):
seq(simplify(A000129(n)), n=0..31); # Peter Luschny, Dec 17 2015

CoefficientList[Series[x/(1 - 2*x - x^2), {x, 0, 60}], x] (* Stefan Steinerberger, Apr 08 2006 *)
Expand[Table[((1 + Sqrt[2])^n - (1 - Sqrt[2])^n)/(2Sqrt[2]), {n, 0, 30}]] (* Artur Jasinski, Dec 10 2006 *)
LinearRecurrence[{2, 1}, {0, 1}, 60] (* Harvey P. Dale, Jan 04 2012 *)
a[ n_] := With[ {s = Sqrt@2}, ((1 + s)^n - (1 - s)^n) / (2 s)] // Simplify; (* Michael Somos, Jun 01 2013 *)
Table[Fibonacci[n, 2], {n, 0, 20}] (* Vladimir Reshetnikov, May 08 2016 *)
Fibonacci[Range[0, 20], 2] (* Eric W. Weisstein, Sep 30 2017 *)
a[ n_] := ChebyshevU[n - 1, I] / I^(n - 1); (* Michael Somos, Oct 30 2021 *)

a[0]:0$
a[1]:1$
a[n]:=2*a[n-1]+a[n-2]$
A000129(n):=a[n]$
makelist(A000129(n),n,0,30); /* Martin Ettl, Nov 03 2012 */

makelist((%i)^(n-1)*ultraspherical(n-1,1,-%i),n,0,24),expand; /* Emanuele Munarini, Mar 07 2018 */

for (n=0, 4000, a=contfracpnqn(vector(n, i, 1+(i>1)))[2, 1]; if (a > 10^(10^3 - 6), break); write("b000129.txt", n, " ", a)); \\ Harry J. Smith, Jun 12 2009

{a(n) = imag( (1 + quadgen( 8))^n )}; /* Michael Somos, Jun 01 2013 */

{a(n) = if( n<0, -(-1)^n, 1) * contfracpnqn( vector( abs(n), i, 1 + (i>1))) [2, 1]}; /* Michael Somos, Jun 01 2013 */

a(n)=([2, 1; 1, 0]^n)[2,1] \\ Charles R Greathouse IV, Mar 04 2014

{a(n) = polchebyshev(n-1, 2, I) / I^(n-1)}; /* Michael Somos, Oct 30 2021 */

from itertools import islice
def A000129_gen(): # generator of terms
    a, b = 0, 1
    yield from [a,b]
    while True:
        a, b = b, a+2*b
        yield b
A000129_list = list(islice(A000129_gen(),20)) # Chai Wah Wu, Jan 11 2022

[lucas_number1(n, 2, -1) for n in range(30)]  # Zerinvary Lajos, Apr 22 2009

G.f.: x/(1 - 2*x - x^2). - Simon Plouffe in his 1992 dissertation.
a(2n+1)=A001653(n). a(2n)=A001542(n). - Ira Gessel, Sep 27 2002
G.f.: Sum_{n >= 0} x^(n+1) *( Product_{k = 1..n} (2*k + x)/(1 + 2*k*x) ) = Sum_{n >= 0} x^(n+1) *( Product_{k = 1..n} (x + 1 + k)/(1 + k*x) ) = Sum_{n >= 0} x^(n+1) *( Product_{k = 1..n} (x + 3 - k)/(1 - k*x) ) may all be proved using telescoping series. - Peter Bala, Jan 04 2015
a(n) = 2*a(n-1) + a(n-2), a(0)=0, a(1)=1.
a(n) = ((1 + sqrt(2))^n - (1 - sqrt(2))^n)/(2*sqrt(2)).
For initial values a(0) and a(1), a(n) = ((a(0)*sqrt(2)+a(1)-a(0))*(1+sqrt(2))^n + (a(0)*sqrt(2)-a(1)+a(0))*(1-sqrt(2))^n)/(2*sqrt(2)). - Shahreer Al Hossain, Aug 18 2019
a(n) = integer nearest a(n-1)/(sqrt(2) - 1), where a(0) = 1. - Clark Kimberling
a(n) = Sum_{i, j, k >= 0: i+j+2k = n} (i+j+k)!/(i!*j!*k!).
a(n)^2 + a(n+1)^2 = a(2n+1) (1999 Putnam examination).
a(2n) = 2*a(n)*A001333(n). - John McNamara, Oct 30 2002
a(n) = ((-i)^(n-1))*S(n-1, 2*i), with S(n, x) := U(n, x/2) Chebyshev's polynomials of the second kind. See A049310. S(-1, x)=0, S(-2, x)= -1.
Binomial transform of expansion of sinh(sqrt(2)x)/sqrt(2). E.g.f.: exp(x)sinh(sqrt(2)x)/sqrt(2). - Paul Barry, May 09 2003
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2k+1)*2^k. - Paul Barry, May 13 2003
a(n-2) + a(n) = (1 + sqrt(2))^(n-1) + (1 - sqrt(2))^(n-1) = A002203(n-1). (A002203(n))^2 - 8(a(n))^2 = 4(-1)^n. - Gary W. Adamson, Jun 15 2003
Unreduced g.f.: x(1+x)/(1 - x - 3x^2 - x^3); a(n) = a(n-1) + 3*a(n-2) + a(n-2). - Mario Catalani (mario.catalani(AT)unito.it), Jul 23 2004
a(n+1) = Sum_{k=0..floor(n/2)} binomial(n-k, k)*2^(n-2k). - Mario Catalani (mario.catalani(AT)unito.it), Jul 23 2004
Apart from initial terms, inverse binomial transform of A052955. - Paul Barry, May 23 2004
a(n)^2 + a(n+2k+1)^2 = A001653(k)*A001653(n+k); e.g., 5^2 + 70^2 = 5*985. - Charlie Marion Aug 03 2005
a(n+1) = Sum_{k=0..n} binomial((n+k)/2, (n-k)/2)*(1+(-1)^(n-k))*2^k/2. - Paul Barry, Aug 28 2005
a(n) = a(n-1) + A001333(n-1) = A001333(n) - a(n-1) = A001109(n)/A001333(n) = sqrt(A001110(n)/A001333(n)^2) = ceiling(sqrt(A001108(n)/2)). - Henry Bottomley, Apr 18 2000
a(n) = F(n, 2), the n-th Fibonacci polynomial evaluated at x=2. - T. D. Noe, Jan 19 2006
Define c(2n) = -A001108(n), c(2n+1) = -A001108(n+1) and d(2n) = d(2n+1) = A001652(n); then ((-1)^n)*(c(n) + d(n)) = a(n). [Proof given by Max Alekseyev.] - Creighton Dement, Jul 21 2005
a(r+s) = a(r)*a(s+1) + a(r-1)*a(s). - Lekraj Beedassy, Sep 03 2006
a(n) = (b(n+1) + b(n-1))/n where {b(n)} is the sequence A006645. - Sergio Falcon, Nov 22 2006
From Miklos Kristof, Mar 19 2007: (Start)
Let F(n) = a(n) = Pell numbers, L(n) = A002203 = companion Pell numbers (A002203):
For a >= b and odd b,  F(a+b) + F(a-b) = L(a)*F(b).
For a >= b and even b, F(a+b) + F(a-b) = F(a)*L(b).
For a >= b and odd b,  F(a+b) - F(a-b) = F(a)*L(b).
For a >= b and even b, F(a+b) - F(a-b) = L(a)*F(b).
F(n+m) + (-1)^m*F(n-m) = F(n)*L(m).
F(n+m) - (-1)^m*F(n-m) = L(n)*F(m).
F(n+m+k) + (-1)^k*F(n+m-k) + (-1)^m*(F(n-m+k) + (-1)^k*F(n-m-k)) = F(n)*L(m)*L(k).
F(n+m+k) - (-1)^k*F(n+m-k) + (-1)^m*(F(n-m+k) - (-1)^k*F(n-m-k)) = L(n)*L(m)*F(k).
F(n+m+k) + (-1)^k*F(n+m-k) - (-1)^m*(F(n-m+k) + (-1)^k*F(n-m-k)) = L(n)*F(m)*L(k).
F(n+m+k) - (-1)^k*F(n+m-k) - (-1)^m*(F(n-m+k) - (-1)^k*F(n-m-k)) = 8*F(n)*F(m)*F(k). (End)
a(n+1)*a(n) = 2*Sum_{k=0..n} a(k)^2 (a similar relation holds for A001333). - Creighton Dement, Aug 28 2007
a(n+1) = Sum_{k=0..n} binomial(n+1,2k+1) * 2^k = Sum_{k=0..n} A034867(n,k) * 2^k = (1/n!) * Sum_{k=0..n} A131980(n,k) * 2^k. - Tom Copeland, Nov 30 2007
Equals row sums of unsigned triangle A133156. - Gary W. Adamson, Apr 21 2008
a(n) (n >= 3) is the determinant of the (n-1) X (n-1) tridiagonal matrix with diagonal entries 2, superdiagonal entries 1 and subdiagonal entries -1. - Emeric Deutsch, Aug 29 2008
a(n) = A000045(n) + Sum_{k=1..n-1} A000045(k)*a(n-k). - Roger L. Bagula and Gary W. Adamson, Sep 07 2008
From Hieronymus Fischer, Jan 02 2009: (Start)
fract((1+sqrt(2))^n) = (1/2)*(1 + (-1)^n) - (-1)^n*(1+sqrt(2))^(-n) = (1/2)*(1 + (-1)^n) - (1-sqrt(2))^n.
See A001622 for a general formula concerning the fractional parts of powers of numbers x > 1, which satisfy x - x^(-1) = floor(x).
a(n) = round((1+sqrt(2))^n/(2*sqrt(2))) for n > 0. (End) [last formula corrected by Josh Inman, Mar 05 2024]
a(n) = ((4+sqrt(18))*(1+sqrt(2))^n + (4-sqrt(18))*(1-sqrt(2))^n)/4 offset 0. - Al Hakanson (hawkuu(AT)gmail.com), Aug 08 2009
If p[i] = Fibonacci(i) and if A is the Hessenberg matrix of order n defined by A[i,j] = p[j-i+1] when i<=j, A[i,j]=-1 when i=j+1, and A[i,j]=0 otherwise, then, for n >= 1, a(n) = det A. - Milan Janjic, May 08 2010
a(n) = 3*a(n-1) - a(n-2) - a(n-3), n > 2. - Gary Detlefs, Sep 09 2010
From Charlie Marion, Apr 13 2011: (Start)
a(n) = 2*(a(2k-1) + a(2k))*a(n-2k) - a(n-4k).
a(n) = 2*(a(2k) + a(2k+1))*a(n-2k-1) + a(n-4k-2). (End)
G.f.: x/(1 - 2*x - x^2) = sqrt(2)*G(0)/4; G(k) = ((-1)^k) - 1/(((sqrt(2) + 1)^(2*k)) - x*((sqrt(2) + 1)^(2*k))/(x + ((sqrt(2) - 1)^(2*k + 1))/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Dec 02 2011
In general, for n > k, a(n) = a(k+1)*a(n-k) + a(k)*a(n-k-1). See definition of Pell numbers and the formula for Sep 04 2008. - Charlie Marion, Jan 17 2012
Sum{n>=1} (-1)^(n-1)/(a(n)*a(n+1)) = sqrt(2) - 1. - Vladimir Shevelev, Feb 22 2013
From Vladimir Shevelev, Feb 24 2013: (Start)
(1) Expression a(n+1) via a(n): a(n+1) = a(n) + sqrt(2*a^2(n) + (-1)^n);
(2) a(n+1)^2 - a(n)*a(n+2) = (-1)^n;
(3) Sum_{k=1..n} (-1)^(k-1)/(a(k)*a(k+1)) = a(n)/a(n+1);
(4) a(n)/a(n+1) = sqrt(2) - 1 + r(n), where |r(n)| < 1/(a(n+1)*a(n+2)). (End)
a(-n) = -(-1)^n * a(n). - Michael Somos, Jun 01 2013
G.f.: G(0)/(2+2*x) - 1/(1+x), where G(k) = 1 + 1/(1 - x*(2*k-1)/(x*(2*k+1) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Aug 10 2013
G.f.: Q(0)*x/2, where Q(k) = 1 + 1/(1 - x*(4*k+2 + x)/( x*(4*k+4 + x) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 30 2013
a(n) = Sum_{r=0..n-1} Sum_{k=0..n-r-1} binomial(r+k,k)*binomial(k,n-k-r-1). - Peter Luschny, Nov 16 2013
a(n) = Sum_{k=1,3,5,...<=n} C(n,k)*2^((k-1)/2). - Vladimir Shevelev, Feb 06 2014
a(2n) = 2*a(n)*(a(n-1) + a(n)). - John Blythe Dobson, Mar 08 2014
a(k*n) = a(k)*a(k*n-k+1) + a(k-1)*a(k*n-k). - Charlie Marion, Mar 27 2014
a(k*n) = 2*a(k)*(a(k*n-k)+a(k*n-k-1)) + (-1)^k*a(k*n-2k). - Charlie Marion, Mar 30 2014
a(n+1) = (1+sqrt(2))*a(n) + (1-sqrt(2))^n. - Art DuPre, Apr 04 2014
a(n+1) = (1-sqrt(2))*a(n) + (1+sqrt(2))^n. - Art DuPre, Apr 04 2014
a(n) = F(n) + Sum_{k=1..n} F(k)*a(n-k), n >= 0 where F(n) the Fibonacci numbers A000045. - Ralf Stephan, May 23 2014
a(n) = round(sqrt(a(2n) + a(2n-1)))/2. - Richard R. Forberg, Jun 22 2014
a(n) = Product_{k divides n} A008555(k). - Tom Edgar, Jan 28 2015
a(n+k)^2 - A002203(k)*a(n)*a(n+k) + (-1)^k*a(n)^2 = (-1)^n*a(k)^2. - Alexander Samokrutov, Aug 06 2015
a(n) = 2^(n-1)*hypergeom([1-n/2, (1-n)/2], [1-n], -1) for n >= 2. - Peter Luschny, Dec 17 2015
a(n+1) = Sum_{k=0..n} binomial(n,k)*2^floor(k/2). - Tony Foster III, May 07 2017
a(n) = exp((i*Pi*n)/2)*sinh(n*arccosh(-i))/sqrt(2). - Peter Luschny, Mar 07 2018
From Rogério Serôdio, Mar 30 2018: (Start)
Some properties:
(1) a(n)^2 - a(n-2)^2 = 2*a(n-1)*(a(n) + a(n-2)) (see A005319);
(2) a(n-k)*a(n+k) = a(n)^2 + (-1)^(n+k+1)*a(k)^2;
(3) Sum_{k=2..n+1} a(k)*a(k-1) = a(n+1)^2 if n is odd, else a(n+1)^2 - 1 if n is even;
(4) a(n) - a(n-2*k+1) = (A077444(k) - 1)*a(n-2*k+1) + a(n-4*k+2);
(5) Sum_{k=n..n+9} a(k) = 41*A001333(n+5). (End)
From Kai Wang, Dec 30 2019: (Start)
a(m+r)*a(n+s) - a(m+s)*a(n+r) = -(-1)^(n+s)*a(m-n)*a(r-s).
a(m+r)*a(n+s) + a(m+s)*a(n+r) = (2*A002203(m+n+r+s) - (-1)^(n+s)*A002203(m-n)*A002203(r-s))/8.
A002203(m+r)*A002203(n+s) - A002203(m+s)*A002203(n+r) = (-1)^(n+s)*8*a(m-n)*a(r-s).
A002203(m+r)*A002203(n+s) - 8*a(m+s)*a(n+r) = (-1)^(n+s)*A002203(m-n)*A002203(r-s).
A002203(m+r)*A002203(n+s) + 8*a(m+s)*a(n+r) = 2*A002203(m+n+r+s)+ (-1)^(n+s)*8*a(m-n)*a(r-s). (End)
From Kai Wang, Jan 12 2020: (Start)
a(n)^2 - a(n+1)*a(n-1) = (-1)^(n-1).
a(n)^2 - a(n+r)*a(n-r) = (-1)^(n-r)*a(r)^2.
a(m)*a(n+1) - a(m+1)*a(n) = (-1)^n*a(m-n).
a(m-n) = (-1)^n (a(m)*A002203(n) - A002203(m)*a(n))/2.
a(m+n) = (a(m)*A002203(n) + A002203(m)*a(n))/2.
A002203(n)^2 - A002203(n+r)*A002203(n-r) = (-1)^(n-r-1)*8*a(r)^2.
A002203(m)*A002203(n+1) - A002203(m+1)*A002203(n) = (-1)^(n-1)*8*a(m-n).
A002203(m-n) = (-1)^(n)*(A002203(m)*A002203(n) - 8*a(m)*a(n) )/2.
A002203(m+n) = (A002203(m)*A002203(n) + 8*a(m)*a(n) )/2. (End)
From Kai Wang, Mar 03 2020: (Start)
Sum_{m>=1} arctan(2/a(2*m+1)) = arctan(1/2).
Sum_{m>=2} arctan(2/a(2*m+1)) = arctan(1/12).
In general, for n > 0,
Sum_{m>=n} arctan(2/a(2*m+1)) = arctan(1/a(2*n)). (End)
a(n) = (A001333(n+3*k) + (-1)^(k-1)*A001333(n-3*k)) / (20*A041085(k-1)) for any k>=1. - Paul Curtz, Jun 23 2021
Sum_{i=0..n} a(i)*J(n-i) = (a(n+1) + a(n) - J(n+2))/2 for J(n) = A001045(n). - Greg Dresden, Jan 05 2022
From Peter Bala, Aug 20 2022: (Start)
Sum_{n >= 1} 1/(a(2*n) + 1/a(2*n)) = 1/2.
Sum_{n >= 1} 1/(a(2*n+1) - 1/a(2*n+1)) = 1/4. Both series telescope - see A075870 and A005319.
Product_{n >= 1} ( 1 + 2/a(2*n) ) = 1 + sqrt(2).
Product_{n >= 2} ( 1 - 2/a(2*n) ) = (1/3)*(1 + sqrt(2)). (End)
G.f. = 1/(1 - Sum_{k>=1} Fibonacci(k)*x^k). - Enrique Navarrete, Dec 17 2023
Sum_{n >=1} 1/a(n) = 1.84220304982752858079237158327980838... - R. J. Mathar, Feb 05 2024
a(n) = ((3^(n+1) + 1)^(n-1) mod (9^(n+1) - 2)) mod (3^(n+1) - 1). - Joseph M. Shunia, Jun 06 2024

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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Markus Kuba and Alois Panholzer, Enumeration formulas for pattern restricted Stirling permutations, Discrete Math. 312 (2012), no. 21, 3179--3194. MR2957938. - From _N. J. A. Sloane_, Sep 25 2012
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Eric Weisstein's World of Mathematics, Vertex Cover.
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

A002605 a(n) = 2*(a(n-1) + a(n-2)), a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 2, 6, 16, 44, 120, 328, 896, 2448, 6688, 18272, 49920, 136384, 372608, 1017984, 2781184, 7598336, 20759040, 56714752, 154947584, 423324672, 1156544512, 3159738368, 8632565760, 23584608256, 64434348032, 176037912576, 480944521216, 1313964867584

Offset: 0

Colin Mallows

Individually, both this sequence and A028859 are convergents to 1 + sqrt(3). Mutually, both sequences are convergents to 2 + sqrt(3) and 1 + sqrt(3)/2. - Klaus E. Kastberg (kastberg(AT)hotkey.net.au), Nov 04 2001
The number of (s(0), s(1), ..., s(n+1)) such that 0 < s(i) < 6 and |s(i) - s(i-1)| <= 1 for i = 1, 2, ..., n + 1, s(0) = 2, s(n+1) = 3. - Herbert Kociemba, Jun 02 2004
The same sequence may be obtained by the following process. Starting a priori with the fraction 1/1, the denominators of fractions built according to the rule: add top and bottom to get the new bottom, add top and 4 times the bottom to get the new top. The limit of the sequence of fractions is sqrt(4). - Cino Hilliard, Sep 25 2005
The Hankel transform of this sequence is [1, 2, 0, 0, 0, 0, 0, 0, 0, ...]. - Philippe Deléham, Nov 21 2007
[1, 3; 1, 1]^n *[1, 0] = [A026150(n), a(n)]. - Gary W. Adamson, Mar 21 2008
(1 + sqrt(3))^n = A026150(n) + a(n)*sqrt(3). - Gary W. Adamson, Mar 21 2008
a(n+1) is the number of ways to tile a board of length n using red and blue tiles of length one and two. - Geoffrey Critzer, Feb 07 2009
Starting with offset 1 = INVERT transform of the Jacobsthal sequence, A001045: (1, 1, 3, 5, 11, 21, ...). - Gary W. Adamson, May 12 2009
Starting with "1" = INVERTi transform of A007482: (1, 3, 11, 39, 139, ...). - Gary W. Adamson, Aug 06 2010
An elephant sequence, see A175654. For the corner squares four A[5] vectors, with decimal values 85, 277, 337 and 340, lead to this sequence (without the leading 0). For the central square these vectors lead to the companion sequence A026150, without the first leading 1. - Johannes W. Meijer, Aug 15 2010
The sequence 0, 1, -2, 6, -16, 44, -120, 328, -896, ... (with alternating signs) is the Lucas U(-2,-2)-sequence. - R. J. Mathar, Jan 08 2013
a(n+1) counts n-walks (closed) on the graph G(1-vertex;1-loop,1-loop,2-loop,2-loop). - David Neil McGrath, Dec 11 2014
Number of binary strings of length 2*n - 2 in the regular language (00+11+0101+1010)*. - Jeffrey Shallit, Dec 14 2015
For n >= 1, a(n) equals the number of words of length n - 1  over {0, 1, 2, 3} in which 0 and 1 avoid runs of odd lengths. - Milan Janjic, Dec 17 2015
a(n+1) is the number of compositions of n into parts 1 and 2, both of two kinds. - Gregory L. Simay, Sep 20 2017
Number of associative, quasitrivial, and order-preserving binary operations on the n-element set {1, ..., n} that have neutral elements. - J. Devillet, Sep 28 2017
(1 + sqrt(3))^n = A026150(n) + a(n)*sqrt(3), for n >= 0; integers in the real quadratic number field Q(sqrt(3)). - Wolfdieter Lang, Feb 10 2018
Starting with 1, 2, 6, 16, ..., number of permutations of length n>0 avoiding the partially ordered pattern (POP) {1>3, 1>4} of length 4. That is, number of length n permutations having no subsequences of length 4 in which the first element is larger than the third and fourth elements. - Sergey Kitaev, Dec 09 2020
a(n) is the number of tilings of a 2 X n board missing one corner cell, with 1 X 1 and L-shaped tiles (where the L-shaped tiles cover 3 squares). Compare to A127864. - Greg Dresden and Yilin Zhu, Jul 17 2025

John Derbyshire, Prime Obsession, Joseph Henry Press, April 2004, p. 16.

Vincenzo Librandi, Table of n, a(n) for n = 0..500
A. Abdurrahman, CM Method and Expansion of Numbers, arXiv:1909.10889 [math.NT], 2019.
Jean-Luc Baril, Nathanaël Hassler, Sergey Kirgizov, and José L. Ramírez, Grand zigzag knight's paths, arXiv:2402.04851 [math.CO], 2024.
Paul Barry, On the Gap-sum and Gap-product Sequences of Integer Sequences, arXiv:2104.05593 [math.CO], 2021.
Paul Barry, Notes on Riordan arrays and lattice paths, arXiv:2504.09719 [math.CO], 2025. See pp. 8, 29.
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.
M. Couceiro, J. Devillet, and J.-L. Marichal, Quasitrivial semigroups: characterizations and enumerations, arXiv:1709.09162 [math.RA], 2017.
M. Diepenbroek, M. Maus, and A. Stoll, Pattern Avoidance in Reverse Double Lists, Preprint 2015. See Table 3.
Sergio Falcón, Binomial Transform of the Generalized k-Fibonacci Numbers, Communications in Mathematics and Applications (2019) Vol. 10, No. 3, 643-651.
Alice L. L. Gao and Sergey Kitaev, On partially ordered patterns of length 4 and 5 in permutations, arXiv:1903.08946 [math.CO], 2019.
Alice L. L. Gao and Sergey Kitaev, On partially ordered patterns of length 4 and 5 in permutations, The Electronic Journal of Combinatorics 26(3) (2019), P3.26.
Dale Gerdemann Bird Flock, Youtube video, 2011.
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=q=2.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 476
D. Jhala, G. P. S. Rathore, and K. Sisodiya, Some Properties of k-Jacobsthal Numbers with Arithmetic Indexes, Turkish Journal of Analysis and Number Theory, 2014, Vol. 2, No. 4, 119-124.
Tanya Khovanova, Recursive Sequences
Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38,5 (2000) 408-419; Eqs. (39), (41) and (45), lhs, m=2.
D. H. Lehmer, On Lucas's test for the primality of Mersenne's numbers, Journal of the London Mathematical Society 1.3 (1935): 162-165. See U_n.
Toufik Mansour and Mark Shattuck, Pattern avoidance in flattened derangements, Disc. Math. Lett. (2025) Vol. 15, 67-74. See p. 74.
Yassine Otmani, The 2-Pascal Triangle and a Related Riordan Array, J. Int. Seq. (2025) Vol. 28, Issue 3, Art. No. 25.3.5. See p. 12.
Alan Prince, Counting parses, Rutgers Optimality Archive, 2010.
Index entries for linear recurrences with constant coefficients, signature (2,2).
Index entries for sequences related to Chebyshev polynomials.
Index entries for Lucas sequences.

First differences are given by A026150.
a(n) = A073387(n, 0), n>=0 (first column of triangle).
Equals (1/3) A083337. First differences of A077846. Pairwise sums of A028860 and abs(A077917).
a(n) = A028860(n)/2 apart from the initial terms.
Row sums of A081577 and row sums of triangle A156710.
The following sequences (and others) belong to the same family: A001333, A000129, A026150, A046717, A015518, A084057, A063727, A002533, A002532, A083098, A083099, A083100, A015519.
Cf. A080953, A052948, A080040, A028859, A030195, A106435, A108898, A125145, A265106, A265107, A265278, A270810, A293005, A293006, A293007.
Cf. A175289 (Pisano periods).
Cf. A002530.
Cf. A127864.

a002605 n = a002605_list !! n
a002605_list =
   0 : 1 : map (* 2) (zipWith (+) a002605_list (tail a002605_list))
-- Reinhard Zumkeller, Oct 15 2011

[Floor(((1 + Sqrt(3))^n - (1 - Sqrt(3))^n)/(2*Sqrt(3))): n in [0..30]]; // Vincenzo Librandi, Aug 18 2011

[n le 2 select n-1 else 2*Self(n-1) + 2*Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 07 2018

a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=2*a[n-1]+2*a[n-2]od: seq(a[n], n=0..33); # Zerinvary Lajos, Dec 15 2008
a := n -> `if`(n<3, n, 2^(n-1)*hypergeom([1-n/2, (1-n)/2], [1-n], -2));
seq(simplify(a(n)), n=0..29); # Peter Luschny, Dec 16 2015

Expand[Table[((1 + Sqrt[3])^n - (1 - Sqrt[3])^n)/(2Sqrt[3]), {n, 0, 30}]] (* Artur Jasinski, Dec 10 2006 *)
a[n_]:=(MatrixPower[{{1,3},{1,1}},n].{{1},{1}})[[2,1]]; Table[a[n],{n,-1,40}] (* Vladimir Joseph Stephan Orlovsky, Feb 19 2010 *)
LinearRecurrence[{2, 2}, {0, 1}, 30] (* Robert G. Wilson v, Apr 13 2013 *)
Round@Table[Fibonacci[n, Sqrt[2]] 2^((n - 1)/2), {n, 0, 20}] (* Vladimir Reshetnikov, Oct 15 2016 *)
nxt[{a_,b_}]:={b,2(a+b)}; NestList[nxt,{0,1},30][[All,1]] (* Harvey P. Dale, Sep 17 2022 *)

Vec(x/(1-2*x-2*x^2)+O(x^99)) \\ Charles R Greathouse IV, Jun 10 2011

A002605(n)=([2,2;1,0]^n)[2,1] \\ M. F. Hasler, Aug 06 2018

[lucas_number1(n,2,-2) for n in range(0, 30)] # Zerinvary Lajos, Apr 22 2009

a = BinaryRecurrenceSequence(2,2)
print([a(n) for n in (0..29)])  # Peter Luschny, Aug 29 2016

a(n) = (-I*sqrt(2))^(n-1)*U(n-1, I/sqrt(2)) where U(n, x) is the Chebyshev U-polynomial. - Wolfdieter Lang
G.f.: x/(1 - 2*x - 2*x^2).
From Paul Barry, Sep 17 2003: (Start)
E.g.f.: x*exp(x)*(sinh(sqrt(3)*x)/sqrt(3) + cosh(sqrt(3)*x)).
a(n) = (1 + sqrt(3))^(n-1)*(1/2 + sqrt(3)/6) + (1 - sqrt(3))^(n-1)*(1/2 - sqrt(3)/6), for n>0.
Binomial transform of 1, 1, 3, 3, 9, 9, ... Binomial transform is A079935. (End)
a(n) = Sum_{k=0..floor(n/2)} binomial(n - k, k)*2^(n - k). - Paul Barry, Jul 13 2004
a(n) = A080040(n) - A028860(n+1). - Creighton Dement, Jan 19 2005
a(n) = Sum_{k=0..n} A112899(n,k). - Philippe Deléham, Nov 21 2007
a(n) = Sum_{k=0..n} A063967(n,k). - Philippe Deléham, Nov 03 2006
a(n) = ((1 + sqrt(3))^n - (1 - sqrt(3))^n)/(2*sqrt(3)).
a(n) = Sum_{k=0..n} binomial(n, 2*k + 1) * 3^k.
Binomial transform of expansion of sinh(sqrt(3)x)/sqrt(3) (0, 1, 0, 3, 0, 9, ...). E.g.f.: exp(x)*sinh(sqrt(3)*x)/sqrt(3). - Paul Barry, May 09 2003
a(n) = (1/3)*Sum_{k=1..5} sin(Pi*k/2)*sin(2*Pi*k/3)*(1 + 2*cos(Pi*k/6))^n, n >= 1. - Herbert Kociemba, Jun 02 2004
a(n+1) = ((3 + sqrt(3))*(1 + sqrt(3))^n + (3 - sqrt(3))*(1 - sqrt(3))^n)/6. - Al Hakanson (hawkuu(AT)gmail.com), Jun 29 2009
Antidiagonals sums of A081577. - J. M. Bergot, Dec 15 2012
G.f.: Q(0)*x/2, where Q(k) = 1 + 1/(1 - x*(4*k + 2 + 2*x)/(x*(4*k + 4 + 2*x) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 30 2013
a(n) = 2^(n - 1)*hypergeom([1 - n/2, (1 - n)/2], [1 - n], -2) for n >= 3. - Peter Luschny, Dec 16 2015
Sum_{k=0..n} a(k)*2^(n-k) = a(n+2)/2 - 2^n. - Greg Dresden, Feb 11 2022
a(n) = 2^floor(n/2) * A002530(n). - Gregory L. Simay, Sep 22 2022
From Peter Bala, May 08 2024: (Start)
G.f.: x/(1 - 2*x - 2*x^2) = Sum_{n >= 0} x^(n+1) *( Product_{k = 1..n} (k + 2*x + 1)/(1 + k*x) )
Also x/(1 - 2*x - 2*x^2) = Sum_{n >= 0} (2*x)^n *( x*Product_{k = 1..n} (m*k + 2 - m + x)/(1 + 2*m*k*x) ) for arbitrary m (both series are telescoping). (End)
a(n) = A127864(n-1) + A127864(n-2). - Greg Dresden and Yilin Zhu, Jul 17 2025

Edited by N. J. A. Sloane, Apr 15 2009

A015518 a(n) = 2a(n-1) + 3a(n-2), with a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 2, 7, 20, 61, 182, 547, 1640, 4921, 14762, 44287, 132860, 398581, 1195742, 3587227, 10761680, 32285041, 96855122, 290565367, 871696100, 2615088301, 7845264902, 23535794707, 70607384120, 211822152361, 635466457082

Offset: 0

Olivier Gérard

Number of walks of length n between any two distinct vertices of the complete graph K_4. - Paul Barry and Emeric Deutsch, Apr 01 2004
For n >= 1, a(n) is the number of integers k, 1 <= k <= 3^(n-1), whose ternary representation ends in an even number of zeros (see A007417). - Philippe Deléham, Mar 31 2004
Form the digraph with matrix A=[0,1,1,1;1,0,1,1;1,1,0,1;1,0,1,1]. A015518(n) corresponds to the (1,3) term of A^n. - Paul Barry, Oct 02 2004
The same sequence may be obtained by the following process. Starting a priori with the fraction 1/1, the denominators of fractions built according to the rule: add top and bottom to get the new bottom, add top and 4 times the bottom to get the new top. The limit of the sequence of fractions is 2. - Cino Hilliard, Sep 25 2005
(A046717(n))^2 + (2*a(n))^2 = A046717(2n). E.g., A046717(3) = 13, 2*a(3) = 14, A046717(6) = 365. 13^2 + 14^2 = 365. - Gary W. Adamson, Jun 17 2006
For n >= 2, number of ordered partitions of n-1 into parts of sizes 1 and 2 where there are two types of 1 (singletons) and three types of 2 (twins). For example, the number of possible configurations of families of n-1 male (M) and female (F) offspring considering only single births and twins, where the birth order of M/F/pair-of-twins is considered and there are three types of twins; namely, both F, both M, or one F and one M - where birth order within a pair of twins itself is disregarded. In particular, for a(3)=7, two children could be either: (1) F, then M; (2) M, then F; (3) F,F; (4) M,M; (5) F,F twins; (6) M,M twins; or (7) M,F twins (emphasizing that birth order is irrelevant here when both/all children are the same gender and when two children are within the same pair of twins). - Rick L. Shepherd, Sep 18 2004
a(n) is prime for n = {2, 3, 5, 7, 13, 23, 43, 281, 359, ...}, where only a(2) = 2 corresponds to a prime of the form (3^k - 1)/4. All prime terms, except a(2) = 2, are the primes of the form (3^k + 1)/4. Numbers k such that (3^k + 1)/4 is prime are listed in A007658. Note that all prime terms have prime indices. Prime terms are listed in A111010. - Alexander Adamchuk, Nov 19 2006
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=-2, A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=charpoly(A,1). - Milan Janjic, Jan 26 2010
Select an odd size subset S from {1,2,...,n}, then select an even size subset from S. - Geoffrey Critzer, Mar 02 2010
a(n) is the number of ternary sequences of length n where the numbers of (0's, 1's) are (even, odd) respectively, and, by symmetry, the number of such sequences where those numbers are (odd, even) respectively. A122983 covers (even, even), and A081251 covers (odd, odd). - Toby Gottfried, Apr 18 2010
An elephant sequence, see A175654. For the corner squares just one A[5] vector, with decimal value 341, leads to this sequence (without the leading 0). For the central square this vector leads to the companion sequence A046717 (without the first leading 1). - Johannes W. Meijer, Aug 15 2010
Let R be the commutative algebra resulting from adjoining the elements of the Klein four-group to the integers (equivalently, K = Z[x,y,z]/{x*y - z, y*z - x, x*z - y, x^2 - 1, y^2 - 1, z^2 - 1}). Then a(n) is equal to the coefficients of x, y, and z in the expansion of (x + y + z)^n. - Joseph E. Cooper III (easonrevant(AT)gmail.com), Nov 06 2010
Pisano period lengths:  1, 2, 2, 4, 4, 2, 6, 8, 2, 4, 10, 4, 6, 6, 4, 16, 16, 2, 18, 4, ... - R. J. Mathar, Aug 10 2012
The ratio a(n+1)/a(n) converges to 3 as n approaches infinity. - Felix P. Muga II, Mar 09 2014
This is a divisibility sequence, also the values of Chebyshev polynomials, and also the number of ways of packing a 2 X n-1 rectangle with dominoes and unit squares. - R. K. Guy, Dec 16 2016
For n>0, gcd(a(n),a(n+1))=1. - Kengbo Lu, Jul 02 2020

John Derbyshire, Prime Obsession, Joseph Henry Press, April 2004, see p. 16.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
A. Abdurrahman, CM Method and Expansion of Numbers, arXiv:1909.10889 [math.NT], 2019.
Jean-Paul Allouche, Jeffrey Shallit, Zhixiong Wen, Wen Wu, and Jiemeng Zhang, Sum-free sets generated by the period-k-folding sequences and some Sturmian sequences, arXiv:1911.01687 [math.CO], 2019.
Katerina Böhmová, Cristina Dalfó, and Clemens Huemer, On cyclic Kautz digraphs, Research Report, 2015.
Garry Bowlin and Matthew G. Brin, Coloring Planar Graphs via Colored Paths in the Associahedra, arXiv preprint arXiv:1301.3984 [math.CO], 2013. - From _N. J. A. Sloane_, Feb 12 2013
AJ Bu and Doron Zeilberger, Using Symbolic Computation to Explore Generalized Dyck Paths and Their Areas, arXiv:2305.09030 [math.CO], 2023.
Ji Young Choi, A Generalization of Collatz Functions and Jacobsthal Numbers, J. Int. Seq., Vol. 21 (2018), Article 18.5.4.
Sergio Falcón, Binomial Transform of the Generalized k-Fibonacci Numbers, Communications in Mathematics and Applications (2019) Vol. 10, No. 3, 643-651.
Dale Gerdemann, Fractal generated from (2,3) recursion, YouTube Video, Dec 05 2014.
Toufik Mansour and Mark Shattuck, Pattern avoidance in flattened derangements, Disc. Math. Lett. (2025) Vol. 15, 67-74. See p. 74.
R. J. Mathar, Counting Walks on Finite Graphs, Nov 2020, Section 2.
Felix Pozon Muga II, Extending the Golden Ratio and the Binet-de Moivre Formula, March 2014.
Index entries for linear recurrences with constant coefficients, signature (2,3).
Index entries for sequences related to Chebyshev polynomials.

a(n) = A080926(n-1) + 1 = (1/3)*A054878(n+1) = (1/3)*abs(A084567(n+1)).
First differences of A033113 and A039300.
Partial sums of A046717.
The following sequences (and others) belong to the same family: A000129, A001333, A002532, A002533, A002605, A015518, A015519, A026150, A046717, A063727, A083098, A083099, A083100, A084057.
Cf. A046717.
Cf. A007658, A111010.
Cf. A001045, A078008, A097073, A115341. - Vladimir Joseph Stephan Orlovsky, Dec 11 2008
Cf. A007482, A059260.

[Round(3^n/4): n in [0..30]]; // Vincenzo Librandi, Jun 24 2011

Table[(3^n-(-1)^n)/4,{n,0,30}] (* Alexander Adamchuk, Nov 19 2006 *)

a(n):= round(3^n/4)$ /* Dimitri Papadopoulos, Nov 28 2023 */

PARI
```
a(n)=round(3^n/4)
```

for n in range(0, 20): print(int((3**n-(-1)**n)/4), end=', ') # Stefano Spezia, Nov 30 2018

Sage
```
[round(3^n/4) for n in range(0,27)]
```

G.f.: x/((1+x)*(1-3*x)).
a(n) = (3^n - (-1)^n)/4 = floor(3^n/4 + 1/2).
a(n) = 3^(n-1) - a(n-1). - Emeric Deutsch, Apr 01 2004
E.g.f.: (exp(3*x) - exp(-x))/4. Second inverse binomial transform of (5^n-1)/4, A003463. Inverse binomial transform for powers of 4, A000302 (when preceded by 0). - Paul Barry, Mar 28 2003
a(n) = Sum_{k=0..floor(n/2)} C(n, 2k+1)*2^(2k). - Paul Barry, May 14 2003
a(n) = Sum_{k=1..n} binomial(n, k)*(-1)^(n+k)*4^(k-1). - Paul Barry, Apr 02 2003
a(n+1) = Sum_{k=0..floor(n/2)} binomial(n-k, k)*2^(n-2*k)*3^k. - Paul Barry, Jul 13 2004
a(n) = U(n-1, i/sqrt(3))(-i*sqrt(3))^(n-1), i^2=-1. - Paul Barry, Nov 17 2003
G.f.: x*(1+x)^2/(1 - 6*x^2 - 8*x^3 - 3*x^4) = x(1+x)^2/characteristic polynomial(x^4*adj(K_4)(1/x)). - Paul Barry, Feb 03 2004
a(n) = sum_{k=0..3^(n-1)} A014578(k) = -(-1)^n*A014983(n) = A051068(3^(n-1)), for n > 0. - Philippe Deléham, Mar 31 2004
E.g.f.: exp(x)*sinh(2*x)/2. - Paul Barry, Oct 02 2004
a(2*n+1) = A054880(n) + 1. - M. F. Hasler, Mar 20 2008
2*a(n) + (-1)^n = A046717(n). - M. F. Hasler, Mar 20 2008
a(n) = ((1+sqrt(4))^n - (1-sqrt(4))^n)/4. - Al Hakanson (hawkuu(AT)gmail.com), Dec 31 2008
a(n) = abs(A014983(n)). - Zerinvary Lajos, May 28 2009
a(n) = round(3^n/4). - Mircea Merca, Dec 28 2010
a(n) = Sum_{k=1,3,5,...} binomial(n,k)*2^(k-1). - Geoffrey Critzer, Mar 02 2010
From Sergei N. Gladkovskii, Jul 19 2012: (Start)
G.f.: G(0)/4 where G(k)= 1 - 1/(9^k - 3*x*81^k/(3*x*9^k - 1/(1 + 1/(3*9^k - 27*x*81^k/(9*x*9^k + 1/G(k+1)))))); (continued fraction).
E.g.f.: G(0)/4 where G(k)= 1 - 1/(9^k - 3*x*81^k/(3*x*9^k - (2*k+1)/(1 + 1/(3*9^k - 27*x*81^k/(9*x*9^k + (2*k+2)/G(k+1)))))); (continued fraction).  (End)
G.f.: G(0)*x/(2*(1-x)), where G(k) = 1 + 1/(1 - x*(4*k-1)/(x*(4*k+3) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013
a(n+1) = Sum_{k = 0..n} A238801(n,k)*2^k. - Philippe Deléham, Mar 07 2014
a(n) = (-1)^(n-1)*Sum_{k=0..n-1} A135278(n-1,k)*(-4)^k = (-1)^(n-1)*Sum_{k=0..n-1} (-3)^k. Equals (-1)^(n-1)*Phi(n,-3), where Phi is the cyclotomic polynomial when n is an odd prime. (For n > 0.) - Tom Copeland, Apr 14 2014
a(n) = 2*A006342(n-1) - n mod 2 if n > 0, a(0)=0. - Yuchun Ji, Nov 30 2018
a(n) = 2*A033113(n-2) + n mod 2 if n > 0, a(0)=0. - Yuchun Ji, Aug 16 2019
a(2*k) = 2*A002452(k), a(2*k+1) = A066443(k). - Yuchun Ji, Aug 14 2019
a(n+1) = 2*Sum_{k=0..n} a(k) if n odd, and 1 + 2*Sum_{k=0..n} a(k) if n even. - Kengbo Lu, May 30 2020
a(n) = F(n) + Sum_{k=1..(n-1)} a(k)*L(n-k), for F(n) and L(n) the Fibonacci and Lucas numbers. - Kengbo Lu and Greg Dresden, Jun 05 2020
From Kengbo Lu, Jun 11 2020: (Start)
a(n) = A002605(n) + Sum_{k = 1..n-2} a(k)*A002605(n-k-1).
a(n) = A006130(n-1) + Sum_{k = 1..n-1} a(k)*A006130(n-k-1). (End)
a(2n) = Sum_{i>=0, j>=0} binomial(n-j-1,i)*binomial(n-i-1,j)* 2^(2n-2i-2j-1)* 3^(i+j). - Kengbo Lu, Jul 02 2020
a(n) = 3*a(n-1) - (-1)^n. - Dimitri Papadopoulos, Nov 28 2023
G.f.: x/((1 + x)*(1 - 3*x)) = Sum_{n >= 0} x^(n+1) * Product_{k = 1..n} (k + 3*x + 1)(1 + k*x) (a telescoping series). Cf. A007482. - Peter Bala, May 08 2024
From Peter Bala, Jun 29 2025: (Start)
For n >= 1, a(n+1) = 2^n * hypergeom([1/2 - (1/2)*n, -(1/2)*n], [-n], -3).
G.f. A(x) = x*exp(Sum_{n >= 1} a(2*n)/a(n)*x^n/n) =  x + 2*x^2 + 7*x^3 + 20*x^4 + ....
sqrt(A(x)/x) is the g.f. of A002426.
The following series telescope:
Sum_{n >= 1} (-3)^n/(a(n)*a(n+1)) = -1; Sum_{n >= 1} (-3)^n/(a(n)*a(n+1)*a(n+2)*a(n+3)) = -1/98.
In general, for k >= 0, Sum_{n >= 1} (-3)^n/(a(n)*a(n+1)*...*a(n+2*k+1)) = -1/((a(1)*a(2)*...*a(2*k+1))*a(2*k+1)).
Sum_{n >= 1} 3^n/(a(n)*a(n+1)*a(n+2)) = 1/4; Sum_{n >= 1} 3^n/(a(n)*a(n+1)*a(n+2)* a(n+3)*a(n+4)) = 1/5600.
In general, for k >= 1, Sum_{n >= 1} 3^n/(a(n)*a(n+1)*...*a(n+2*k)) = 1/((a(1)*a(2)*...*a(2*k))*a(2*k)). (End)

More terms from Emeric Deutsch, Apr 01 2004

Edited by Ralf Stephan, Aug 30 2004

cp's OEIS Frontend

A083694 a(n) = 2*A002532(n).

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A111012 Primes in A002532.

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A000129 Pell numbers: a(0) = 0, a(1) = 1; for n > 1, a(n) = 2*a(n-1) + a(n-2).

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

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A002605 a(n) = 2*(a(n-1) + a(n-2)), a(0) = 0, a(1) = 1.

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A015518 a(n) = 2a(n-1) + 3a(n-2), with a(0)=0, a(1)=1.

A046717 a(n) = 2a(n-1) + 3a(n-2), a(0) = a(1) = 1.

A026150 a(0) = a(1) = 1; a(n+2) = 2a(n+1) + 2a(n).

A063727 a(n) = 2a(n-1) + 4a(n-2), a(0)=1, a(1)=2.