A001076 -id:A001076 - 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

A000351 Powers of 5: a(n) = 5^n.

Original entry on oeis.org

1, 5, 25, 125, 625, 3125, 15625, 78125, 390625, 1953125, 9765625, 48828125, 244140625, 1220703125, 6103515625, 30517578125, 152587890625, 762939453125, 3814697265625, 19073486328125, 95367431640625, 476837158203125, 2384185791015625, 11920928955078125

Offset: 0

N. J. A. Sloane

Same as Pisot sequences E(1, 5), L(1, 5), P(1, 5), T(1, 5). Essentially same as Pisot sequences E(5, 25), L(5, 25), P(5, 25), T(5, 25). See A008776 for definitions of Pisot sequences.
a(n) has leading digit 1 if and only if n = A067497 - 1. - Lekraj Beedassy, Jul 09 2002
With interpolated zeros 0, 1, 0, 5, 0, 25, ... (g.f.: x/(1 - 5*x^2)) second inverse binomial transform of Fibonacci(3n)/Fibonacci(3) (A001076). Binomial transform is A085449. - Paul Barry, Mar 14 2004
Sums of rows of the triangles in A013620 and A038220. - Reinhard Zumkeller, May 14 2006
Sum of coefficients of expansion of (1 + x + x^2 + x^3 + x^4)^n. a(n) is number of compositions of natural numbers into n parts less than 5. a(2) = 25 there are 25 compositions of natural numbers into 2 parts less than 5. - Adi Dani, Jun 22 2011
The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n >= 1, a(n) equals the number of 5-colored compositions of n such that no adjacent parts have the same color. - Milan Janjic, Nov 17 2011
Numbers n such that sigma(5n) = 5n + sigma(n). In fact we have this theorem: p is a prime if and only if all solutions of the equation sigma(p*x) = p*x + sigma(x) are powers of p. - Jahangeer Kholdi, Nov 23 2013
From Doug Bell, Jun 22 2015: (Start)
Empirical observation: Where n is an odd multiple of 3, let x = (a(n) + 1)/9 and let y be the decimal expansion of x/a(n); then y*(x+1)/x + 1 = y rotated to the left.
Example:
  a(3) = 125;
  x = (125 + 1)/9 = 14;
  y = 112, which is the decimal expansion of 14/125 = 0.112;
  112*(14 + 1)/14 + 1 = 121 = 112 rotated to the left.
(End)
a(n) is the number of n-digit integers that contain only odd digits (A014261). - Bernard Schott, Nov 12 2022
Number of pyramids in the Sierpinski fractal square-based pyramid at the n-th step, while A279511 gives the corresponding number of vertices (see IREM link with drawings). - Bernard Schott, Nov 29 2022

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n=0..100
O. M. Cain, The Exceptional Selfcondensability of Powers of Five, arXiv:1910.13829 [math.HO], 2019.
Peter J. Cameron, Sequences realized by oligomorphic permutation groups, J. Integ. Seqs. Vol. 3 (2000), #00.1.5.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 270
IREM Paris-Nord, La pyramide de Sierpinski (in French).
Tanya Khovanova, Recursive Sequences
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Yash Puri and Thomas Ward, Arithmetic and growth of periodic orbits, J. Integer Seqs., Vol. 4 (2001), #01.2.1.
Eric Weisstein's World of Mathematics, Box Fractal
Index entries for linear recurrences with constant coefficients, signature (5).

Cf. A009969 (even bisection), A013710 (odd bisection), A005054 (first differences), A003463 (partial sums).
Cf. A006495, A006496, A159991, A001021.
Cf. A001076, A008776, A013620, A038220, A067497, A085449, A014261.
Sierpinski fractal square-based pyramid: A020858 (Hausdorff dimension), A279511 (number of vertices), this sequence (number of pyramids).

a000351 = (5 ^)
a000351_list = iterate (* 5) 1  -- Reinhard Zumkeller, Oct 31 2012

[5^n : n in [0..30]]; // Wesley Ivan Hurt, Sep 27 2016

[ seq(5^n,n=0..30) ];
A000351:=-1/(-1+5*z); # Simon Plouffe in his 1992 dissertation

Table[5^n, {n, 0, 30}] (* Stefan Steinerberger, Apr 06 2006 *)
5^Range[0, 30] (* Harvey P. Dale, Aug 22 2011 *)

makelist(5^n,n,0,20); /* Martin Ettl, Dec 27 2012 */

a(n)=5^n \\ Charles R Greathouse IV, Jun 10 2011

def a(n): return 5**n
print([a(n) for n in range(24)]) # Michael S. Branicky, Nov 12 2022

(List.fill(50)(5: BigInt)).scanLeft(1: BigInt)( * ) // Alonso del Arte, May 31 2019

a(n) = 5^n.
a(0) = 1; a(n) = 5*a(n-1) for n > 0.
G.f.: 1/(1 - 5*x).
E.g.f.: exp(5*x).
a(n) = A006495(n)^2 + A006496(n)^2.
a(n) = A159991(n) / A001021(n). - Reinhard Zumkeller, May 02 2009
From Bernard Schott, Nov 12 2022: (Start)
Sum_{n>=0} 1/a(n) = 5/4.
Sum_{n>=0} (-1)^n/a(n) = 5/6. (End)
a(n) = Sum_{k=0..n} C(2*n+1,n-k)*A000045(2*k+1). - Vladimir Kruchinin, Jan 14 2025

A006190 a(n) = 3*a(n-1) + a(n-2), with a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 3, 10, 33, 109, 360, 1189, 3927, 12970, 42837, 141481, 467280, 1543321, 5097243, 16835050, 55602393, 183642229, 606529080, 2003229469, 6616217487, 21851881930, 72171863277, 238367471761, 787274278560, 2600190307441, 8587845200883, 28363725910090

Offset: 0

N. J. A. Sloane

Denominators of continued fraction convergents to (3+sqrt(13))/2. - Benoit Cloitre, Jun 14 2003
a(n) and A006497(n) occur in pairs: (a,b): (1,3), (3,11), (10,36), (33,119), (109,393), ... such that b^2 - 13a^2 = 4(-1)^n. - Gary W. Adamson, Jun 15 2003
Form the 4-node graph with matrix A = [1,1,1,1; 1,1,1,0; 1,1,0,1; 1,0,1,1]. Then this sequence counts the walks of length n from the vertex with degree 5 to one (any) of the other vertices. - Paul Barry, Oct 02 2004
a(n+1) is the binomial transform of A006138. - Paul Barry, May 21 2006
a(n+1) is the diagonal sum of the exponential Riordan array (exp(3x),x). - Paul Barry, Jun 03 2006
Number of paths in the right half-plane from (0,0) to the line x=n-1, consisting of steps U=(1,1), D=(1,-1), h=(1,0) and H=(2,0). Example: a(3)=10 because we have hh, H, UD, DU, hU, Uh, UU, hD, Dh and DD. - Emeric Deutsch, Sep 03 2007
Equals INVERT transform of A000129. Example: a(5) = 109 = (29, 12, 5, 2, 1) dot (1, 1, 3, 10, 33) = (29 + 12 + 15 + 20 + 33). - Gary W. Adamson, Aug 06 2010
For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 3's along the main diagonal, 1's along the superdiagonal and subdiagonal, and 0's everywhere else. - John M. Campbell, Jul 08 2011
These numbers could also be called "bronze Fibonacci numbers". Indeed, 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; analogously, for Pell numbers (A000129), or "silver Fibonacci numbers", P(n+1) = ceiling(delta*a(n)), if n is even and P(n+1) = floor(delta*a(n)), if n is odd, where delta = delta_S = 1 + sqrt(2) is the silver ratio. Here, for n >= 1, we have a(n+1) = ceiling(c*a(n)), if n is even and a(n+1) = floor(c*a(n)), if n is odd, where c = (3 + sqrt(13))/2 is the bronze ratio (cf. comment in A098316). - Vladimir Shevelev, Feb 23 2013
Let p(n,x) denote the Fibonacci polynomial, defined by p(1,x) = 1, p(2,x) = x, p(n,x) = x*p(n-1,x) + p(n-2,x). Let q(n,x) be the numerator polynomial of the rational function p(n, x + 1 + 1/x). Then q(n,1) = a(n). - Clark Kimberling, Nov 04 2013
The (1,1)-entry of the matrix A^n where A = [0,1,0; 1,2,1; 1,1,2]. - David Neil McGrath, Jul 18 2014
a(n+1) counts closed walks on K2, containing three loops on the other vertex. Equivalently the (1,1)-entry of A^(n+1) where the adjacency matrix of digraph is A = (0,1; 1,3). - David Neil McGrath, Oct 29 2014
For n >= 1, a(n) equals the number of words of length n-1 on alphabet {0,1,2,3} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
With offset 1 is the INVERTi transform of A001076. - Gary W. Adamson, Jul 24 2015
Apart from the initial 0, this is the p-INVERT transform of (1,0,1,0,1,0,...) for p(S) = 1 - 3 S. See A291219. - Clark Kimberling, Sep 02 2017
From Rogério Serôdio, Mar 30 2018: (Start)
This is a divisibility sequence (i.e., if n|m then a(n)|a(m)).
gcd(a(n),a(n+k)) = a(gcd(n, k)) for all positive integers n and k. (End)
Numbers of straight-chain fatty acids involving oxo and/or hydroxy groups, if cis-/trans isomerism and stereoisomerism are neglected. - Stefan Schuster, Apr 04 2018
Number of 3-compositions of n restricted to odd parts (and allowed zeros); see Hopkins & Ouvry reference. - Brian Hopkins, Aug 17 2020
From Michael A. Allen, Jan 25 2023: (Start)
Also called the 3-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence.
a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are 3 kinds of squares available. (End)
a(n) is the number of compositions of n when there are P(k) sorts of parts k, with k,n >= 1, P(k) = A000129(k) is the k-th Pell number (see example below). - Enrique Navarrete, Dec 15 2023

			From _Enrique Navarrete_, Dec 15 2023: (Start)
From the comment on compositions with Pell number of parts, A000129(k), there are A000129(1)=1 type of 1, A000129(2)=2 types of 2, A000129(3)=5 types of 3, A000129(4)=12 types of 4, A000129(5)=29 types of 5 and A000129(6)=70 types of 6.
The following table gives the number of compositions of n=6:
Composition, number of such compositions, number of compositions of this type:
  6,              1,     70;
  5+1,            2,     58;
  4+2,            2,     48;
  3+3,            1,     25;
  4+1+1,          3,     36;
  3+2+1,          6,     60;
  2+2+2,          1,      8;
  3+1+1+1,        4,     20;
  2+2+1+1,        6,     24;
  2+1+1+1+1,      5,     10;
  1+1+1+1+1+1,    1,      1;
for a total of a(6)=360 compositions of n=6. (End).

H. L. Abbott and D. Hanson, A lattice path problem, Ars Combin., 6 (1978), 163-178.
A. Brousseau, Fibonacci and Related Number Theoretic Tables. Fibonacci Association, San Jose, CA, 1972, p. 128.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
L.-N. Machaut, Query 3436, L'Intermédiaire des Mathématiciens, 16 (1909), 62-63. - N. J. A. Sloane, Mar 08 2022

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..200 from T. D. Noe)
H. L. Abbott and D. Hanson, A lattice path problem, Ars Combin., 6 (1978), 163-178. (Annotated scanned copy)
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
Dorin Andrica, Ovidiu Bagdasar, and George Cătălin Tųrcąs, On some new results for the generalised Lucas sequences, An. Şt. Univ. Ovidius Constanţa (Romania, 2021) Vol. 29, No. 1, 17-36.
Ayoub B. Ayoub, Fibonacci-like sequences and Pell equations, The College Mathematics Journal, Vol. 38 (2007), pp. 49-53.
Paul Barry, Notes on Riordan arrays and lattice paths, arXiv:2504.09719 [math.CO], 2025. See pp. 8, 29.
Daniel Birmajer, Juan B. Gil, and Michael D. Weiner, On the Enumeration of Restricted Words over a Finite Alphabet, J. Int. Seq. 19 (2016) # 16.1.3, Example 8.
Henrique F. da Cruz, Ilda Inácio, and Rogério Serôdio, Convertible subspaces that arise from different numberings of the vertices of a graph, Ars Mathematica Contemporanea (2019) Vol. 16, No. 2, 473-486.
Colin Defant, Meeting Covered Elements in nu-Tamari Lattices, arXiv:2104.03890 [math.CO], 2021.
Sergio Falcon, On The Generating Functions of the Powers of the K-Fibonacci Numbers, Scholars Journal of Engineering and Technology (SJET), 2014; 2 (4C):669-675.
Sergio Falcon, The k-Fibonacci difference sequences, Chaos, Solitons & Fractals, Volume 87, June 2016, Pages 153-157.
M. C. Firengiz and A. Dil, Generalized Euler-Seidel method for second order recurrence relations, Notes on Number Theory and Discrete Mathematics, Vol. 20, 2014, No. 4, 21-32.
Juan B. Gil and Aaron Worley, Generalized metallic means, arXiv:1901.02619 [math.NT], 2019.
Brian Hopkins and Stéphane Ouvry, Combinatorics of Multicompositions, arXiv:2008.04937 [math.CO], 2020.
A. F. Horadam, Generating identities for generalized Fibonacci and Lucas triples, Fib. Quart., 15 (1977), 289-292.
Haruo Hosoya, What Can Mathematical Chemistry Contribute to the Development of Mathematics?, HYLE--International Journal for Philosophy of Chemistry, Vol. 19, No.1 (2013), pp. 87-105.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 158
Milan Janjić, Hessenberg Matrices and Integer Sequences, J. Int. Seq. 13 (2010) # 10.7.8, section 3.
Milan Janjić, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Tanya Khovanova, Recursive Sequences
W. F. Klostermeyer, M. E. Mays, L. Soltes and G. Trapp, A Pascal rhombus, Fibonacci Quarterly, 35 (1976), 318-328.
Pablo Lam-Estrada, Myriam Rosalía Maldonado-Ramírez, José Luis López-Bonilla, and Fausto Jarquín-Zárate, The sequences of Fibonacci and Lucas for each real quadratic fields Q(Sqrt(d)), arXiv:1904.13002 [math.NT], 2019.
Prabha Sivaraman Nair and Rejikumar Karunakaran, On k-Fibonacci Brousseau Sums, J. Int. Seq. (2024) Art. No. 24.6.4. See p. 2.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
S. Schuster, M. Fichtner and S. Sasso, Use of Fibonacci numbers in lipidomics - Enumerating various classes of fatty acids, Sci. Rep., 7 (2017) 39821.
Kai Wang, On k-Fibonacci Sequences And Infinite Series List of Results and Examples, 2020.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (3,1).

Row sums of Pascal's rhombus (A059317). Also row sums of triangle A054456(n, m).
Sequences with g.f. 1/(1-k*x-x^2) or x/(1-k*x-x^2): A000045 (k=1), A000129 (k=2), this sequence (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).
Cf. A006497, A052906, A175182 (Pisano periods), A201001 (prime subsequence), A092936 (squares).
Cf. A243399.

a:=[0,1];; for n in [3..30] do a[n]:=3*a[n-1]+a[n-2]; od; a; # Muniru A Asiru, Mar 31 2018

a006190 n = a006190_list !! n
a006190_list = 0 : 1 : zipWith (+) (map (* 3) $ tail a006190_list) a006190_list
-- Reinhard Zumkeller, Feb 19 2011

[ n eq 1 select 0 else n eq 2 select 1 else 3*Self(n-1)+Self(n-2): n in [1..30] ]; // Vincenzo Librandi, Aug 19 2011

a[0]:=0: a[1]:=1: for n from 2 to 35 do a[n]:= 3*a[n-1]+a[n-2] end do: seq(a[n],n=0..30); # Emeric Deutsch, Sep 03 2007
A006190:=-1/(-1+3*z+z**2); # Simon Plouffe in his 1992 dissertation, without the leading 0
seq(combinat[fibonacci](n,3),n=0..30); # R. J. Mathar, Dec 07 2011

a[n_] := (MatrixPower[{{1, 3}, {1, 2}}, n].{{1}, {1}})[[2, 1]]; Table[ a[n], {n, -1, 24}] (* Robert G. Wilson v, Jan 13 2005 *)
LinearRecurrence[{3,1},{0,1},30] (* or *) CoefficientList[Series[x/ (1-3x-x^2), {x,0,30}], x] (* Harvey P. Dale, Apr 20 2011 *)
Table[If[n==0, a1=1; a0=0, a2=a1; a1=a0; a0=3*a1+a2], {n, 0, 30}] (* Jean-François Alcover, Apr 30 2013 *)
Table[Fibonacci[n, 3], {n, 0, 30}] (* Vladimir Reshetnikov, May 08 2016 *)

a(n)=if(n<1,0,contfracpnqn(vector(n,i,2+(i>1)))[2,1])

a(n)=([1,3;1,2]^n)[2,1] \\ Charles R Greathouse IV, Mar 06 2014

concat([0],Vec(x/(1-3*x-x^2)+O(x^30))) \\ Joerg Arndt, Apr 30 2013

[lucas_number1(n,3,-1) for n in range(0, 30)] # Zerinvary Lajos, Apr 22 2009

G.f.: x/(1 - 3*x - x^2).
From Benoit Cloitre, Jun 14 2003: (Start)
a(3*n) = 2*A041019(5*n-1), a(3*n+1) = A041019(5*n), a(3*n+2) = A041019(5*n+3).
a(2*n) = 3*A004190(n-1); a(3*n) = 10*A041613(n-1) for n >= 1. (End)
From Gary W. Adamson, Jun 15 2003: (Start)
a(n-1) + a(n+1) = A006497(n).
A006497(n)^2 - 13*a(n)^2 = 4(-1)^n. (End)
a(n) = U(n-1, (3/2)i)(-i)^(n-1), i^2 = -1. - Paul Barry, Nov 19 2003
a(n) = Sum_{k=0..n-1} binomial(n-k-1,k)*3^(n-2*k-1). - Paul Barry, Oct 02 2004
a(n) = F(n, 3), the n-th Fibonacci polynomial evaluated at x=3.
Let M = {{0, 1}, {1, 3}}, v[1] = {0, 1}, v[n] = M.v[n - 1]; then a(n) = Abs[v[n][[1]]]. - Roger L. Bagula, May 29 2005 [Or a(n) = [M^(n+1)]{1,1}. - _L. Edson Jeffery, Aug 27 2013]
From Paul Barry, May 21 2006: (Start)
a(n+1) = Sum_{k=0..n} Sum_{j=0..n-k} C(k,j)*C(n-j,k)*2^(k-j).
a(n) = Sum_{k=0..n} Sum_{j=0..n-k} C(k,j)*C(n-j,k)*2^(n-j-k).
a(n+1) = Sum_{k=0..floor(n/2)} C(n-k,k)*3^(n-2*k).
a(n) = Sum_{k=0..n} C(k,n-k)*3^(2*k-n). (End)
E.g.f.: exp(3*x/2)*sinh(sqrt(13)*x/2)/(sqrt(13)/2). - Paul Barry, Jun 03 2006
a(n) = (ap^n - am^n)/(ap - am), with ap = (3 + sqrt(13))/2, am = (3 - sqrt(13))/2.
Let C = (3 + sqrt(13))/2 = exp arcsinh(3/2) = 3.3027756377... Then C^n, n > 0 = a(n)*(1/C) + a(n+1). Let X = the 2 X 2 matrix [0, 1; 1, 3]. Then X^n = [a(n-1), a(n); a(n), a(n+1)]. - Gary W. Adamson, Dec 21 2007
1/3 = 3/(1*10) + 3/(3*33) + 3/(10*109) + 3/(33*360) + 3/(109*1189) + ... . - Gary W. Adamson, Mar 16 2008
a(n) = ((3 + sqrt(13))^n - (3 - sqrt(13))^n)/(2^n*sqrt(13)). - Al Hakanson (hawkuu(AT)gmail.com), Jan 12 2009
a(p) == 13^((p-1)/2) mod p, for odd primes p. - Gary W. Adamson, Feb 22 2009
From Johannes W. Meijer, Jun 12 2010: (Start)
Limit_{k->oo} a(n+k)/a(k) = (A006497(n) + a(n)*sqrt(13))/2.
Limit_{n->oo} A006497(n)/a(n) = sqrt(13). (End)
Sum_{k>=1} (-1)^(k-1)/(a(k)*a(k+1)) = (sqrt(13)-3)/2. - Vladimir Shevelev, Feb 23 2013
From Vladimir Shevelev, Feb 24 2013: (Start)
(1) Expression a(n+1) via a(n): a(n+1) = (3*a(n) + sqrt(13*a(n)^2 + 4*(-1)^n))/2;
(2) a^2(n+1) - 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(13)-3)/2 + r(n), where |r(n)| < 1/(a(n+1)*a(n+2)). (End)
a(n) = sqrt(13*(A006497(n))^2 + (-1)^(n-1)*52)/13. - Vladimir Shevelev, Mar 13 2013
Sum_{n >= 1} 1/( a(2*n) + 1/a(2*n) ) = 1/3; Sum_{n >= 1} 1/( a(2*n + 1) - 1/a(2*n + 1) ) = 1/9. - Peter Bala, Mar 26 2015
From Rogério Serôdio, Mar 30 2018: (Start)
Some properties:
(1) a(n)*a(n+1) = 3*Sum_{k=1..n} a(k)^2;
(2) a(n)^2 + a(n+1)^2 = a(2*n+1);
(3) a(n)^2 - a(n-2)^2 = 3*a(n-1)*(a(n) + a(n-2));
(4) a(m*(p+1)) = a(m*p)*a(m+1) + a(m*p-1)*a(m);
(5) a(n-k)*a(n+k) = a(n)^2 + (-1)^(n+k+1)*a(k)^2;
(6) a(2*n) = a(n)*(3*a(n) + 2*a(n-1));
(7) 3*Sum_{k=2..n+1} a(k)*a(k-1) is equal to a(n+1)^2 if n odd, and is equal to a(n+1)^2 - 1 if n is even;
(8) a(n) = alpha(k)*a(n-2*k+1) + a(n-4*k+2), where alpha(k) = ap^(2*k-1) + am^(2*k-1), with ap = (3 + sqrt(13))/2, am = (3 - sqrt(13))/2;
(9) 131|Sum_{k=n..n+9} a(k), for all positive n. (End)
From Kai Wang, Feb 10 2020: (Start)
a(n)^2 - a(n+r)*a(n-r) = (-1)^(n-r)*a(r)^2 - Catalan's identity.
arctan(1/a(2n)) - arctan(1/a(2n+2)) = arctan(a(2)/a(2n+1)).
arctan(1/a(2n)) = Sum_{m>=n} arctan(a(2)/a(2m+1)).
The same formula holds for Fibonacci numbers and Pell numbers. (End)
a(n+2) = 3^(n+1) + Sum_{k=0..n} a(k)*3^(n-k). - Greg Dresden and Gavron Campbell, Feb 22 2022
G.f. = 1/(1-Sum_{k>=1} P(k)*x^k), P(k)=A000129(k) (with a(0)=1). - Enrique Navarrete, Dec 17 2023
G.f.: x/(1 - 3*x - x^2) = Sum_{n >= 0} x^(n+1) *( Product_{k = 1..n} (m*k + 3 - m + x)/(1 + m*k*x) ) for arbitrary m (a telescoping series). - Peter Bala, May 08 2024
Sum_{n>=0} a(n)/phi^(3*n) = 1, where phi = A001622 is the golden ratio. - Diego Rattaggi, Apr 07 2025

Second formula corrected by Johannes W. Meijer, Jun 02 2010

A015448 a(0) = 1, a(1) = 1, and a(n) = 4*a(n-1) + a(n-2) for n >= 2.

Original entry on oeis.org

1, 1, 5, 21, 89, 377, 1597, 6765, 28657, 121393, 514229, 2178309, 9227465, 39088169, 165580141, 701408733, 2971215073, 12586269025, 53316291173, 225851433717, 956722026041, 4052739537881, 17167680177565, 72723460248141, 308061521170129, 1304969544928657, 5527939700884757

Offset: 0

Olivier Gérard

If one deletes the leading 0 in A084326, takes the inverse binomial transform, and adds a(0)=1 in front, one obtains this sequence here. - Al Hakanson (hawkuu(AT)gmail.com), May 02 2009
For n >= 1, row sums of triangle
  m |k=0     1     2     3     4     5     6     7
====+=============================================
  0 |  1
  1 |  1     4
  2 |  1     4    16
  3 |  1     8    16    64
  4 |  1     8    48    64   256
  5 |  1    12    48   256   256  1024
  6 |  1    12    96   256  1280  1024  4096
  7 |  1    16    96   640  1280  6144  4096 16384
which is triangle for numbers 4^k*C(m,k) with duplicated diagonals. - Vladimir Shevelev, Apr 12 2012
a(n) = a(n;-2) = 3^n*Sum_{k=0..n} binomial(n,k)*F(k+1)*(-2/3)^k, where a(n;d), n=0,1,...,d, denotes the delta-Fibonacci numbers defined in comments to A000045 (see also the papers of Witula et al.). We note that (see A033887) F(3n+1) = 3^n*a(n,2/3) = Sum_{k=0..n} binomial(n,k)*F(k-1)*(-2/3)^k, which implies F(3n+1) + 3^(-n)*a(n) = Sum_{k=0..n} binomial(n,k)*L(k)*(-2/3)^k, where L(k) denotes the k-th Lucas number. - Roman Witula, Jul 12 2012
a(n+1) is (for n >= 0) the number of length-n strings of 5 letters {0,1,2,3,4} with no two adjacent nonzero letters identical. The general case (strings of L letters) is the sequence with g.f. (1+x)/(1-(L-1)*x-x^2). - Joerg Arndt, Oct 11 2012
Starting with offset 1 the sequence is the INVERT transform of (1, 4, 4*3, 4*3^2, 4*3^3, ...); i.e., of A003946: (1, 4, 12, 36, 108, ...). - Gary W. Adamson, Aug 06 2016
a(n+1) equals the number of quinary sequences of length n such that no two consecutive terms differ by 3. - David Nacin, May 31 2017

T. D. Noe, Table of n, a(n) for n = 0..200
Joerg Arndt, Matters Computational (The Fxtbook), pp. 313-315.
Mohammad K. Azarian, Fibonacci Identities as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 38, 2012, pp. 1871-1876 (see Corollary 1 (vi)).
Gary Detlefs and Wolfdieter Lang, Improved Formula for the Multi-Section of the Linear Three-Term Recurrence Sequence, arXiv:2304.12937 [math.CO], 2023.
Amelia Gilson, Hadley Killen, Steven J. Miller, Nadia Razek, Joshua M. Siktar, and Liza Sulkin, Zeckendorf's Theorem Using Indices in an Arithmetic Progression, arXiv:2005.10396 [math.NT], 2020.
Edyta Hetmaniok, Bozena Piatek, and Roman Wituła, Binomials Transformation Formulae of Scaled Fibonacci Numbers, Open Math. 15 (2017), 477-485.
I. M. Gessel and Ji Li, Compositions and Fibonacci identities, J. Int. Seq. 16 (2013) 13.4.5.
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Tanya Khovanova, Recursive Sequences
Bahar Kuloğlu, Engin Özkan, and Marin Marin, Fibonacci and Lucas Polynomials in n-gon, An. Şt. Univ. Ovidius Constanţa (Romania 2023) Vol. 31, No 2, 127-140.
Roman Witula, Binomials transformation formulae of scaled Lucas numbers, Demonstratio Math. 46 (2013), 15-27.
R. Witula and Damian Slota, delta-Fibonacci numbers, Appl. Anal. Discr. Math 3 (2009) 310-329, MR2555042
Index entries for linear recurrences with constant coefficients, signature (4,1).

Cf. A001076, A147722 (INVERT transform), A109499 (INVERTi transform), A154626 (Binomial transform), A086344 (inverse binomial transform), A003946, A049310.

Cf. A000032, A000045, A014445, A014448, A033887, A046854, A055830, A055870, A084326, A167808.

[Fibonacci(3*n-1): n in [0..40]]; // Vincenzo Librandi, Jul 04 2015

with(combinat): a:=n->fibonacci(n,4)-3*fibonacci(n-1,4): seq(a(n), n=1..23); # Zerinvary Lajos, Apr 04 2008

Fibonacci/@(3*Range[0,30]-1) (* Vladimir Joseph Stephan Orlovsky, Mar 01 2010 *)
LinearRecurrence[{4,1},{1,1},30] (* Harvey P. Dale, May 15 2019 *)

a[0]:1$
a[1]:1$
a[n]:=4*a[n-1]+a[n-2]$
A015448(n):=a[n]$
makelist(A015448(n),n,0,30); /* Martin Ettl, Nov 03 2012 */

a(n) = fibonacci(3*n-1); \\ Altug Alkan, Dec 10 2015

a(n) = Fibonacci(3n-1) = ( (1+sqrt(5))*(2-sqrt(5))^n - (1-sqrt(5))*(2+sqrt(5))^n )/ (2*sqrt(5)).
O.g.f.: (1-3*x)/(1-4*x-x^2). - Len Smiley, Dec 09 2001
a(n) = Sum_{k=0..n} 3^k*A055830(n,k). - Philippe Deléham, Oct 18 2006
a(n) = upper left term in the 2 X 2 matrix [1,2; 2,3]^n. - Gary W. Adamson, Mar 02 2008
[a(n), A001076(n)] = [1,4; 1,3]^n * [1,0]. - Gary W. Adamson, Mar 21 2008
a(n) = A167808(3*n-1) for n > 0. - Reinhard Zumkeller, Nov 12 2009
a(n) = Fibonacci(3n+1) mod Fibonacci(3n), n > 0.
a(n) = (A000032(3*n)-Fibonacci(3*n))/2 = (A014448(n)-A014445(n))/2.
For n >= 2, a(n) = F_n(4) + F_(n+1)(4), where F_n(x) is a 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) = A001076(n+1) - 3*A001076(n). - R. J. Mathar, Jul 12 2012
From Gary Detlefs and Wolfdieter Lang, Aug 20 2012: (Start)
a(n) = (5*F(n)^3 + 5*F(n-1)^3 + 3*(-1)^n*F(n-2))/2,
a(n) = (F(n+1)^3 + 2*F(n)^3 - F(n-2)^3)/2, n >= 0, with F(-1) = 1 and F(-2) = -1. Second line from first one with 3*(-1)^n* F(n-2) = F(n-1)^3 - 4*F(n-2)^3 - F(n-3)^3 (in Koshy's book, p. 89, 32. (with a - sign) and 33. For the Koshy reference see A000045) and the F^3 recurrence (see row n=4 of A055870, or Koshy p. 87, 1.). First line from the preceding R. J. Mathar formula with F(3*n) = 5*F(n)^3 + 3*(-1)^n*F(n) (Koshy p. 89, 46.) and the above mentioned formula, Koshy's 32. and 33., with n -> n+2 in order to eliminate F(n+1)^3. (End)
For n > 0, a(n) = L(n-1)*L(n)*F(n) + F(n+1)*(-1)^n with L(n)=A000032(n). - J. M. Bergot, Dec 10 2015
For n > 1, a(n)^2 is the denominator of continued fraction [4,4,...,4, 6, 4,4,...4], which has n-1 4's before, and n-1 4's after, the middle 6. - Greg Dresden, Sep 18 2019
From Gary Detlefs and Wolfdieter Lang, Mar 06 2023: (Start)
a(n) = A001076(n) + A001076(n-1), with A001076(-1) = 1. See the R. J. Mathar formula above.
a(n+1) = i^n*(S(n-1,-4*i) - i*S(n-2,-4*i)), for n >= 0, with i = sqrt(-1), and the Chebyshev S-polynomials (see A049310) with S(n, -1) = 0. From the simplified Fibonacci trisection formula for {F(3*n+2)}_{n>=0}. (End)
a(n) = Sum_{k=0..n} A046854(n-1,k)*4^k. - R. J. Mathar, Feb 10 2024
E.g.f.: exp(2*x)*(5*cosh(sqrt(5)*x) - sqrt(5)*sinh(sqrt(5)*x))/5. - Stefano Spezia, Jun 03 2024

A027941 a(n) = Fibonacci(2*n + 1) - 1.

Original entry on oeis.org

0, 1, 4, 12, 33, 88, 232, 609, 1596, 4180, 10945, 28656, 75024, 196417, 514228, 1346268, 3524577, 9227464, 24157816, 63245985, 165580140, 433494436, 1134903169, 2971215072, 7778742048, 20365011073, 53316291172, 139583862444, 365435296161, 956722026040

Offset: 0

Clark Kimberling

Also T(2n+1,n+1), T given by A027935. Also first row of Inverse Stolarsky array.
Third diagonal of array defined by T(i, 1)=T(1, j)=1, T(i, j)=Max(T(i-1, j)+T(i-1, j-1); T(i-1, j-1)+T(i, j-1)). - Benoit Cloitre, Aug 05 2003
Number of Schroeder paths of length 2(n+1) having exactly one up step starting at an even height (a Schroeder path is a lattice path starting from (0,0), ending at a point on the x-axis, consisting only of steps U=(1,1) (up steps), D=(1,-1) (down steps) and H=(2,0) (level steps) and never going below the x-axis). Schroeder paths are counted by the large Schroeder numbers (A006318). Example: a(1)=4 because among the six Schroeder paths of length 4 only the paths (U)HD, (U)UDD, H(U)D, (U)DH have exactly one U step that starts at an even height (shown between parentheses). - Emeric Deutsch, Dec 19 2004
Also: smallest number not writeable as the sum of fewer than n positive Fibonacci numbers. E.g., a(5)=88 because it is the smallest number that needs at least 5 Fibonacci numbers: 88 = 55 + 21 + 8 + 3 + 1. - Johan Claes, Apr 19 2005 [corrected for offset and clarification by Mike Speciner, Sep 19 2023] In general, a(n) is the sum of n positive Fibonacci numbers as a(n) = Sum_{i=1..n} A000045(2*i). See A001076 when negative Fibonacci numbers can be included in the sum. - Mike Speciner, Sep 24 2023
Except for first term, numbers a(n) that set a new record in the number of Fibonacci numbers needed to sum up to n. Position of records in sequence A007895. - Ralf Stephan, May 15 2005
Successive extremal petal bends beta(n) = a(n-2). See the Ring Lemma of Rodin and Sullivan in K. Stephenson, Introduction to Circle Packing (Cambridge U. P., 2005), pp. 73-74 and 318-321. - David W. Cantrell (DWCantrell(AT)sigmaxi.net)
a(n+1)= AAB^(n)(1), n>=1, with compositions of Wythoff's complementary A(n):=A000201(n) and B(n)=A001950(n) sequences. See the W. Lang link under A135817 for the Wythoff representation of numbers (with A as 1 and B as 0 and the argument 1 omitted). E.g., 4=`110`, 12=`1100`, 33=`11000`, 88=`110000`, ..., in Wythoff code. AA(1)=1=a(1) but for uniqueness reason 1=A(1) in Wythoff code. - N. J. A. Sloane, Jun 29 2008
Start with n. Each n generates a sublist {n-1,n-1,n-2,..,1}. Each element of each sublist also generates a sublist. Add numbers in all terms. For example, 3->{2,2,1} and both 2->{1,1}, so a(3) = 3 + 2 + 2 + 1 + 1 + 1 + 1 + 1 = 12. - Jon Perry, Sep 01 2012
For n>0: smallest number such that the inner product of Zeckendorf binary representation and its reverse equals n: A216176(a(n)) = n, see also A189920. - Reinhard Zumkeller, Mar 10 2013
Also, numbers m such that 5*m*(m+2)+1 is a square. - Bruno Berselli, May 19 2014
Also, number of nonempty submultisets of multisets of weight n that span an initial interval of integers (see 2nd example). - Gus Wiseman, Feb 10 2015
From Robert K. Moniot, Oct 04 2020: (Start)
Including a(-1):=0, consecutive terms (a(n-1),a(n))=(u,v) or (v,u) give all points on the hyperbola u^2-u+v^2-v-4*u*v=0 with both coordinates nonnegative integers.  Note that this follows from identifying (1,u+1,v+1) with the Markov triple (1,Fibonacci(2n-1),Fibonacci(2n+1)).  See A001519 (comments by Robert G. Wilson, Oct 05 2005, and Wolfdieter Lang, Jan 30 2015).
Let T(n) denote the n-th triangular number. If i, j are any two successive elements of the above sequence then (T(i-1) + T(j-1))/T(i+j-1) = 3/5. (End)

			a(5) = 88 = 2*33 + 12 + 4 + 1 + 5. a(6) = 232 = 2*88 + 33 + 12 + 4 + 1 + 6. - _Jon Perry_, Sep 01 2012
a(4) = 33 counts all nonempty submultisets of the last row: [1][2][3][4], [11][12][13][14][22][23][24][33][34], [111][112][113][122][123][124][133][134][222][223][233][234], [1111][1112][1122][1123][1222][1223][1233][1234]. - _Gus Wiseman_, Feb 10 2015

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 12.

T. D. Noe, Table of n, a(n) for n = 0..200
R. C. Alperin, A nonlinear recurrence and its relations to Chebyshev polynomials, Fib. Q., Vol. 58, No. 2 (2020), 140-142.
Russ Euler and Jawad Sadek, Problem B-912, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 39, No. 1 (2001), p. 85; From a Product to a Sum, Solution to Problem B-912 by Charles K. Cook, ibid., Vol. 39, No. 5 (2001), pp. 468-469.
Sela Fried, Even-up words and their variants, arXiv:2505.14196 [math.CO], 2025. See p. 7.
Clark Kimberling, Interspersions.
Clark Kimberling, Interspersions and dispersions, Proceedings of the American Mathematical Society, 117 (1993) 313-321.
Ioana-Claudia Lazăr, Lucas sequences in t-uniform simplicial complexes, arXiv:1904.06555 [math.GR], 2019.
R. J. Mathar, Paving rectangular regions with rectangular tiles: tatami and non-tatami tilings, arXiv:1311.6135 [math.CO], 2013, Table 60 (doubled).
Luis A. Medina and Armin Straub, On multiple and infinite log-concavity, Annals of Combinatorics, Vol. 20, No. 1 (2016), pp. 125-138; arXiv preprint, arXiv:1405.1765 [math.CO], 2014; preprint, 2014.
László Németh, Hyperbolic Pascal pyramid, arXiv:1511.02067 [math.CO], 2015 (2nd line of Table 1 is 3*a(n-2)).
László Németh, Pascal pyramid in the space H^2×R, arXiv:1701.06022 [math.CO], 2016. See bn in Table 1 p.10.
N. J. A. Sloane, Classic Sequences.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (4,-4,1).

Cf. A000045, A035507, A001906, A006318, A000032.
Related to partial sums of Fibonacci(k*n) over n: A000071, A099919, A058038, A138134, A053606; this sequence is the case k=2.
Cf. A212336 for more sequences with g.f. of the type 1/(1 - k*x + k*x^2 - x^3).
Cf. A000225 (sublist connection).
Cf. A258993 (row sums, n > 0), A000967.

a027941 = (subtract 1) . a000045 . (+ 1) . (* 2)
-- Reinhard Zumkeller, Mar 10 2013

[Fibonacci(2*n+1)-1: n in [0..30]]; // Vincenzo Librandi, Apr 18 2011

with(combinat): seq(fibonacci(2*n+1)-1,n=1..27); # Emeric Deutsch, Dec 19 2004
a:=n->sum(binomial(n+k+1,2*k), k=0..n): seq(a(n), n=-1..26); # Zerinvary Lajos, Oct 02 2007

Table[Fibonacci[2*n+1]-1,{n,0,17}] (* Vladimir Joseph Stephan Orlovsky, Jul 21 2008 *)
LinearRecurrence[{4,-4,1},{0,1,4},40] (* Harvey P. Dale, Aug 17 2021 *)

a(n):=sum(binomial(n+1,k+1)*fib(k),k,0,n); /* Vladimir Kruchinin, Oct 14 2016 */

a(n)=fibonacci(2*n+1)-1 \\ Charles R Greathouse IV, Nov 20 2012

concat(0, Vec(x/((1-x)*(1-3*x+x^2)) + O(x^40))) \\ Colin Barker, Jun 03 2016

a(n) = Sum_{i=1..n} binomial(n+i, n-i). - Benoit Cloitre, Oct 15 2002
G.f.: Sum_{k>=1} x^k/(1-x)^(2*k+1). - Benoit Cloitre, Apr 21 2003
a(n) = Sum_{k=1..n} F(2*k), i.e., partial sums of A001906. - Benoit Cloitre, Oct 27 2003
a(n) = Sum_{k=0..n-1} U(k, 3/2) = Sum_{k=0..n-1} S(k, 3), with S(k, 3) = A001906(k+1). - Paul Barry, Nov 14 2003
G.f.: x/((1-x)*(1-3*x+x^2)) = x/(1-4*x+4*x^2-x^3).
a(n) = 4*a(n-1) - 4*a(n-2) + a(n-3) with n>=2, a(-1)=0, a(0)=0, a(1)=1.
a(n) = 3*a(n-1) - a(n-2) + 1 with n>=1, a(-1)=0, a(0)=0.
a(n) = Sum_{k=1..n} F(k)*L(k), where L(k) = Lucas(k) = A000032(k) = F(k-1) + F(k+1). - Alexander Adamchuk, May 18 2007
a(n) = 2*a(n-1) + (Sum_{k=1..n-2} a(k)) + n. - Jon Perry, Sep 01 2012
Sum {n >= 1} 1/a(n) = 3 - phi, where phi = 1/2*(1 + sqrt(5)) is the golden ratio. The ratio of adjacent terms r(n) := a(n)/a(n-1) satisfies the recurrence r(n+1) = (4*r(n) - 1)/(r(n) + 1) for n >= 2. - Peter Bala, Dec 05 2013
a(n) = S(n, 3) - S(n-1, 3) - 1, n >= 0, with Chebyshev's S-polynomials (see A049310), where S(-1, x) = 0. - Wolfdieter Lang, Aug 28 2014
a(n) = -1 + (2^(-1-n)*((3-sqrt(5))^n*(-1+sqrt(5)) + (1+sqrt(5))*(3+sqrt(5))^n)) / sqrt(5). - Colin Barker, Jun 03 2016
E.g.f.: (sqrt(5)*sinh(sqrt(5)*x/2) + 5*cosh(sqrt(5)*x/2))*exp(3*x/2)/5 - exp(x). - Ilya Gutkovskiy, Jun 03 2016
a(n) = Sum_{k=0..n} binomial(n+1,k+1)*Fibonacci(k). - Vladimir Kruchinin, Oct 14 2016
a(n) = Sum_{k=0..n-1} Sum_{i=0..n-1} C(k+i+1,k-i). - Wesley Ivan Hurt, Sep 21 2017
a(n)*a(n-2) = a(n-1)*(a(n-1) - 1) for n>1. - Robert K. Moniot, Aug 23 2020
a(n) = Sum_{k=1..n} C(2*n-k,k). - Wesley Ivan Hurt, Dec 22 2020
a(n) = Sum_{k = 1..2*n+2} (-1)^k*Fibonacci(k). - Peter Bala, Nov 14 2021
a(n) = (2*cosh((1 + 2*n)*arccsch(2)))/sqrt(5) - 1. - Peter Luschny, Nov 21 2021
a(n) = F(n + (n mod 2)) * L(n+1 - (n mod 2)), where L(n) = A000032(n) and F(n) = A000045(n) (Euler and Sadek, 2001). - Amiram Eldar, Jan 13 2022

More terms from James Sellers, Sep 08 2000

Paul Barry's Nov 14 2003 formula, recurrences and g.f. corrected for offset 0 and index link for Chebyshev polynomials added by Wolfdieter Lang, Aug 28 2014

A002530 a(n) = 4*a(n-2) - a(n-4) for n > 1, a(n) = n for n = 0, 1.

Original entry on oeis.org

0, 1, 1, 3, 4, 11, 15, 41, 56, 153, 209, 571, 780, 2131, 2911, 7953, 10864, 29681, 40545, 110771, 151316, 413403, 564719, 1542841, 2107560, 5757961, 7865521, 21489003, 29354524, 80198051, 109552575, 299303201, 408855776, 1117014753, 1525870529, 4168755811

Offset: 0

N. J. A. Sloane

Denominators of continued fraction convergents to sqrt(3), for n >= 1.
Also denominators of continued fraction convergents to sqrt(3) - 1. See A048788 for numerators. - N. J. A. Sloane, Dec 17 2007. Convergents are 1, 2/3, 3/4, 8/11, 11/15, 30/41, 41/56, 112/153, ...
Consider the mapping f(a/b) = (a + 3*b)/(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, 2/1, 5/3, 7/4, 19/11, ... converging to 3^(1/2). Sequence contains the denominators. The same mapping for N, i.e., f(a/b) = (a + Nb)/(a + b) gives fractions converging to N^(1/2). - Amarnath Murthy, Mar 22 2003
Sqrt(3) = 2/2 + 2/3 + 2/(3*11) + 2/(11*41) + 2/(41*153) + 2/(153*571), ...; the sum of the first 6 terms of this series = 1.7320490367..., while sqrt(3) = 1.7320508075... - Gary W. Adamson, Dec 15 2007
From Clark Kimberling, Aug 27 2008: (Start)
  Related convergents (numerator/denominator):
  lower principal convergents: A001834/A001835
  upper principal convergents: A001075/A001353
  intermediate convergents: A005320/A001075
  principal and intermediate convergents: A143642/A140827
  lower principal and intermediate convergents: A143643/A005246. (End)
Row sums of triangle A152063 = (1, 3, 4, 11, ...). - Gary W. Adamson, Nov 26 2008
From Alois P. Heinz, Apr 13 2011: (Start)
Also number of domino tilings of the 3 X (n-1) rectangle with upper left corner removed iff n is even.  For n=4 the 4 domino tilings of the 3 X 3 rectangle with upper left corner removed are:
   . ._. . ._. . ._. . ._.
   .|__| .|__| .| | | .|___|
   | |_| | | | | | ||| |_| |
   ||__| |||_| ||__| |_|_|  (End)
This is the sequence of Lehmer numbers u_n(sqrt(R),Q) with the parameters R = 2 and Q = -1. It is a strong divisibility sequence, that is, gcd(a(n),a(m)) = a(gcd(n,m)) for all natural numbers n and m. - Peter Bala, Apr 18 2014
2^(-floor(n/2))*(1 + sqrt(3))^n = A002531(n) + a(n)*sqrt(3); integers in the real quadratic number field Q(sqrt(3)). - Wolfdieter Lang, Feb 11 2018
Let T(n) = 2^(n mod 2), U(n) = a(n), V(n) = A002531(n), x(n) = V(n)/U(n). Then T(n*m) * U(n+m) = U(n)*V(m) + U(m)*V(n), T(n*m) * V(n+m) = 3*U(n)*U(m) + V(m)*V(n), x(n+m) = (3 + x(n)*x(m))/(x(n) + x(m)). - Michael Somos, Nov 29 2022

			Convergents to sqrt(3) are: 1, 2, 5/3, 7/4, 19/11, 26/15, 71/41, 97/56, 265/153, 362/209, 989/571, 1351/780, 3691/2131, ... = A002531/A002530 for n >= 1.
1 + 1/(1 + 1/(2 + 1/(1 + 1/2))) = 19/11 so a(5) = 11.
G.f. = x + x^2 + 3*x^3 + 4*x^4 + 11*x^5 + 15*x^6 + 41*x^7 + ... - _Michael Somos_, Mar 18 2022

Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.
Russell Lyons, A bird's-eye view of uniform spanning trees and forests, in Microsurveys in Discrete Probability, AMS, 1998.
I. Niven and H. S. Zuckerman, An Introduction to the Theory of Numbers. 2nd ed., Wiley, NY, 1966, p. 181.
Murat Sahin and Elif Tan, Conditional (strong) divisibility sequences, Fib. Q., 56 (No. 1, 2018), 18-31.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
A. Tarn, Approximations to certain square roots and the series of numbers connected therewith, Mathematical Questions and Solutions from the Educational Times, 1 (1916), 8-12.

Harry J. Smith, Table of n, a(n) for n = 0..2000
Peter Bala, Notes on 2-periodic continued fractions and Lehmer sequences
Paul Barry, Symmetric Third-Order Recurring Sequences, Chebyshev Polynomials, and Riordan Arrays, JIS 12 (2009) 09.8.6.
Mario Catalani, Sequences related to convergents to square root of rationals, arXiv:math/0305270 [math.NT], 2003.
Marcia Edson, Scott Lewis and Omer Yayenie, The k-periodic Fibonacci sequence and an extended Binet's formula, INTEGERS 11 (2011) #A32.
Aviezri S. Fraenkel, Jonathan Levitt, and Michael Shimshoni, Characterization of the set of values f(n)=[n alpha], n=1,2,..., Discrete Math. 2 (1972), no.4, 335-345.
Clark Kimberling, Best lower and upper approximates to irrational numbers, Elemente der Mathematik, 52 (1997) 122-126.
Clark Kimberling, Matrix Transformations of Integer Sequences, J. Integer Seqs., Vol. 6, 2003.
MacTutor, D'Arcy Thompson on Greek irrationals
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Albert Tarn, Approximations to certain square roots and the series of numbers connected therewith [Annotated scanned copy]
D'Arcy Thompson, Excess and Defect: Or the Little More and the Little Less, Mind, New Series, Vol. 38, No. 149 (Jan., 1929), pp. 43-55 (13 pages). See page 48.
Hein van Winkel, Q-quadrangles inscribed in a circle, 2014. See Table 1. [Reference from Antreas Hatzipolakis, Jul 14 2014]
Russell Walsmith, CL-Chemy Transforms Fibonacci-Type Sequences to Arrays
E. W. Weisstein, MathWorld: Lehmer Number
Index entries for "core" sequences
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (0,4,0,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A002531 (numerators of convergents to sqrt(3)), A048788, A003297.
Bisections: A001353 and A001835.
Cf. A152063.
Cf. A108412, A049310.
Analog for sqrt(m): A000129 (m=2), A001076 (m=5), A041007 (m=6), A041009 (m=7), A041011 (m=8), A005668 (m=10), A041015 (m=11), A041017 (m=12), ..., A042935 (m=999), A042937 (m=1000).

I:=[0,1,1,3]; [n le 4 select I[n] else 4*Self(n-2) - Self(n-4): n in [1..50]]; // G. C. Greubel, Feb 25 2019

a := proc(n) option remember; if n=0 then 0 elif n=1 then 1 elif n=2 then 1 elif n=3 then 3 else 4*a(n-2)-a(n-4) fi end; [ seq(a(i),i=0..50) ];
A002530:=-(-1-z+z**2)/(1-4*z**2+z**4); # conjectured (correctly) by Simon Plouffe in his 1992 dissertation

Join[{0},Table[Denominator[FromContinuedFraction[ContinuedFraction[Sqrt[3],n]]], {n,1,50}]] (* Stefan Steinerberger, Apr 01 2006 *)
Join[{0},Denominator[Convergents[Sqrt[3],50]]] (* or *) LinearRecurrence[ {0,4,0,-1},{0,1,1,3},50] (* Harvey P. Dale, Jan 29 2013 *)
a[ n_] := If[n<0, -(-1)^n, 1] SeriesCoefficient[ x*(1+x-x^2)/(1-4*x^2+x^4), {x, 0, Abs@n}]; (* Michael Somos, Apr 18 2019 *)
a[ n_] := ChebyshevU[n-1, Sqrt[-1/2]]*Sqrt[2]^(Mod[n, 2]-1)/I^(n-1) //Simplify; (* Michael Somos, Nov 29 2022 *)

{a(n) = if( n<0, -(-1)^n * a(-n), contfracpnqn(vector(n, i, 1 + (i>1) * (i%2)))[2, 1])}; /* Michael Somos, Jun 05 2003 */

{ for (n=0, 50, a=contfracpnqn(vector(n, i, 1+(i>1)*(i%2)))[2, 1]; write("b002530.txt", n, " ", a); ); } \\ Harry J. Smith, Jun 01 2009

my(w=quadgen(12)); A002530(n)=real((2+w)^(n\/2)*if(bittest(n,0),1-w/3,w/3));
apply(A002530, [0..30]) \\ M. F. Hasler, Nov 04 2019

from functools import cache
@cache
def a(n): return [0, 1, 1, 3][n] if n < 4 else 4*a(n-2) - a(n-4)
print([a(n) for n in range(36)]) # Michael S. Branicky, Nov 13 2022

(x*(1+x-x^2)/(1-4*x^2+x^4)).series(x, 50).coefficients(x, sparse=False) # G. C. Greubel, Feb 25 2019

G.f.: x*(1 + x - x^2)/(1 - 4*x^2 + x^4).
a(n) = 4*a(n-2) - a(n-4). [Corrected by László Szalay, Feb 21 2014]
a(n) = -(-1)^n * a(-n) for all n in Z, would satisfy the same recurrence relation. - Michael Somos, Jun 05 2003
a(2*n) = a(2*n-1) + a(2*n-2), a(2*n+1) = 2*a(2*n) + a(2*n-1).
From Benoit Cloitre, Dec 15 2002: (Start)
a(2*n) = ((2 + sqrt(3))^n - (2 - sqrt(3))^n)/(2*sqrt(3)).
a(2*n) = A001353(n).
a(2*n-1) = ceiling((1 + 1/sqrt(3))/2*(2 + sqrt(3))^n) = ((3 + sqrt(3))^(2*n - 1) + (3 - sqrt(3))^(2*n - 1))/6^n.
a(2*n-1) = A001835(n). (End)
a(n+1) = Sum_{k=0..floor(n/2)} binomial(n - k, k) * 2^floor((n - 2*k)/2). - Paul Barry, Jul 13 2004
a(n) = Sum_{k=0..floor(n/2)} binomial(floor(n/2) + k, floor((n - 1)/2 - k))*2^k. - Paul Barry, Jun 22 2005
G.f.: (sqrt(6) + sqrt(3))/12*Q(0), where Q(k) = 1 - a/(1 + 1/(b^(2*k) - 1 - b^(2*k)/(c + 2*a*x/(2*x - g*m^(2*k)/(1 + a/(1 - 1/(b^(2*k + 1) + 1 - b^(2*k + 1)/(h - 2*a*x/(2*x + g*m^(2*k + 1)/Q(k + 1)))))))))). - Sergei N. Gladkovskii, Jun 21 2012
a(n) = (alpha^n - beta^n)/(alpha - beta) for n odd, and a(n) = (alpha^n - beta^n)/(alpha^2 - beta^2) for n even, where alpha = 1/2*(sqrt(2) + sqrt(6)) and beta = (1/2)*(sqrt(2) - sqrt(6)). Cf. A108412. - Peter Bala, Apr 18 2014
a(n) = (-sqrt(2)*i)^n*S(n, sqrt(2)*i)*2^(-floor(n/2)) = A002605(n)*2^(-floor(n/2)), n >= 0, with i = sqrt(-1) and S the Chebyshev polynomials (A049310). - Wolfdieter Lang, Feb 10 2018
a(n+1)*a(n+2) - a(n+3)*a(n) = (-1)^n, n >= 0. - Kai Wang, Feb 06 2020
E.g.f.: sinh(sqrt(3/2)*x)*(sinh(x/sqrt(2)) + sqrt(2)*cosh(x/sqrt(2)))/sqrt(3). - Stefano Spezia, Feb 07 2020
a(n) = ((1 + sqrt(3))^n - (1 - sqrt(3))^n)/(2*2^floor(n/2))/sqrt(3) = A002605(n)/2^floor(n/2). - Robert FERREOL, Apr 13 2023

Definition edited by M. F. Hasler, Nov 04 2019

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A175183 Pisano period of the 4-Fibonacci numbers A001076.

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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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A006190 a(n) = 3*a(n-1) + a(n-2), with a(0)=0, a(1)=1.