A001519 -id:A001519 - cp's OEIS frontend

A014445 Even Fibonacci numbers; or, Fibonacci(3*n).

0, 2, 8, 34, 144, 610, 2584, 10946, 46368, 196418, 832040, 3524578, 14930352, 63245986, 267914296, 1134903170, 4807526976, 20365011074, 86267571272, 365435296162, 1548008755920, 6557470319842, 27777890035288, 117669030460994, 498454011879264, 2111485077978050

Offset: 0

Mohammad K. Azarian

a(n) = 3^n*b(n;2/3) = -b(n;-2), but we have 3^n*a(n;2/3) = F(3n+1) = A033887 and a(n;-2) = F(3n-1) = A015448, where a(n;d) and b(n;d), n=0,1,...,d, denote the so-called delta-Fibonacci numbers (the argument "d" of a(n;d) and b(n;d) is abbreviation of the symbol "delta") defined by the following equivalent relations: (1 + d*((sqrt(5) - 1)/2))^n = a(n;d) + b(n;d)*((sqrt(5) - 1)/2) equiv. a(0;d)=1, b(0;d)=0, a(n+1;d) = a(n;d) + d*b(n;d), b(n+1;d) = d*a(n;d) + (1-d)b(n;d) equiv. a(0;d)=a(1;d)=1, b(0;1)=0, b(1;d)=d, and x(n+2;d) + (d-2)*x(n+1;d) + (1-d-d^2)*x(n;d) = 0 for every n=0,1,...,d, and x=a,b equiv. a(n;d) = Sum_{k=0..n} C(n,k)*F(k-1)*(-d)^k, and b(n;d) = Sum_{k=0..n} C(n,k)*(-1)^(k-1)*F(k)*d^k equiv. a(n;d) = Sum_{k=0..n} C(n,k)*F(k+1)*(1-d)^(n-k)*d^k, and b(n;d) = Sum_{k=1..n} C(n;k)*F(k)*(1-d)^(n-k)*d^k. The sequences a(n;d) and b(n;d) for special values d are connected with many known sequences: A000045, A001519, A001906, A015448, A020699, A033887, A033889, A074872, A081567, A081568, A081569, A081574, A081575, A163073 (see also the papers of Witula et al.). - Roman Witula, Jul 12 2012
For any odd k, Fibonacci(k*n) = sqrt(Fibonacci((k-1)*n) * Fibonacci((k+1)*n) + Fibonacci(n)^2). - Gary Detlefs, Dec 28 2012
The ratio of consecutive terms approaches the continued fraction 4 + 1/(4 + 1/(4 +...)) = A098317. - Hal M. Switkay, Jul 05 2020

			G.f. = 2*x + 8*x^2 + 34*x^3 + 144*x^4 + 610*x^5 + 2584*x^6 + 10946*x^7 + ...

Arthur T. Benjamin and Jennifer J. Quinn,, Proofs that really count: the art of combinatorial proof, M.A.A., 2003, id. 232.

T. D. Noe, Table of n, a(n) for n = 0..200
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 (v)).
H. H. Ferns, Problem B-115, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 5, No. 2 (1967), p. 202; Identities for F_{kn} and L{kn}, Solution to Problem B-115 by Stanley Rabinowitz, ibid., Vol. 6, No. 1 (1968), pp. 92-93.
Ira M. Gessel and Ji Li, Compositions and Fibonacci identities, J. Int. Seq., Vol. 16 (2013), Article 13.4.5.
Edyta Hetmaniok, Bozena Piatek, and Roman Wituła, Binomials Transformation Formulae of Scaled Fibonacci Numbers, Open Math., Vol. 15 (2017), pp. 477-485.
Tanya Khovanova, Recursive Sequences.
Ron Knott, Mathematics of the Fibonacci Series.
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.
Peter McCalla and Asamoah Nkwanta, Catalan and Motzkin Integral Representations, arXiv:1901.07092 [math.NT], 2019.
Michael Z. Spivey and Laura L. Steil, The k-Binomial Transforms and the Hankel Transform, Journal of Integer Sequences, Vol. 9 (2006), Article 06.1.1.
Roberto Tauraso, Some Congruences for Central Binomial Sums Involving Fibonacci and Lucas Numbers, JIS 19 (2016) # 16.5.4
Roman Witula and Damian Slota, delta-Fibonacci numbers, Appl. Anal. Discr. Math., Vol. 3, No. 2 (2009), pp. 310-329, MR2555042.
Roman Witula, Binomials transformation formulae of scaled Lucas numbers, Demonstratio Math., Vol. 46 (2013), pp. 15-27.
Index entries for linear recurrences with constant coefficients, signature (4,1).

Cf. A000032, A000045, A001076, A049310, A111125.
Cf. A001519, A001906, A015448, A020699, A033887, A033889, A074872, A081567, A081568, A081569, A081574, A081575, A098317, A163073.
First differences of A099919.
Third column of array A102310.

[Fibonacci(3*n): n in [0..30]]; // Vincenzo Librandi, Apr 18 2011

a:= n-> (<<0|1>, <1|1>>^(3*n))[1,2]:
seq(a(n), n=0..25);  # Alois P. Heinz, Feb 03 2023

Table[Fibonacci[3n], {n,0,30}] (* Stefan Steinerberger, Apr 07 2006 *)
LinearRecurrence[{4,1},{0,2},30] (* Harvey P. Dale, Nov 14 2021 *)
Table[ SeriesCoefficient[2*x/(1 - 4*x - x^2), {x, 0, n}], {n, 0, 20}] (* Nikolaos Pantelidis, Feb 02 2023 *)

numlib::fibonacci(3*n) $ n = 0..30; // Zerinvary Lajos, May 09 2008

a(n)=fibonacci(3*n) \\ Charles R Greathouse IV, Oct 25 2012

[fibonacci(3*n) for n in range(0, 30)] # Zerinvary Lajos, May 15 2009

a(n) = Sum_{k=0..n} binomial(n, k)*F(k)*2^k. - Benoit Cloitre, Oct 25 2003
From Lekraj Beedassy, Jun 11 2004: (Start)
a(n) = 4*a(n-1) + a(n-2), with a(-1) = 2, a(0) = 0.
a(n) = 2*A001076(n).
a(n) = (F(n+1))^3 + (F(n))^3 - (F(n-1))^3. (End)
a(n) = Sum_{k=0..floor((n-1)/2)} C(n, 2*k+1)*5^k*2^(n-2*k). - Mario Catalani (mario.catalani(AT)unito.it), Jul 22 2004
a(n) = Sum_{k=0..n} F(n+k)*binomial(n, k). - Benoit Cloitre, May 15 2005
O.g.f.: 2*x/(1 - 4*x - x^2). - R. J. Mathar, Mar 06 2008
a(n) = second binomial transform of (2,4,10,20,50,100,250). This is 2* (1,2,5,10,25,50,125) or 5^n (offset 0): *2 for the odd numbers or *4 for the even. The sequences are interpolated. Also a(n) = 2*((2+sqrt(5))^n - (2-sqrt(5))^n)/sqrt(20). - Al Hakanson (hawkuu(AT)gmail.com), May 02 2009
a(n) = 3*F(n-1)*F(n)*F(n+1) + 2*F(n)^3, F(n)=A000045(n). - Gary Detlefs, Dec 23 2010
a(n) = (-1)^n*3*F(n) + 5*F(n)^3, n >= 0. See the D. Jennings formula given in a comment on A111125, where also the reference is given. - Wolfdieter Lang, Aug 31 2012
With L(n) a Lucas number, F(3*n) = F(n)*(L(2*n) + (-1)^n) = (L(3*n+1) + L(3*n-1))/5 starting at n=1. - J. M. Bergot, Oct 25 2012
a(n) = sqrt(Fibonacci(2*n)*Fibonacci(4*n) + Fibonacci(n)^2). - Gary Detlefs, Dec 28 2012
For n > 0, a(n) = 5*F(n-1)*F(n)*F(n+1) - 2*F(n)*(-1)^n. - J. M. Bergot, Dec 10 2015
a(n) = -(-1)^n * a(-n) for all n in Z. - Michael Somos, Nov 15 2018
a(n) = (5*Fibonacci(n)^3 + Fibonacci(n)*Lucas(n)^2)/4 (Ferns, 1967). - Amiram Eldar, Feb 06 2022
a(n) = 2*i^(n-1)*S(n-1,-4*i), with i = sqrt(-1), and the Chebyshev S-polynomials (see A049310) with S(-1, x) = 0. From the simplified trisection formula. - Gary Detlefs and Wolfdieter Lang, Mar 04 2023
E.g.f.: 2*exp(2*x)*sinh(sqrt(5)*x)/sqrt(5). - Stefano Spezia, Jun 03 2024
a(n) = 2*F(n) + 3*Sum_{k=0..n-1} F(3*k)*F(n-k). - Yomna Bakr and Greg Dresden, Jun 10 2024

A122367 Dimension of 3-variable non-commutative harmonics (twisted derivative) of order n. The dimension of the space of non-commutative polynomials of degree n in 3 variables which are killed by all symmetric differential operators (where for a monomial w, d_{xi} ( xi w ) = w and d_{xi} ( xj w ) = 0 for i != j).

Original entry on oeis.org

1, 2, 5, 13, 34, 89, 233, 610, 1597, 4181, 10946, 28657, 75025, 196418, 514229, 1346269, 3524578, 9227465, 24157817, 63245986, 165580141, 433494437, 1134903170, 2971215073, 7778742049, 20365011074, 53316291173, 139583862445, 365435296162, 956722026041

Offset: 0

Mike Zabrocki, Aug 30 2006

Essentially identical to A001519.
From Matthew Lehman, Jun 14 2008: (Start)
Number of monotonic rhythms using n time intervals of equal duration (starting with n=0).
Representationally, let 0 be an interval which is "off" (rest),
1 an interval which is "on" (beep),
1 1 two consecutive "on" intervals (beep, beep),
1 0 1 (beep, rest, beep) and
1-1 two connected consecutive "on" intervals (beeeep).
For f(3)=13:
  0 0 0, 0 0 1, 0 1 0, 0 1 1, 0 1-1, 1 0 0, 1 0 1,
  1 1 0, 1-1 0, 1 1 1, 1 1-1, 1-1 1, 1-1-1.
(End)
Equivalent to the number of one-dimensional graphs of n nodes, subject to the condition that a node is either 'on' or 'off' and that any two neighboring 'on' nodes can be connected. - Matthew Lehman, Nov 22 2008
Sum_{n>=0} arctan(1/a(n)) = Pi/2. - Jaume Oliver Lafont, Feb 27 2009
This is the limit sequence for certain generalized Pell numbers. - Gregory L. Simay, Oct 21 2024

			a(1) = 2 because x1-x2, x1-x3 are both of degree 1 and are killed by the differential operator d_x1 + d_x2 + d_x3.
a(2) = 5 because x1*x2 - x3*x2, x1*x3 - x2*x3, x2*x1 - x3*x1, x1*x1 - x2*x1 - x2*x2 + x1*x2, x1*x1 - x3*x1 - x3*x3 + x3*x1 are all linearly independent and are killed by d_x1 + d_x2 + d_x3, d_x1 d_x1 + d_x2 d_x2 + d_x3 d_x3 and Sum_{j = 1..3} (d_xi d_xj, i).

Colin Barker, Table of n, a(n) for n = 0..1000
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 (ii)).
Paul Barry and A. Hennessy, The Euler-Seidel Matrix, Hankel Matrices and Moment Sequences, J. Int. Seq. 13 (2010) # 10.8.2, Example 13.
N. Bergeron, C. Reutenauer, M. Rosas, and M. Zabrocki, Invariants and Coinvariants of the Symmetric Group in Noncommuting Variables, arXiv:math/0502082 [math.CO], 2005; Canad. J. Math. 60 (2008), no. 2, 266-296
C. Chevalley, Invariants of finite groups generated by reflections, Amer. J. Math. 77 (1955), 778-782.
I. M. Gessel and Ji Li, Compositions and Fibonacci identities, J. Int. Seq. 16 (2013) 13.4.5.
Tanya Khovanova, Recursive Sequences
Ron Knott, Pi and the Fibonacci numbers. - _Jaume Oliver Lafont_, Feb 27 2009
Diego Marques and Alain Togbé, On the sum of powers of two consecutive Fibonacci numbers, Proc. Japan Acad. Ser. A Math. Sci., Volume 86, Number 10 (2010), 174-176.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory, Vol. 7, No. 5 (2011), pp. 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences, Integers, Volume 12A (2012), The John Selfridge Memorial Volume.
M. C. Wolf, Symmetric functions of noncommutative elements, Duke Math. J. 2 (1936), 626-637.
Index entries for linear recurrences with constant coefficients, signature (3,-1).

Cf. A001519, A048575, A055105, A055107, A087903, A074664, A008277, A106729, A112340, A122368, A122369, A122370, A122371, A122372.

Cf. similar sequences listed in A238379.

[Fibonacci(2*n+1): n in [0..40]]; // Vincenzo Librandi, Jul 04 2015

a:=n->if n=0 then 1; elif n=1 then 2 else 3*a(n-1)-a(n-2); fi;
A122367List := proc(m) local A, P, n; A := [1,2]; P := [2];
for n from 1 to m - 2 do P := ListTools:-PartialSums([op(A), P[-1]]);
A := [op(A), P[-1]] od; A end: A122367List(30); # Peter Luschny, Mar 24 2022

Table[Fibonacci[2 n + 1], {n, 0, 30}] (* Vincenzo Librandi, Jul 04 2015 *)

Vec((1-x)/(1-3*x+x^2) + O(x^50)) \\ Michel Marcus, Jul 04 2015

G.f.: (1-q)/(1-3*q+q^2). More generally, (Sum_{d=0..n} (n!/(n-d)!*q^d)/Product_{r=1..d} (1 - r*q)) / (Sum_{d=0..n} q^d/Product_{r=1..d} (1 - r*q)) where n=3.
a(n) = 3*a(n-1) - a(n-2) with a(0) = 1, a(1) = 2.
a(n) = Fibonacci(2n+1) = A000045(2n+1). - Philippe Deléham, Feb 11 2009
a(n) = (2^(-1-n)*((3-sqrt(5))^n*(-1+sqrt(5)) + (1+sqrt(5))*(3+sqrt(5))^n)) / sqrt(5). - Colin Barker, Oct 14 2015
a(n) = Sum_{k=0..n} Sum_{i=0..n} binomial(k+i-1, k-i). - Wesley Ivan Hurt, Sep 21 2017
a(n) = A048575(n-1) for n >= 1. - Georg Fischer, Nov 02 2018
a(n) = Fibonacci(n)^2 + Fibonacci(n+1)^2. - Michel Marcus, Mar 18 2019
a(n) = Product_{k=1..n} (1 + 4*cos(2*k*Pi/(2*n+1))^2). - Seiichi Manyama, Apr 30 2021
From J. M. Bergot, May 27 2022: (Start)
a(n) = F(n)*L(n+1) + (-1)^n where L(n)=A000032(n) and F(n)=A000045(n).
a(n) = (L(n)^2 + L(n)*L(n+2))/5 - (-1)^n.
a(n) = 2*(area of a triangle with vertices at (L(n-1), L(n)), (F(n+1), F(n)), (L(n+1), L(n+2))) - 5*(-1)^n for n > 1. (End)
G.f.: (1-x)/(1-3x+x^2) = 1/(1-2x-x^2-x^3-x^4-...) - Gregory L. Simay, Oct 21 2024
E.g.f.: exp(3*x/2)*(5*cosh(sqrt(5)*x/2) + sqrt(5)*sinh(sqrt(5)*x/2))/5. - Stefano Spezia, Nov 07 2024
From Peter Bala, May 04 2025: (Start)
a(n) = sqrt(2/5) * sqrt( 1 - T(2*n+1, - 3/2) ), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind.
a(2*n+1/2) = sqrt(5)*a(n)^2 - 2/sqrt(5).
a(3*n+1) = 5*a(n)^3 - 3*a(n); hence a(n) divides a(3*n+1).
a(4*n+3/2) = 5^(3/2)*a(n)^4 - 4*sqrt(5)*a(n)^2 + 2/sqrt(5).
a(5*n+2) = (5^2)*a(n)^5 - 5*5*a(n)^3 + 5*a(n); hence a(n) divides a(5*n+2).
See A034807 for the unsigned coefficients [1, 2; 1, 3; 1, 4, 2; 1, 5, 5; ...].
In general, for k >= 0, a(k*n + (k-1)/2) = a(-1/2) * T(k, a(n)/a(-1/2)), where a(n) = (2^(-1-n)*((3 - sqrt(5))^n *(-1 + sqrt(5)) + (1 + sqrt(5))*(3 + sqrt(5))^n)) / sqrt(5), as given above, and a(-1/2) = 2/sqrt(5).
The aerated sequence [b(n)]n>=1 = [1, 0, 2, 0, 5, 0, 13, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -5, Q = 1 of the 3-parameter family of divisibility sequences found by Williams and Guy.
Sum_{n >= 1} 1/(a(n) - 1/a(n)) = 1 (telescoping series: for n >= 1, 1/(a(n) - 1/a(n)) = 1/A001906(n) - 1/A001906(n+1).) (End)

A007805 a(n) = Fibonacci(6*n + 3)/2.

Original entry on oeis.org

1, 17, 305, 5473, 98209, 1762289, 31622993, 567451585, 10182505537, 182717648081, 3278735159921, 58834515230497, 1055742538989025, 18944531186571953, 339945818819306129, 6100080207560938369

Offset: 0

James A. Raymond, Clark Kimberling

Hypotenuse (z) of Pythagorean triples (x,y,z) with |2x-y|=1.
x(n) := 2*A049629(n) and y(n) := a(n), n >= 0, give all positive solutions of the Pell equation x^2 - 5*y^2 = -1. See the Gregory V. Richardson formula, where his x is the y here and A075796(n+1) = x(n). - Wolfdieter Lang, Jun 20 2013
Positive numbers n such that 5*n^2 - 1 is a square (A075796(n+1)^2). - Gregory V. Richardson, Oct 13 2002

G. C. Greubel, Table of n, a(n) for n = 0..795 (terms 0..100 from T. D. Noe)
A. J. C. Cunningham, Binomial Factorisations, Vols. 1-9, Hodgson, London, 1923-1929. See Vol. 1, page xxxv.
Tanya Khovanova, Recursive Sequences
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Hyperbola, Forum Geometricorum (2019) Vol. 19, 11-16.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences Integers, Volume 12A (2012) The John Selfridge Memorial Volume
Index entries for linear recurrences with constant coefficients, signature (18,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A000045.
Row 18 of array A094954.
Row 2 of array A188647.
Cf. similar sequences listed in A238379.

a007805 = (`div` 2) . a000045 . (* 3) . (+ 1) . (* 2)
-- Reinhard Zumkeller, Mar 26 2013

I:=[1, 17]; [n le 2 select I[n] else 18*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 19 2017

seq(combinat:-fibonacci(6*n+3)/2, n=0..30); # Robert Israel, Sep 10 2014

LinearRecurrence[{18, -1}, {1, 17}, 50] (* Sture Sjöstedt, Nov 29 2011 *)
Table[Fibonacci[6n+3]/2, {n, 0, 20}] (* Harvey P. Dale, Dec 17 2011 *)
CoefficientList[Series[(1-x)/(1-18*x+x^2), {x, 0, 50}], x] (* G. C. Greubel, Dec 19 2017 *)

a(n)=fibonacci(6*n+3)/2 \\ Edward Jiang, Sep 09 2014

x='x+O('x^30); Vec((1-x)/(1-18*x+x^2)) \\ G. C. Greubel, Dec 19 2017

G.f.: (1-x)/(1-18*x+x^2).
a(n) = 18*a(n-1) - a(n-2), n > 1, a(0)=1, a(1)=17.
a(n) = A134495(n)/2 = A001076(2n+1).
a(n+1) = 9*a(n) + 4*sqrt(5*a(n)^2-1). - Richard Choulet, Aug 30 2007, Dec 28 2007
a(n) = ((2+sqrt(5))^(2*n+1) - (2-sqrt(5))^(2*n+1))/(2*sqrt(5)). - Dean Hickerson, Dec 09 2002
a(n) ~ (1/10)*sqrt(5)*(sqrt(5) + 2)^(2*n+1). - Joe Keane (jgk(AT)jgk.org), May 15 2002
Limit_{n->infinity} a(n)/a(n-1) = 8*phi + 5 = 9 + 4*sqrt(5). - Gregory V. Richardson, Oct 13 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i); then a(n) = q(n, 16). - Benoit Cloitre, Dec 06 2002
a(n) = 19*a(n-1)- 19*a(n-2) + a(n-3); f(x) = (sqrt(5)/10)*((2+sqrt(5))*(9+4*sqrt(5))^(x-1) - (2-sqrt(5))*(9-4*sqrt(5))^(x-1)). - Antonio Alberto Olivares, May 15 2008
a(n) = 17*a(n-1) + 17*a(n-2) - a(n-3). - Antonio Alberto Olivares, Jun 19 2008
a(n) = b(n+1) - b(n), n >= 0, with b(n) := F(6*n)/F(6) = A049660(n). First differences. See the o.g.f.s. - Wolfdieter Lang, 2012
a(n) = S(n,18) - S(n-1,18) with the Chebyshev S-polynomials (A049310). - Wolfdieter Lang, Jun 20 2013
Sum_{n >= 1} 1/( a(n) - 1/a(n) ) = 1/4^2. Compare with A001519 and A097843. - Peter Bala, Nov 29 2013
a(n) = 9*a(n-1) + 8*A049629(n-1), n >= 1, a(0) = 1. This is just the rewritten Chebyshev S(n, 18) recurrence. - Wolfdieter Lang, Aug 26 2014
From Peter Bala, Mar 23 2015: (Start)
a(n) = ( Fibonacci(6*n + 6 - 2*k) - Fibonacci(6*n + 2*k) )/( Fibonacci(6 - 2*k) - Fibonacci(2*k) ), for k an arbitrary integer.
a(n) = ( Fibonacci(6*n + 6 - 2*k - 1) + Fibonacci(6*n + 2*k + 1) )/( Fibonacci(6 - 2*k - 1) + Fibonacci(2*k + 1) ), for k an arbitrary integer.
The aerated sequence (b(n)) n>=1 = [1, 0, 17, 0, 305, 0, 5473, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -20, Q = 1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials. (End)
a(n) = sqrt(2 + (9-4*sqrt(5))^(1+2*n) + (9+4*sqrt(5))^(1+2*n))/(2*sqrt(5)). - Gerry Martens, Jun 04 2015

Better description and more terms from Michael Somos

A096270 Fixed point of the morphism 0->01, 1->011.

Original entry on oeis.org

0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0

Offset: 0

N. J. A. Sloane, Jun 22 2004

This is another version of the Fibonacci word A005614.
(With offset 1) for k>0, a(ceiling(k*phi^2))=0 and a(floor(k*phi^2))=1 where phi=(1+sqrt(5))/2 is the Golden ratio. - Benoit Cloitre, Apr 01 2006
(With offset 1) for n>1 a(A000045(n)) = (1-(-1)^n)/2.
Equals the Fibonacci word A005614 with an initial zero.
Also the Sturmian word of slope phi (cf. A144595). - N. J. A. Sloane, Jan 13 2009
More precisely: (a(n)) is the inhomogeneous Sturmian word of slope phi-1 and intercept 0: a(n) = floor((n+1)*(phi-1)) - floor(n*(phi-1)), n >= 0. - Michel Dekking, May 21 2018
The ratio of number of 1's to number of 0's tends to the golden ratio (1+sqrt(5))/2 = 1.618... - Zak Seidov, Feb 15 2012

J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003.

A.H.M. Smeets, Table of n, a(n) for n = 0..20000
Scott Balchin and Dan Rust, Computations for Symbolic Substitutions, Journal of Integer Sequences, Vol. 20 (2017), Article 17.4.1.
J-P. Borel and François Laubie, Construction de mots de Christoffel, Comptes rendus de l'Académie des sciences. Série 1, Mathématique 313.8 (1991): 483-485.
N. J. A. Sloane, Families of Essentially Identical Sequences, Mar 24 2021 (Includes this sequence)
Richard Southwell and Jianwei Huang, Complex Networks from Simple Rewrite Systems, arXiv preprint arXiv:1205.0596 [cs.SI], 2012.
Index entries for sequences that are fixed points of mappings

Cf. A003849, A096268, A001519. See A005614, A114986 for other versions.

The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A000201 as the parent: A000201, A001030, A001468, A001950, A003622, A003842, A003849, A004641, A005614, A014675, A022342, A088462, A096270, A114986, A124841. - N. J. A. Sloane, Mar 11 2021

[-1+Floor(n*(1+Sqrt(5))/2)-Floor((n-1)*(1+Sqrt(5))/2): n in [1..100]]; // Wesley Ivan Hurt, Aug 29 2022

Nest[ Function[l, {Flatten[(l /. {0 -> {0, 1}, 1 -> {0, 1, 1}})]}], {0}, 6] (* Robert G. Wilson v, Feb 04 2005 *)

a(n)=-1+floor(n*(1+sqrt(5))/2)-floor((n-1)*(1+sqrt(5))/2) \\ Benoit Cloitre, Apr 01 2006

from math import isqrt
def A096270(n): return (n+1+isqrt(5*(n+1)**2)>>1)-(n+isqrt(5*n**2)>>1)>>1 # Chai Wah Wu, Aug 29 2022

Conjecture: a(n) is given recursively by a(1)=0 and, for n>1, by a(n)=1 if n=F(2k+1) and a(n)=a(n-F(2k+1)) otherwise, where F(2k+1) is the largest odd-indexed Fibonacci number smaller than or equal to n. (This has been confirmed for more than nine million terms.) The odd-indexed bisection of the Fibonacci numbers (A001519) is {1, 2, 5, 13, 34, 89, ...}. So by the conjecture, we would expect that a(30) = a(30-13) = a(17) = a(17-13) = a(4) = a(4-2) = a(2) = 1, which is in fact correct. - John W. Layman, Jun 29 2004
From Michel Dekking, Apr 13 2016: (Start)
Proof of the above conjecture:
Let g be the morphism above: g(0)=01, g(1)=011. Then g^n(0) has length F(2n+1), and (a(n)) starts with g^n(0) for all n>0. Obviously g^n(0) ends in 1 for all n, proving the first part of the conjecture.
We extend the semigroup of words with letters 0 and 1 to the free group, adding the inverses 0*:=0^{-1} and 1*:=1^{-1}. Easy observation: for any word w one has g(w1)= g(w0)1. We claim that for all n>1 one has g^n(0)=u(n)v(n)v(n)0*1, where u(n)=g(u(n-1))0 and v(n)=0*g(v(n-1))0. The recursion starts with u(2)=0, v(2)=10. Indeed: g^2(0)=01011=u(2)v(2)v(2)0*1. Induction step:
  g^{n+1}(0)=g(g^n(0))= g(u(n)v(n)v(n)0*1)= g(u(n)v(n)v(n))1= g(u(n))00*g(v(n))00*g(v(n))00*1=u(n+1)v(n+1)v(n+1)0*1.
Since v(n) has length F(2n-1), which is the largest odd-indexed Fibonacci number smaller than or equal to m for all m between F(2n-1) and F(2n+1), the claim proves the second part of the conjecture. (End)
(With offset 1) a(n) = -1 + floor(n*phi) - floor((n-1)*phi) where phi=(1+sqrt(5))/2 so a(n) = -1 + A082389(n). - Benoit Cloitre, Apr 01 2006

More terms from John W. Layman, Jun 29 2004

A024718 a(n) = (1/2)(1 + Sum_{k=0..n} binomial(2k, k)).

Original entry on oeis.org

1, 2, 5, 15, 50, 176, 638, 2354, 8789, 33099, 125477, 478193, 1830271, 7030571, 27088871, 104647631, 405187826, 1571990936, 6109558586, 23782190486, 92705454896, 361834392116, 1413883873976, 5530599237776, 21654401079326, 84859704298202, 332818970772254

Offset: 0

Clark Kimberling

nonn

Total number of leaves in all rooted ordered trees with at most n edges. - Michael Somos, Feb 14 2006
Also: Number of UH-free Schroeder paths of semilength n with horizontal steps only at level less than two [see Yan]. - R. J. Mathar, May 24 2008
Hankel transform is A010892. - Paul Barry, Apr 28 2009
Binomial transform of A005773. - Philippe Deléham, Dec 13 2009
Number of vertices all of whose children are leaves in all ordered trees with n+1 edges. Example: a(3) = 15; for an explanation see David Callan's comment in A001519. - Emeric Deutsch, Feb 12 2015

			G.f. = 1 + 2*x + 5*x^2 + 15*x^3 + 50*x^4 + 176*x^5 + 638*x^6 + ...

Michael De Vlieger, Table of n, a(n) for n = 0..1664
Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
Sherry H. F. Yan, Schröder Paths and Pattern Avoiding Partitions, arXiv:0805.2465 [math.CO], 2008-2009.

Partial sums of A088218.
Bisection of A086905.
Second column of triangle A102541.
Cf. A000108, A006134, A024494, A079309.

A024718 := n -> (binomial(2*n, n)*hypergeom([1, -n], [1/2 - n], 1/4) + 1)/2:
seq(simplify(A024718(n)), n = 0..26); # Peter Luschny, Dec 15 2024

Table[Sum[Binomial[2k-1,k-1],{k,0,n}],{n,0,100}] (* Emanuele Munarini, May 18 2018 *)

a(n) = (1 + sum(k=0, n, binomial(2*k, k)))/2; \\ Michel Marcus, May 18 2018

a(n) = A079309(n) + 1.
G.f.: 1/((1 - x)*(2 - C)), where C = g.f. for the Catalan numbers A000108. - N. J. A. Sloane, Aug 30 2002
Given g.f. A(x), then x * A(x - x^2) is the g.f. of A024494. - Michael Somos, Feb 14 2006
G.f.: (1 + 1 / sqrt(1 - 4*x)) / (2 - 2*x). - Michael Somos, Feb 14 2006
D-finite with recurrence: n*a(n) - (5*n-2)*a(n-1) + 2*(2*n-1)*a(n-2) = 0. - R. J. Mathar, Dec 02 2012
Remark: The above recurrence is true (it can be easily proved by differentiating the generating function). Notice that it is the same recurrence satisfied by the partial sums of the central binomial coefficients (A006134). - Emanuele Munarini, May 18 2018
0 = a(n)*(16*a(n+1) - 22*a(n+2) + 6*a(n+3)) + a(n+1)*(-18*a(n+1) + 27*a(n+2) - 7*a(n+3)) + a(n+2)*(-3*a(n+2) + a(n+3)) for all n in Z if a(n) = 1/2 for n < 0. - Michael Somos, Apr 23 2014
a(n) = ((1 - I/sqrt(3))/2 - binomial(2*n+1, n)*hypergeom([n+3/2, 1], [n+2], 4)). - Peter Luschny, May 18 2018
a(n) = [x^n] 1/((1-x+x^2) * (1-x)^n). - Seiichi Manyama, Apr 06 2024

Name edited by Petros Hadjicostas, Aug 04 2020

A124758 Product of the parts of the compositions in standard order.

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 2, 1, 4, 3, 4, 2, 3, 2, 2, 1, 5, 4, 6, 3, 6, 4, 4, 2, 4, 3, 4, 2, 3, 2, 2, 1, 6, 5, 8, 4, 9, 6, 6, 3, 8, 6, 8, 4, 6, 4, 4, 2, 5, 4, 6, 3, 6, 4, 4, 2, 4, 3, 4, 2, 3, 2, 2, 1, 7, 6, 10, 5, 12, 8, 8, 4, 12, 9, 12, 6, 9, 6, 6, 3, 10, 8, 12, 6, 12, 8, 8, 4, 8, 6, 8, 4, 6, 4, 4, 2, 6, 5, 8, 4, 9, 6

Offset: 0

Franklin T. Adams-Watters, Nov 06 2006

The standard order of compositions is given by A066099.

A composition of n is a finite sequence of positive integers summing to n. The k-th composition in standard order (row k of A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. - Gus Wiseman, Apr 03 2020

			Composition number 11 is 2,1,1; 2*1*1 = 2, so a(11) = 2.
The table starts:
  1
  1
  2 1
  3 2 2 1
  4 3 4 2 3 2 2 1
  5 4 6 3 6 4 4 2 4 3 4 2 3 2 2 1
The 146-th composition in standard order is (3,3,2), with product 18, so a(146) = 18. - _Gus Wiseman_, Apr 03 2020

Alois P. Heinz, Rows n = 0..14, flattened
George Beck and Karl Dilcher, A Matrix Related to Stern Polynomials and the Prouhet-Thue-Morse Sequence, arXiv:2106.10400 [math.CO], 2021.

Cf. A066099, A118851, A011782 (row lengths), A001906 (row sums).
The lengths of standard compositions are given by A000120.
The version for prime indices is A003963.
The version for binary indices is A096111.
Taking the sum instead of product gives A070939.
The sum of binary indices is A029931.
The sum of prime indices is A056239.
Taking GCD instead of product gives A326674.
Positions of first appearances are A331579.
Cf. A001519, A011782, A048793, A114994, A124767, A228351, A272919, A329369 (similar recurrence), A333218, A333219.

stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
Table[Times@@stc[n],{n,0,100}] (* Gus Wiseman, Apr 03 2020 *)

For a composition b(1),...,b(k), a(n) = Product_{i=1}^k b(i).
a(A164894(n)) = a(A246534(n)) = n!. - Gus Wiseman, Apr 03 2020
a(A233249(n)) = a(A333220(n)) = A003963(n). - Gus Wiseman, Apr 03 2020
From Mikhail Kurkov, Jul 11 2021: (Start)
Some conjectures:
a(2n+1) = a(n) for n >= 0.
a(2n) = (1 + 1/A001511(n))*a(n) = 2*a(n) + a(n - 2^f(n)) - a(2n - 2^f(n)) for n > 0 with a(0)=1 where f(n) = A007814(n).
From the 1st formula for a(2n) we get a(4n+2) = 2*a(n), a(4n) = 2*a(2n) - a(n).
Sum_{k=0..2^n - 1} a(k) = A001519(n+1) for n >= 0.
a((4^n - 1)/3) = A011782(n) for n >= 0.
a(2^m*(2^n - 1)) = m + 1 for n > 0, m >= 0. (End)

A001517 Bessel polynomials y_n(x) (see A001498) evaluated at 2.

Original entry on oeis.org

1, 3, 19, 193, 2721, 49171, 1084483, 28245729, 848456353, 28875761731, 1098127402131, 46150226651233, 2124008553358849, 106246577894593683, 5739439214861417731, 332993721039856822081, 20651350143685984386753

Offset: 0

N. J. A. Sloane

Numerators of successive convergents to e using continued fraction 1 + 2/(1 + 1/(6 + 1/(10 + 1/(14 + 1/(18 + 1/(22 + 1/26 + ...)))))).

Number of ways to use the elements of {1,...,k}, n <= k <= 2n, once each to form a collection of n lists, each having length 1 or 2. - Bob Proctor, Apr 18 2005, Jun 26 2006

L. Euler, 1737.
I. S. Gradshteyn and I. M. Ryzhik, Tables of Integrals, Series and Products, 6th ed., Section 0.126, p. 2.
J. Riordan, Combinatorial Identities, Wiley, 1968, p. 77.
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).

Robert Israel, Table of n, a(n) for n = 0..334 (first 101 terms from T. D. Noe)
P. Bala, A note on the Catalan transform of a sequence.
J. W. L. Glaisher, On Lambert's proof of the irrationality of Pi and on the irrationality of certain other quantities, Reports of British Assoc. Adv. Sci., 1871, pp. 16-18.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 131.
D. H. Lehmer, Arithmetical periodicities of Bessel functions, Annals of Mathematics, 33 (1932): 143-150. The sequence is on page 149.
D. H. Lehmer, Review of various tables by P. Pederson, Math. Comp., 2 (1946), 68-69.
W. Mlotkowski, A. Romanowicz, A family of sequences of binomial type, Probability and Mathematical Statistics, Vol. 33, Fasc. 2 (2013), pp. 401-408.
R. A. Proctor, Let's Expand Rota's Twelvefold Way for Counting Partitions!, arXiv:math/0606404 [math.CO], 2006-2007.
J. Riordan, Letter to N. J. A. Sloane, Jul. 1968.
J. Riordan, Letter, Jul 06 1978.
N. J. A. Sloane, Letter to J. Riordan, Nov. 1970.
Index entries for related partition-counting sequences
Index entries for sequences related to Bessel functions or polynomials

Essentially the same as A080893.
a(n) = A099022(n)/n!.
Partial sums: A105747.
Replace "lists" with "sets" in comment: A001515.
Cf. A001515, A001518, A002119, A053556, A053557, A243593.

A:= gfun:-rectoproc({a(n) = (4*n-2)*a(n-1) + a(n-2),a(0)=1,a(1)=3},a(n),remember):
map(A, [$0..20]); # Robert Israel, Jul 22 2015
f:=proc(n) option remember; if n = 0 then 1 elif n=1 then 3 else f(n-2)+(4*n-2)*f(n-1); fi; end;
[seq(f(n), n=0..20)]; # N. J. A. Sloane, May 09 2016
seq(simplify(KummerU(-n, -2*n, 1)), n = 0..16); # Peter Luschny, May 10 2022

Table[(2k)! Hypergeometric1F1[-k, -2k, 1]/k!, {k, 0, 10}] (* Vladimir Reshetnikov, Feb 16 2011 *)

PARI
```
a(n)=sum(k=0,n,(n+k)!/k!/(n-k)!)
```

A001517 = lambda n: hypergeometric([-n, n+1], [], -1)
[simplify(A001517(n)) for n in (0..16)] # Peter Luschny, Oct 17 2014

a(n) = Sum_{k=0..n} (n+k)!/(k!*(n-k)!) = (e/Pi)^(1/2) K_{n+1/2}(1/2).
D-finite with recurrence a(n) = (4*n-2)*a(n-1) + a(n-2), n >= 2.
a(n) = (1/n!)*Sum_{k=0..n} (-1)^(n+k)*binomial(n,k)*A000522(n+k). - Vladeta Jovovic, Sep 30 2006
E.g.f. (for offset 1): exp(x*c(x)), where c(x)=(1-sqrt(1-4*x))/(2*x) (cf. A000108). - Vladimir Kruchinin, Aug 10 2010
G.f.: 1/Q(0), where Q(k) = 1 - x - 2*x*(k+1)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 17 2013
a(n) = (1/n!)*Integral_{x>=0} (x*(1 + x))^n*exp(-x) dx. Expansion of exp(x) in powers of y = x*(1 - x): exp(x) = 1 + y + 3*y^2/2! + 19*y^3/3! + 193*y^4/4! + 2721*y^5/5! + .... - Peter Bala, Dec 15 2013
a(n) = exp(1/2) / sqrt(Pi) * BesselK(n+1/2, 1/2). - Vaclav Kotesovec, Mar 15 2014
a(n) ~ 2^(2*n+1/2) * n^n / exp(n-1/2). - Vaclav Kotesovec, Mar 15 2014
a(n) = hypergeom([-n, n+1], [], -1). - Peter Luschny, Oct 17 2014
From G. C. Greubel, Aug 16 2017: (Start)
a(n) = (1/2)_{n} * 4^n * hypergeometric1f1(-n; -2*n; 1).
G.f.: (1/(1-t))*hypergeometric2f0(1, 1/2; -; 4*t/(1-t)^2). (End)
a(n) = Sum_{k=0..n} binomial(n,k)*binomial(n+k,k)*k!. - Ilya Gutkovskiy, Nov 24 2017
a(n) = KummerU(-n, -2*n, 1). - Peter Luschny, May 10 2022

More terms from Vladeta Jovovic, Apr 03 2000

Additional comments from Michael Somos, Jul 15 2002

A094954 Array T(k,n) read by antidiagonals. G.f.: x(1-x)/(1-kx+x^2), k>1.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 5, 1, 1, 4, 11, 13, 1, 1, 5, 19, 41, 34, 1, 1, 6, 29, 91, 153, 89, 1, 1, 7, 41, 169, 436, 571, 233, 1, 1, 8, 55, 281, 985, 2089, 2131, 610, 1, 1, 9, 71, 433, 1926, 5741, 10009, 7953, 1597, 1, 1, 10, 89, 631, 3409, 13201, 33461, 47956, 29681

Offset: 1

Ralf Stephan, May 31 2004

Also, values of polynomials with coefficients in A098493 (see Fink et al.). See A098495 for negative k.

Number of dimer tilings of the graph S_{k-1} X P_{2n-2}.

			1,1,1,1,1,1,1,1,1,1,1,1,1,1, ...
1,2,5,13,34,89,233,610,1597, ...
1,3,11,41,153,571,2131,7953, ...
1,4,19,91,436,2089,10009,47956, ...
1,5,29,169,985,5741,33461,195025, ...
1,6,41,281,1926,13201,90481,620166, ...

Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contributions to Discrete Mathematics, Vol 3, No 2 (2008), pp. 76-114.
Elizabeth Wilmer, A note on Stephan's conjecture 87
Elizabeth Wilmer, A note on Stephan's conjecture 87 [cached copy]

Rows are first differences of rows in array A073134.
Rows 2-14 are A000012, A001519, A079935/A001835, A004253, A001653, A049685, A070997, A070998, A072256, A078922, A077417, A085260, A001570. Other rows: A007805 (k=18), A075839 (k=20), A077420 (k=34), A078988 (k=66).
Columns include A028387. Diagonals include A094955, A094956. Antidiagonal sums are A094957.

max = 14; row[k_] := Rest[ CoefficientList[ Series[ x*(1-x)/(1-k*x+x^2), {x, 0, max}], x]]; t = Table[ row[k], {k, 2, max+1}]; Flatten[ Table[ t[[k-n+1, n]], {k, 1, max}, {n, 1, k}]] (* Jean-François Alcover, Dec 27 2011 *)

PARI
```
T(k,n)=polcoeff(x*(1-x)/(1-k*x+x*x),n)
```

Recurrence: T(k, 1) = 1, T(k, 2) = k-1, T(k, n) = kT(k, n-1) - T(k, n-2).
For n>3, T(k, n) = [k(k-2) + T(k, n-1)T(k, n-2)] / T(k, n-3).
T(k, n+1) = S(n, k) - S(n-1, k) = U(n, k/2) - U(n-1, k/2), with S, U = Chebyshev polynomials of second kind.
T(k+2, n+1) = Sum[i=0..n, k^(n-i) * C(2n-i, i)] (from comments by Benoit Cloitre).

A147703 Triangle [1,1,1,0,0,0,...] DELTA [1,0,0,0,...] with Deléham DELTA defined in A084938.

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 5, 9, 5, 1, 13, 27, 20, 7, 1, 34, 80, 73, 35, 9, 1, 89, 234, 252, 151, 54, 11, 1, 233, 677, 837, 597, 269, 77, 13, 1, 610, 1941, 2702, 2225, 1199, 435, 104, 15, 1, 1597, 5523, 8533, 7943, 4956, 2158, 657, 135, 17, 1

Offset: 0

Paul Barry, Nov 10 2008

Equal to A062110*A007318 when A062110 is regarded as a triangle read by rows.

			Triangle begins
   1;
   1,   1;
   2,   3,   1;
   5,   9,   5,   1;
  13,  27,  20,   7,  1;
  34,  80,  73,  35,  9,  1;
  89, 234, 252, 151, 54, 11, 1;

Indranil Ghosh, Rows 0..100, flattened

Row sums are A006012. Diagonal sums are A147704.

Cf. A062110, A007318, A206800, A001519, A321620.

# The function RiordanSquare is defined in A321620:
RiordanSquare(1 / (1 - x / (1 - x / (1 - x))), 10); # Peter Luschny, Jan 26 2020

nmax=9; Flatten[CoefficientList[Series[CoefficientList[Series[(1 - 2*x)/(1 - (3 + y)*x + (1 + y)*x^2), {x, 0, nmax}], x], {y, 0, nmax}], y]] (* Indranil Ghosh, Mar 11 2017 *)

Riordan array ((1-2x)/(1-3x+x^2), x(1-x)/(1-3x+x^2)).
Sum_{k=0..n} T(n,k)*x^k = A152239(n), A152223(n), A152185(n), A152174(n), A152167(n), A152166(n), A152163(n), A000007(n), A001519(n), A006012(n), A081704(n), A082761(n), A147837(n), A147838(n), A147839(n), A147840(n), A147841(n), for x = -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8 respectively. - Philippe Deléham, Dec 01 2008
G.f.: (1-2*x)/(1-(3+y)*x+(1+y)*x^2). - Philippe Deléham, Nov 26 2011
T(n,k) = 3*T(n-1,k) + T(n-1,k-1) - T(n-2,k) - T(n-2,k-1), for n > 1. - Philippe Deléham, Feb 12 2012
The Riordan square of the odd indexed Fibonacci numbers A001519. - Peter Luschny, Jan 26 2020

A026671 Number of lattice paths from (0,0) to (n,n) with steps (0,1), (1,0) and, when on the diagonal, (1,1).

Original entry on oeis.org

1, 3, 11, 43, 173, 707, 2917, 12111, 50503, 211263, 885831, 3720995, 15652239, 65913927, 277822147, 1171853635, 4945846997, 20884526283, 88224662549, 372827899079, 1576001732485, 6663706588179, 28181895551161, 119208323665543, 504329070986033, 2133944799315027

Offset: 0

Clark Kimberling, Miklos Bona

1, 1, 3, 11, 43, 173, ... is the unique sequence for which both the Hankel transform of the sequence itself and the Hankel transform of its left shift are the powers of 2 (A000079). For example, det[{{1, 1, 3}, {1, 3, 11}, {3, 11, 43}}] = det[{{1, 3, 11}, {3, 11, 43}, {11, 43, 173}}] = 4. - David Callan, Mar 30 2007
From Paul Barry, Jan 25 2009: (Start)
a(n) is the image of F(2n+2) under the Catalan matrix (1,xc(x)) where c(x) is the g.f. of A000108.
The sequence 1,1,3,... is the image of A001519 under (1,xc(x)). This sequence has g.f. given by 1/(1-x-2x^2/(1-3x-x^2/(1-2x-x^2/(1-2x-x^2/(1-... (continued fraction). (End)
Binomial transform of A111961. - Philippe Deléham, Feb 11 2009
From Paul Barry, Nov 03 2010: (Start)
The sequence 1,1,3,... has g.f. 1/(1-x/sqrt(1-4x)), INVERT transform of A000984.
It is an eigensequence of the sequence array for A000984. (End)

L. W. Shapiro and C. J. Wang, Generating identities via 2 X 2 matrices, Congressus Numerantium, 205 (2010), 33-46.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Jean-Christophe Aval, Adrien Boussicault and Sandrine Dasse-Hartaut, The tree structure in staircase tableaux, arXiv:1109.4907 [math.CO], 2011-2013.
Cyril Banderier, Markus Kuba, and Michael Wallner, Analytic Combinatorics of Composition schemes and phase transitions with mixed Poisson distributions, arXiv:2103.03751 [math.PR], 2021.
Miklos Bona, The permutation classes equinumerous to the smooth class, Electron. J. Combin., 5 (1998), no. 1, Research Paper 31, 12 pp.
David Callan and Toufik Mansour, Five subsets of permutations enumerated as weak sorting permutations, arXiv:1602.05182 [math.CO], 2016.
Aoife Hennessy, A Study of Riordan Arrays with Applications to Continued Fractions, Orthogonal Polynomials and Lattice Paths, Ph. D. Thesis, Waterford Institute of Technology, Oct. 2011.
Milan Janjić, Pascal Matrices and Restricted Words, J. Int. Seq., Vol. 21 (2018), Article 18.5.2.
J. W. Layman, The Hankel Transform and Some of its Properties, J. Integer Sequences, 4 (2001), #01.1.5.
Huyile Liang, Jeffrey Remmel and Sainan Zheng, Stieltjes moment sequences of polynomials, arXiv:1710.05795 [math.CO], 2017, see page 16.

a(n) = T(2n-1, n-1), T given by A026736.
a(n) = T(2n, n), T given by A026670.
a(n) = T(2n+1, n+1), T given by A026725.
Row sums of triangle A054335.
Cf. A026781, A001076, A176280.

a:=[3,11,43];; for n in [4..30] do a[n]:=(2*(4*n-3)*a[n-1] - 3*(5*n-8)*a[n-2] - 2*(2*n-3)*a[n-3])/n; od; Concatenation([1], a); # G. C. Greubel, Jul 16 2019

R:=PowerSeriesRing(Rationals(), 30); Coefficients(R!( 1/(Sqrt(1-4*x)-x) )); // G. C. Greubel, Jul 16 2019

Table[SeriesCoefficient[1/(Sqrt[1-4*x]-x),{x,0,n}],{n,0,30}] (* Vaclav Kotesovec, Oct 08 2012 *)

{a(n)= if(n<0, 0, polcoeff( 1/(sqrt(1 -4*x +x*O(x^n)) -x), n))} /* Michael Somos, Apr 20 2007 */

my(x='x+O('x^66)); Vec( 1/(sqrt(1-4*x)-x) ) \\ Joerg Arndt, May 04 2013

(1/(sqrt(1-4*x)-x)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jul 16 2019

From Wolfdieter Lang, Mar 21 2000: (Start)
G.f.: 1/(sqrt(1-4*x)-x).
a(n) = Sum_{i=1..n} a(i-1)*binomial(2*(n-i), n-i) + binomial(2*n, n), n >= 1, a(0)=1. (End)
G.f.: 1/(1 -x -2*x*c(x)) where c(x) = g.f. for Catalan numbers A000108. - Michael Somos, Apr 20 2007
From Paul Barry, Jan 25 2009: (Start)
G.f.: 1/(1 - 3xc(x) + x^2*c(x)^2);
G.f.: 1/(1-3x-2x^2/(1-2x-x^2/(1-2x-x^2/(1-2x-x^2/(1-... (continued fraction).
a(0) = 1, a(n) = Sum_{k=0..n} (k/(2n-k))*C(2n-k,n-k)*F(2k+2). (End)
a(n) = Sum_{k=0..n} A039599(n,k) * A000045(k+2). - Philippe Deléham, Feb 11 2009
From Paul Barry, Feb 08 2009: (Start)
G.f.: 1/(1-x/(1-2x/(1-x/(1-x/(1-x/(1-x/(1-x/(1-... (continued fraction);
G.f. of 1,1,3,... is 1/(1-x-2x/(1-x/(1-x/(1-x/(1-... (continued fraction). (End)
From Gary W. Adamson, Jul 14 2011: (Start)
a(n) = the upper left term in M^n, M = the infinite square production matrix:
  3, 2, 0, 0, 0, 0, ...
  1, 1, 1, 0, 0, 0, ...
  1, 1, 1, 1, 0, 0, ...
  1, 1, 1, 1, 1, 0, ...
  1, 1, 1, 1, 1, 1, ...
  ... (End)
From Vaclav Kotesovec, Oct 08 2012: (Start)
D-finite with recurrence: n*a(n) = 2*(4*n-3)*a(n-1) - 3*(5*n-8)*a(n-2) - 2*(2*n-3)*a(n-3).
a(n) ~ (2+sqrt(5))^n/sqrt(5). (End)
a(n) = Sum_{k=0..n+1} 4^(n+1-k) * binomial(n-k/2,n+1-k). - Seiichi Manyama, Mar 30 2025
From Peter Luschny, Mar 30 2025: (Start)
a(n) = 4^n*(binomial(n-1/2, n)*hypergeom([1, (1-n)/2, -n/2], [1/2, 1/2-n], -1/4) + hypergeom([(1-n)/2, 1-n/2], [1-n], -1/4)/4) for n > 0.
a(n) = A001076(n) + A176280(n). (End)

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A014445 Even Fibonacci numbers; or, Fibonacci(3*n).

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A007805 a(n) = Fibonacci(6*n + 3)/2.

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A096270 Fixed point of the morphism 0->01, 1->011.

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A024718 a(n) = (1/2)*(1 + Sum_{k=0..n} binomial(2*k, k)).

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A124758 Product of the parts of the compositions in standard order.

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A024718 a(n) = (1/2)(1 + Sum_{k=0..n} binomial(2k, k)).