A104449 -id:A104449 - cp's OEIS frontend

A000285 a(0) = 1, a(1) = 4, and a(n) = a(n-1) + a(n-2) for n >= 2.

1, 4, 5, 9, 14, 23, 37, 60, 97, 157, 254, 411, 665, 1076, 1741, 2817, 4558, 7375, 11933, 19308, 31241, 50549, 81790, 132339, 214129, 346468, 560597, 907065, 1467662, 2374727, 3842389, 6217116, 10059505, 16276621, 26336126, 42612747, 68948873, 111561620, 180510493, 292072113, 472582606

Offset: 0

N. J. A. Sloane

a(n-1) = Sum_{k=0..ceiling((n-1)/2)} P(4;n-1-k,k), n >= 1, with a(-1)=3. These are the sums over the SW-NE diagonals in P(4;n,k), the (4,1) Pascal triangle A093561. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs. Also SW-NE diagonal sums in the Pascal (1,3) triangle A095660.
In general, for a Fibonacci sequence beginning with 1,b we have a(n) = (2^(-1-n)*((1-sqrt(5))^n*(1+sqrt(5)-2b) + (1+sqrt(5))^n*(-1+sqrt(5)+2b)))/sqrt(5). In this case we have b=4. - Herbert Kociemba, Dec 18 2011
Pisano period lengths: 1, 3, 8, 6, 20, 24, 16, 12, 24, 60, 5, 24, 28, 48, 40, 24, 36, 24, 18, 60, ... - R. J. Mathar, Aug 10 2012
a(n) = number of independent vertex subsets (i.e., the Merrifield-Simmons index) of the tree obtained from the path tree P_{n-1} by attaching two pendant edges to one of its endpoints (n >= 2). Example: if n=3, then we have the star tree with edges ab, ac, ad; it has 9 independent vertex subsets: empty, a, b, c, d, bc, cd, bd, bcd.
For n >= 2, the number a(n-1) is the dimension of a commutative Hecke algebra of type D_n with independent parameters. See Theorem 1.4 and Corollary 1.5 in the link "Hecke algebras with independent parameters". - Jia Huang, Jan 20 2019
For n >= 1, a(n) is the number of edge covers of the tadpole graph T_{3,n-1} with T_{3,0} interpreted as just the cycle C_3. Example: If n=2, we have C_3 and P_1 joined by a bridge, which is just the triangle with a pendant, and this graph has 5 edge covers. In general, because of the path portion of the graph, the number of edge covers of T{3,n-1} satisfies the same recurrence as Fibonacci sequence and it starts with 4,5. - Feryal Alayont, Aug 27 2023
Eswarathasan (1978) called these numbers "pseudo-Fibonacci numbers", and proved that 1, 4, and 9 are the only squares in this sequence. If the recurrence is extended to negative indices, then there is only one more square, a(-9) = 81. Eswarathasan (1979) proved that none of the terms (even with negative indices) are twice a square. - Amiram Eldar, Mar 09 2024
For n>2, a(n) + (-1)^ceiling(n/2) is the number of ways to tile this strip of length n-1, with a central staircase, using unit squares and dominoes:
           _
         ||
 _____|_||________
|||_|||_|||_|||. - Greg Dresden and Runhe Zhang, Sep 07 2024

			G.f. = 1 + 4*x + 5*x^2 + 9*x^3 + 14*x^4 + 23*x^5 + 37*x^6 + 60*x^7 + ...

Richard E. Merrifield and Howard E. Simmons, Topological Methods in Chemistry, Wiley, New York, 1989. pp. 131.
Joe Roberts, Lure of the Integers, Math. Assoc. America, 1992, p. 224.
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..500
Ben Adenbaum, Jennifer Elder, Pamela E. Harris, and J. Carlos Martínez Mori, Boolean intervals in the weak Bruhat order of a finite Coxeter group, arXiv:2403.07989 [math.CO], 2024. See pp. 2, 10.
Brandon Avila and Tanya Khovanova, Free Fibonacci Sequences, arXiv preprint arXiv:1403.4614 [math.NT], 2014 and J. Int. Seq. 17 (2014) # 14.8.5.
Alfred Brousseau, Seeking the lost gold mine or exploring Fibonacci factorizations, Fib. Quart., 3 (1965), 129-130.
Alfred Brousseau, Fibonacci and Related Number Theoretic Tables, Fibonacci Association, San Jose, CA, 1972. See p. 53.
A. Eswarathasan, On Square Pseudo-Fibonacci Numbers, Fibonacci Quarterly, Vol. 16, No. 4 (1978), pp. 310-314.
A. Eswarathasan, On Pseudo-Fibonacci Numbers of the Form 2S^2, Where S is an Integer, Fibonacci Quarterly, Vol. 17, No. 2 (1979), pp. 142-147.
Jia Huang, Hecke algebras with independent parameters, arXiv preprint arXiv:1405.1636 [math.RT], 2014; Journal of Algebraic Combinatorics 43 (2016) 521-551.
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.
José L. Ramírez, Gustavo N. Rubiano, and Rodrigo de Castro, A Generalization of the Fibonacci Word Fractal and the Fibonacci Snowflake, arXiv preprint arXiv:1212.1368 [cs.DM], 2012.
Index entries for linear recurrences with constant coefficients, signature (1,1).

Essentially the same as A104449, which only has A104449(0)=3 prefixed.
Cf. A090888, A101220, A109754, A091157 (subsequence of primes).
Cf. A013655, A131775.

F:=Fibonacci;; List([0..40], n-> F(n+2) +2*F(n) ); # G. C. Greubel, Nov 08 2019

a000285 n = a000285_list !! n
a000285_list = 1 : 4 : zipWith (+) a000285_list (tail a000285_list)
-- Reinhard Zumkeller, Apr 28 2011

a0:=1; a1:=4; [GeneralizedFibonacciNumber(a0,a1,n): n in [0..30]]; // Bruno Berselli, Feb 12 2013

with(combinat):a:=n->2*fibonacci(n)+fibonacci(n+2): seq(a(n), n=0..34);

LinearRecurrence[{1,1},{1,4},40] (* or *) Table[(3*LucasL[n]- Fibonacci[n])/2,{n,40}] (* Harvey P. Dale, Jul 18 2011 *)
a[ n_]:= Fibonacci[n] + LucasL[n+1]; (* Michael Somos, May 28 2014 *)

a[0]:1$ a[1]:4$ a[n]:=a[n-1]+a[n-2]$ makelist(a[n],n,0,30); /* Martin Ettl, Oct 25 2012 */

Vec((1+3*x)/(1-x-x^2)+O(x^40)) \\ Charles R Greathouse IV, Nov 20 2012

f=fibonacci; [f(n+2) +2*f(n) for n in (0..40)] # G. C. Greubel, Nov 08 2019

G.f.: (1+3*x)/(1-x-x^2). - Simon Plouffe in his 1992 dissertation
Row sums of A131775 starting (1, 4, 5, 9, 14, 23, ...). - Gary W. Adamson, Jul 14 2007
a(n) = 2*Fibonacci(n) + Fibonacci(n+2). - Zerinvary Lajos, Oct 05 2007
a(n) = ((1+sqrt(5))^n - (1-sqrt(5))^n)/(2^n*sqrt(5)) + (3/2)* ((1+sqrt(5))^(n-1) - (1-sqrt(5))^(n-1))/(2^(n-2)*sqrt(5)). Offset 1. a(3)=5. - Al Hakanson (hawkuu(AT)gmail.com), Jan 14 2009
a(n) = 3*Fibonacci(n+2) - 2*Fibonacci(n+1). - Gary Detlefs, Dec 21 2010
a(n) = A104449(n+1). - Michael Somos, Apr 07 2012
From Michael Somos, May 28 2014: (Start)
a(n) = A101220(3, 0, n+1).
a(n) = A109754(3, n+1).
a(k) = A090888(2, k-1), for k > 0.
a(-1 - n) = (-1)^n * A013655(n).
a(n) = Fibonacci(n) + Lucas(n+1), see Mathematica field. (End)
11*Fibonacci(n+1) = a(n+3) - a(n-2) = 3*a(n-1) + 2*a(n). - Manfred Arens and Michel Marcus, Jul 14 2014
a(n) = (9*F(n) + F(n-3))/2. - J. M. Bergot, Jul 15 2017
a(n-1) = 3 * A000045(n) + A000045(n+1). - R. J. Mathar, Feb 14 2024
E.g.f.: (cosh(x/2) + sinh(x/2))*(5*cosh(sqrt(5)*x/2) + 7*sqrt(5)*sinh(sqrt(5)*x/2))/5. - Stefano Spezia, Dec 31 2024

A013655 a(n) = F(n+1) + L(n), where F(n) and L(n) are Fibonacci and Lucas numbers, respectively.

Original entry on oeis.org

3, 2, 5, 7, 12, 19, 31, 50, 81, 131, 212, 343, 555, 898, 1453, 2351, 3804, 6155, 9959, 16114, 26073, 42187, 68260, 110447, 178707, 289154, 467861, 757015, 1224876, 1981891, 3206767, 5188658, 8395425, 13584083, 21979508, 35563591, 57543099, 93106690, 150649789, 243756479

Offset: 0

Mohammad K. Azarian

Apart from initial term, same as A001060.
Pisano period lengths same as for A001060. - R. J. Mathar, Aug 10 2012
The beginning of this sequence is the only sequence of four consecutive primes in a Fibonacci-type sequence. - Franklin T. Adams-Watters, Mar 21 2015
(a(2*k), a(2*k+1)) give for k >= 0 the proper positive solutions of one of two families (or classes) of solutions (x, y) of the indefinite binary quadratic form x^2 + x*y - y^2 of discriminant 5 representing 11. The other family of such solutions is given by (x2, y2) = (b(2*k), b(2*k+1)) with b = A104449. See the formula in terms of Chebyshev S polynomials S(n, 3) = A001906(n+1) below, which follows from the fundamental solution (3, 2) by applying positive powers of the automorphic matrix A^k = Matrix([A(k), B(k)], [B(k), A(k+1)]), with A(k) = S(k-1, 3) - S(k-2, 3) and B(k) = S(k-1, 3). See also A089270 with the Alfred Brousseau link with D = 11. - Wolfdieter Lang, May 28 2019
For n>1, a(n) is the number of ways to tile a strip of length n+1 with black and white squares and white dominos, where there must be exactly one black square and it must appear amongst the first three cells. - Greg Dresden and Emma Li, Aug 24 2024

G. C. Greubel, Table of n, a(n) for n = 0..2500
Mark W. Coffey, James L. Hindmarsh, Matthew C. Lettington, and John Pryce, On Higher Dimensional Interlacing Fibonacci Sequences, Continued Fractions and Chebyshev Polynomials, arXiv:1502.03085 [math.NT], 2015 (see p. 31).
Rigoberto Flórez, Robinson A. Higuita, and Antara Mukherjee, The Geometry of some Fibonacci Identities in the Hosoya Triangle, arXiv:1804.02481 [math.NT], 2018.
Tanya Khovanova, Recursive Sequences
Shaoxiong Yuan, Generalized Identities of Certain Continued Fractions, arXiv:1907.12459 [math.NT], 2019.
Index entries for linear recurrences with constant coefficients, signature (1,1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A000045, A001060, A001906, A089270, A104449.

[2*Fibonacci(n-3)+Fibonacci(n): n in [2..41]]; // Vincenzo Librandi, Apr 16 2011

[GeneralizedFibonacciNumber(3, 2, n): n in [0..39]]; // Arkadiusz Wesolowski, Mar 16 2016

with(combinat): a:=n->2*fibonacci(n-1)+fibonacci(n+2): seq(a(n), n=0..40); # Zerinvary Lajos, Oct 05 2007

LinearRecurrence[{1, 1}, {3, 2}, 40] (* or *)
Table[Fibonacci[n + 1] + LucasL[n], {n, 0, 40}] (* or *)
Table[Fibonacci[n + 3] + Fibonacci[n - 3] - 3*Fibonacci[n], {n,2,40}] (* Bruno Berselli, Dec 30 2016 *)

a(n)=([0,1; 1,1]^n*[3;2])[1,1] \\ Charles R Greathouse IV, Sep 24 2015

a(n)=2*fibonacci(n-3) + fibonacci(n) \\ Charles R Greathouse IV, Sep 24 2015

a(n) = A000045(n+1) + A000032(n).
a(n) = a(n-1) + a(n-2).
a(n) = F(n+3) - F(n-2) for n>1, where F=A000045. - Gerald McGarvey, Jul 10 2004
a(n) = 2*F(n-3) + F(n) for n>1. - Zerinvary Lajos, Oct 05 2007
G.f.: (3-x)/(1-x-x^2). - Philippe Deléham, Nov 19 2008
a(n) = Sum_{k = n-3..n+1} F(k). - Gary Detlefs, Dec 30 2012
a(n) = ((3*sqrt(5)+1)*(((1+sqrt(5))/2)^n)+(3*sqrt(5)-1)*(((1-sqrt(5))/2)^n))/(2*sqrt(5)). - Bogart B. Strauss, Jul 19 2013
a(n) = F(n+3) + F(n-3) - 3*F(n) for n>1. - Bruno Berselli, Dec 29 2016
Bisection: a(2*k) = 3*S(k, 3) - 4*S(k-1, 3), a(2*k+1) = 2*S(k, 3) + S(k-1, 3), for k >= 0, with the Chebyshev S(n, 3) polynomials from A001906(n+1) for n >= -1. - Wolfdieter Lang, May 28 2019
a(3n + 2)/a(3n - 1) = continued fraction 4,4,4,...,4,-5 (that's n 4's followed by a single -5). - Greg Dresden and Shaoxiong Yuan, Jul 16 2019
E.g.f.: ((- 1 + 3*sqrt(5))*exp((1/2)*(1 - sqrt(5))*x) + (1 + 3*sqrt(5))*exp((1/2)*(1 + sqrt(5))*x))/(2*sqrt(5)). - Stefano Spezia, Jul 17 2019
a(n) = (F(3n+1) - F(n+1)^3)/(F(n)^2) for n>1, where F(n) = A000045(n). - Michael Tulskikh, Jul 22 2020
a(n) = 3 * Sum_{k=0..n-2} A168561(n-2,k) + 2 * Sum_{k=0..n-1} A168561(n-1,k), n>0. - R. J. Mathar, Feb 14 2024

Definition corrected by Gary Detlefs, Dec 30 2012

A001060 a(n) = a(n-1) + a(n-2) with a(0)=2, a(1)=5. Sometimes called the Evangelist Sequence.

Original entry on oeis.org

2, 5, 7, 12, 19, 31, 50, 81, 131, 212, 343, 555, 898, 1453, 2351, 3804, 6155, 9959, 16114, 26073, 42187, 68260, 110447, 178707, 289154, 467861, 757015, 1224876, 1981891, 3206767, 5188658, 8395425, 13584083, 21979508, 35563591, 57543099, 93106690, 150649789

Offset: 0

N. J. A. Sloane

Literally the same as A013655(n+1), since A001060(-1) = A013655(0) = 3. - Eric W. Weisstein, Jun 30 2017
Used by the Sofia Gubaidulina and other composers. - Ian Stewart, Jun 07 2012
From a(2) on, sums of five consecutive Fibonacci numbers; the subset of primes is essentially in A153892. - R. J. Mathar, Mar 24 2010
Pisano period lengths: 1, 3, 8, 6, 20, 24, 16, 12, 24, 60, 10, 24, 28, 48, 40, 24, 36, 24, 18, 60, ... (is this A001175?). - R. J. Mathar, Aug 10 2012
Also the number of independent vertex sets and vertex covers in the (n+1)-pan graph. - Eric W. Weisstein, Jun 30 2017
From Wajdi Maaloul, Jun 10 2022: (Start)
For n > 0, a(n) is the number of ways to tile the figure below with squares and dominoes (a strip of length n+1 that contains a vertical strip of height 3 in its second tile). For instance, a(4) is the number of ways to tile this figure (of length 5) with squares and dominoes.
     _
    |_|
   ||_______
  |||_|||_|
(End)

R. V. Jean, Mathematical Approach to Pattern and Form in Plant Growth, Wiley, 1984. See p. 5.
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).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Alfred Brousseau, Seeking the lost gold mine or exploring Fibonacci factorizations, Fib. Quart., 3 (1965), 129-130.
Alfred Brousseau, Fibonacci and Related Number Theoretic Tables, Fibonacci Association, San Jose, CA, 1972. See p. 52.
Paul Coleman, An Introduction to the Music of Sofia Gubaidulina
Tanya Khovanova, Recursive Sequences
Casey Mongoven, Fibonacci Pitch Sequences. - _Ian Stewart_, Jun 07 2012
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
Eric Weisstein's World of Mathematics, Independent Vertex Set
Eric Weisstein's World of Mathematics, Pan Graph
Eric Weisstein's World of Mathematics, Vertex Cover
Index entries for linear recurrences with constant coefficients, signature (1,1).

Cf. A000032, A000045, A000285, A104449.

Apart from initial term, same as A013655.

F:=Fibonacci;; List([0..40], n-> F(n+4) - F(n-1) ); # G. C. Greubel, Sep 19 2019

I:=[2,5]; [n le 2 select I[n] else Self(n-1)+Self(n-2): n in [1..50]]; // Vincenzo Librandi, Jan 16 2012

a0:=2; a1:=5; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..35]]; // Bruno Berselli, Feb 12 2013

with(combinat): a:= n-> 2*fibonacci(n)+fibonacci(n+3): seq(a(n), n=0..40); # Zerinvary Lajos, Oct 05 2007
A001060:=-(2+3*z)/(-1+z+z**2); # conjectured by Simon Plouffe in his 1992 dissertation

Table[Fibonacci[n+4] -Fibonacci[n-1], {n, 0, 50}] (* Vladimir Joseph Stephan Orlovsky, Nov 23 2009 *)
LinearRecurrence[{1,1}, {2,5}, 50] (* Vincenzo Librandi, Jan 16 2012 *)
Table[Fibonacci[n+2] + LucasL[n+1], {n, 0, 40}] (* Eric W. Weisstein, Jun 30 2017 *)
CoefficientList[Series[(2+3x)/(1-x-x^2), {x, 0, 40}], x] (* Eric W. Weisstein, Sep 22 2017 *)

a(n)=6*fibonacci(n)+fibonacci(n-3) \\ Charles R Greathouse IV, Jul 14 2017

a(n)=([0,1; 1,1]^n*[2;5])[1,1] \\ Charles R Greathouse IV, Jul 14 2017

f=fibonacci; [f(n+4) - f(n-1) for n in (0..40)] # G. C. Greubel, Sep 19 2019

a(n) = 2*Fibonacci(n) + Fibonacci(n+3). - Zerinvary Lajos, Oct 05 2007
a(n) = Fibonacci(n+4) - Fibonacci(n-1) for n >= 1. - Ian Stewart, Jun 07 2012
a(n) = Fibonacci(n) + 2*Fibonacci(n+2) = 5*Fibonacci(n) + 2*Fibonacci(n-1). The ratio r(n) := a(n+2)/a(n) satisfies the recurrence r(n+1) = (2*r(n) - 1)/(r(n) - 1). If M denotes the 2 X 2 matrix [2, -1; 1, -1] then [a(n+2), a(n)] = M^n[2, -1]. - Peter Bala, Dec 06 2013
a(n) = 6*F(n) + F(n-3), for F(n)=A000045. - J. M. Bergot, Jul 14 2017
a(n) = -(-1)^n*A000285(-2-n) = -(-1)^n*A104449(-1-n) for all n in Z. - Michael Somos, Oct 28 2018
E.g.f.: 2*exp(x/2)*(5*cosh(sqrt(5)*x/2) + 4*sqrt(5)*sinh(sqrt(5)*x/2))/5. - Stefano Spezia, May 26 2025

More terms from James Sellers, May 04 2000

A356988 a(n) = n - a^[2](n - a^[3](n-1)) with a(1) = 1, where a^[2](n) = a(a(n)) and a^[3](n) = a(a(a(n))).

Original entry on oeis.org

1, 1, 2, 3, 3, 4, 5, 5, 6, 7, 8, 8, 8, 9, 10, 11, 12, 13, 13, 13, 13, 14, 15, 16, 17, 18, 19, 20, 21, 21, 21, 21, 21, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 34, 34, 34, 34, 34, 34, 34, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 55, 55, 55, 55, 55

Offset: 1

Peter Bala, Sep 08 2022

This is the second sequence in a family of nested-recurrent sequences with apparently similar structure defined as follows. Given a sequence s = {s(n) : n >= 1} we define the k-th iterated sequence s^[k] by putting s^[1](n) = s(n) and setting s^[k](n) = s^[k-1](s(n)) for k >= 2. For k >= 1, we define a nested-recurrent sequence, dependent on k, by putting u(1) = 1 and setting u(n) = n - u^[k](n - u^[k+1](n-1)) for n >= 2. This is the case k = 2. For other cases see A006165 (k = 1), A356989 (k = 3) and A356990 (k = 4).
The sequence is slow, that is, for n >= 1, a(n+1) - a(n) is either 0 or 1. The sequence is unbounded.
The line graph of the sequence {a(n)} thus consists of a series of plateaus (where the value of the ordinate a(n) remains constant as n increases) joined by lines of slope 1.
The sequence of plateau heights begins 3, 5, 8, 13, 21, 34, 55, ..., the Fibonacci numbers A000045.
The plateaus start at abscissa values n = 4, 7, 11, 18, 29, 47, 76, ..., the Lucas numbers A000032, and finish at abscissa values n = 5, 8, 13, 21, 34, 55, 89, ..., the Fibonacci numbers. The sequence of plateau lengths 1, 1, 2, 3, 5, 8, 13, ... is thus the Fibonacci sequence.
The iterated sequences{a^[k](n) : n >= 1}, k = 2, 3,..., share similar properties to the present sequence. See the Example section below.

			Related sequences:
1) The square of the sequence: {a^[2](n) : n >= 1} = {a(a(n)) : n >= 1}. The first few terms are
  1, 1, 1, 2, 2, 3, 3, 3, 4, 5, 5, 5, 5, 6, 7, 8, 8, 8, 8, 8, 8, 9, 10, 11, 12, 13, 13, 13, 13, 13, 13, 13, 13, 13, 14, 15, 16, 17, 18, 19, 20, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 35, 36, 37, 38, 39, ...
The sequence is slow. The line graph of the sequence has plateaus of height Fibonacci(k), k >= 2, starting at abscissa value 2*Fibonacci(k) and ending at abscissa Fibonacci(k+2).
2) The cube of the sequence: {a^[3](n) : n >= 1} = {a(a(a(n))) : n >= 1}. The first few terms are
  1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 3, 3, 3, 4, 5, 5, 5, 5, 5, 5, 5, 6, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 10, 11, 12, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 14, 15, 16, 17, 18, 19, 20, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 22, 23, 24, 25, 26, 27, ...
The line graph of the sequence has plateaus of height Fibonacci(k), k >= 2, starting at abscissa value 3*Fibonacci(k) and ending at abscissa Fibonacci(k+3).

Peter Bala, Notes on A356988

Cf. A000032, A000045, A001611, A006165, A022114, A022120, A022367, A104449, A356989, A356990, A356991 - A356999.

a := proc(n) option remember; if n = 1 then 1 else n - a(a(n - a(a(a(n-1))))) end if; end proc:
seq(a(n), n = 1..100);

a(n+1) - a(n) = 0 or 1.
The terms of the sequence are completely determined by the following two results:
a) for n >= 2, a(L(n-1) + j) = F(n) for 0 <= j <= F(n-3), where F(n) = A000045(n), the n-th Fibonacci number with F(-1) = 1 and L(n) = A000032(n), the n-th Lucas number;
b) for n >= 2, a(F(n+1) + j) = F(n) + j for 0 <= j <= F(n-1).
Hence a(F(n+2)) = a(F(n+1)) + a(F(n)) for n >= 2 and a(L(n+2)) = a(L(n+1)) + a(L(n)) for n >= 0.
a(2*F(n)) = Lucas(n-1) for n >= 2;
a(3*F(n)) = 2*F(n) for n >= 1;
a(4*F(n)) = F(n+2) for n >= 2;
a(5*F(n)) = 4*F(n) - F(n-1) = A022120(n-2) for n >= 2.
a(2*L(n)) = F(n) + 3*F(n-1) = A104449(n) for n >= 0;
a(3*L(n)) = F(n+3) for n >= 3;
a(4*L(n)) = F(n+4) - L(n-3) = A022114(n-1) for n >= 3;
a(5*L(n)) = 11*F(n-1) + F(n-4) = A022367(n-1) for n >= 4.
For n >= 1, m >= 2, a(F(m*n)) = F(m*n-1) and a(L(m*n)) = F(m*n+1). Hence
a(L(m*n)) + a(F(m*n)) = L(m*n) and a(L(m*n)) - a(F(m*n)) = F(m*n).
Conjectures:
1) a(n) + a^[2](n - a^[2](n - a^[2](n))) = n for n >= 2.
2) If k >= 2 and  m = 2*k - 1 then a(m*n - a(k*n)) = a(m*n - a(m*n - a(m*n - a(k*n)))).

A054486 Expansion of (1+2x)/(1-3x+x^2).

Original entry on oeis.org

1, 5, 14, 37, 97, 254, 665, 1741, 4558, 11933, 31241, 81790, 214129, 560597, 1467662, 3842389, 10059505, 26336126, 68948873, 180510493, 472582606, 1237237325, 3239129369, 8480150782, 22201322977, 58123818149, 152170131470, 398386576261, 1042989597313

Offset: 0

Barry E. Williams, May 06 2000

Binomial transform of A000285. - R. J. Mathar, Oct 26 2011

			G.f. = 1 + 5*x + 14*x^2 + 37*x^3 + 97*x^4 + 254*x^5 + 665*x^6 + 1741*x^7 + ...

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 122-125, 194-196.

G. C. Greubel, Table of n, a(n) for n = 0..1000
I. Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969), pp. 181-193.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (3,-1).

Cf. A000032, A000045, A000285, A002878, A100545, A104449, A228208.

F:=Fibonacci;; List([0..30], n-> F(2*n+2) +2*F(2*n) ); # G. C. Greubel, Nov 08 2019

R:=PowerSeriesRing(Integers(), 30); Coefficients(R!( (1+2*x)/(1-3*x+x^2)) ); // Marius A. Burtea, Nov 05 2019

a:=[1,5]; [n le 2 select a[n] else 3*Self(n-1)-Self(n-2): n in [1..30]]; // Marius A. Burtea, Nov 05 2019

with(combinat); f:=fibonacci; seq(f(2*n+2)+2*f(2*n), n=0..30); # G. C. Greubel, Nov 08 2019

CoefficientList[Series[(2*z+1)/(z^2-3*z+1), {z, 0, 30}], z] (* Vladimir Joseph Stephan Orlovsky, Jul 15 2011 *)
a[ n_]:= 3 Fibonacci[2n] + Fibonacci[2n+1]; (* Michael Somos, Mar 17 2015 *)
LinearRecurrence[{3,-1},{1,5},40] (* Harvey P. Dale, Apr 24 2019 *)

Vec((1+2*x)/(1-3*x+x^2)+O(x^99)) \\ Charles R Greathouse IV, Jul 15 2011

{a(n) = 3*fibonacci(2*n) + fibonacci(2*n+1)}; /* Michael Somos, Mar 17 2015 */

f=fibonacci; [f(2*n+2) +2*f(2*n) for n in (0..30)] # G. C. Greubel, Nov 08 2019

a(n) = 3*a(n-1) - a(n-2), a(0)=1, a(1)=5.
a(n) = (5*(((3+sqrt(5))/2)^n - ((3-sqrt(5))/2)^n) - (((3+sqrt(5))/2)^(n-1) - ((3-sqrt(5))/2)^(n-1)))/sqrt(5).
a(n) + 7*A001519(n) = A005248(n). - Creighton Dement, Oct 30 2004
a(n) = Lucas(2*n+1) + Fibonacci(2*n) = A002878(n) + A001906(n) = A025169(n-1) + A001906(n+1).
a(n) = (-1)^n*Sum_{k = 0..n} A238731(n,k)*(-6)^k. - Philippe Deléham, Mar 05 2014
0 = -11 + a(n)^2 - 3*a(n)*a(n+1) + a(n+1)^2 for all n in Z. - Michael Somos, Mar 17 2015
a(n) = -2*F(n)^2 + 6*F(n)*F(n+1) + F(n+1)^2 for all n in Z where F = Fibonacci. - Michael Somos, Mar 17 2015
a(n) = 3*F(2*n) + F(2*n+1) for all n in Z where F = Fibonacci. - Michael Somos, Mar 17 2015
a(n) = -A100545(-2-n) for all n in Z. - Michael Somos, Mar 17 2015
a(n) = A000285(2*n) = A228208(2*n+1) = A104449(2*n+1) for all n in Z. - Michael Somos, Mar 17 2015
From Klaus Purath, Nov 05 2019: (Start)
a(n) = (a(n-m) + a(n+m))/Lucas(2*m), m <= n.
a(n) = sum of 2*m+1 consecutive terms starting with a(n-m) divided by Lucas(2*m+1), m <= n.
a(n) = alternating sum of 2*m+1 consecutive terms starting with a(n-m) divided by Fibonacci(2*m+1), m <= n.
a(n) + a(n+1) = sum of 2*m+2 consecutive terms starting with a(n-m) divided by Fibonacci(2*m+2), m <= n.
a(n) + a(n+1) = (a(n-m) + a(n+m+1))/Fibonacci(2*m+1), m <= n.
The following formulas are extended to negative indexes:
a(n) = 3*Fibonacci(2*n+1) - Fibonacci(2*n-3).
a(n) = (Fibonacci(2*n+5) - 3* Fibonacci(2*n-1))/2.
a(n) = (4*Lucas(2*n+2) - Lucas(2*n-4))/5.
a(n) = Fibonacci(2*n+5) - 4*Fibonacci(2*n+1).
a(n) = (5*Fibonacci(2*n+5) - Fibonacci(2*n-7))/12. (End)
E.g.f.: exp(-(1/2)*(-3+sqrt(5))*x)*(-7 + sqrt(5) + (7 + sqrt(5))*exp(sqrt(5)*x))/(2*sqrt(5)). - Stefano Spezia, Nov 19 2019
a(n) = 3*n + 1 + Sum_{k=1..n} k*a(n-k). - Yu Xiao, Jun 20 2020

"a(1)=5", not "a(0)=5" from Dan Nielsen (nielsed(AT)uah.edu), Sep 10 2009

A192878 Constant term in the reduction by (x^2 -> x + 1) of the polynomial p(n,x) given in Comments.

Original entry on oeis.org

3, 0, 4, 5, 18, 42, 115, 296, 780, 2037, 5338, 13970, 36579, 95760, 250708, 656357, 1718370, 4498746, 11777875, 30834872, 80726748, 211345365, 553309354, 1448582690, 3792438723, 9928733472, 25993761700, 68052551621, 178163893170, 466439127882, 1221153490483

Offset: 0

Clark Kimberling, Jul 11 2011

The polynomial p(n,x) is defined by p(0,x) = 1, p(1,x) = x + 1, and p(n,x) = x*p(n-1,x) + 2*(x^2)*p(n-1,x) + 1. See A192872.

For n>0, also the coefficient of x in the reduction x^2 -> x + 1 of the polynomial A000285(n-1)*x^(n-1). - R. J. Mathar, Jul 12 2011

			The first six polynomials and reductions:
p(0,x) = 3 -> 3
p(1,x) = x -> x
p(2,x) = 4*x^2 -> 4 + 4*x
p(3,x) = 5*x^3 -> 5 + 10*x
p(4,x) = 9*x^4 -> 18 + 27*x
p(5,x) = 14*x^5 -> 42 + 27*x
In general, p(n,x) = A104449(n)*x^n -> A192878(n) + A192879(n)*x.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A000285, A192872, A192879.

a:=[3,0,4];; for n in [4..30] do a[n]:=2*a[n-1]+2*a[n-2]-a[n-3]; od; a; # G. C. Greubel, Jan 08 2019

m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (3-2*x^2-6*x)/((1+x)*(x^2-3*x+1)) )); // G. C. Greubel, Jan 08 2019

q = x^2; s = x + 1; z = 25;
p[0, x_] := 3; p[1, x_] := x;
p[n_, x_] := p[n - 1, x]*x + p[n - 2, x]*x^2;
Table[Expand[p[n, x]], {n, 0, 7}]
reduce[{p1_, q_, s_, x_}] :=
FixedPoint[(s PolynomialQuotient @@ #1 +
       PolynomialRemainder @@ #1 &)[{#1, q, x}] &, p1]
t = Table[reduce[{p[n, x], q, s, x}], {n, 0, z}];
u1 = Table[Coefficient[Part[t, n], x, 0], {n, 1, z}]  (* A192878 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}]  (* A192879 *)
FindLinearRecurrence[u1]
FindLinearRecurrence[u2]
LinearRecurrence[{2,2,-1}, {3,0,4}, 30] (* G. C. Greubel, Jan 08 2019 *)

a(n) = round((2^(-n)*(7*(-2)^n-(-4+sqrt(5))*(3+sqrt(5))^n+(3-sqrt(5))^n*(4+sqrt(5))))/5) \\ Colin Barker, Sep 29 2016

Vec((3-2*x^2-6*x)/((1+x)*(x^2-3*x+1)) + O(x^40)) \\ Colin Barker, Sep 29 2016

((3-2*x^2-6*x)/((1+x)*(x^2-3*x+1))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jan 08 2019

a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
G.f.: ( 3 - 2*x^2 - 6*x ) / ( (1+x)*(1 - 3*x + x^2) ). - R. J. Mathar, May 07 2014
a(n) = (2^(-n)*(7*(-2)^n - (-4+sqrt(5))*(3+sqrt(5))^n + (3-sqrt(5))^n*(4+sqrt(5))))/5. - Colin Barker, Sep 29 2016

A192879 Coefficient of x in the reduction by (x^2 -> x + 1) of the polynomial p(n,x) given in Comments.

Original entry on oeis.org

0, 1, 4, 10, 27, 70, 184, 481, 1260, 3298, 8635, 22606, 59184, 154945, 405652, 1062010, 2780379, 7279126, 19057000, 49891873, 130618620, 341963986, 895273339, 2343856030, 6136294752, 16065028225, 42058789924, 110111341546, 288275234715, 754714362598

Offset: 0

Clark Kimberling, Jul 11 2011

The polynomial p(n,x) is defined by p(0,x) = 1, p(1,x) = x+1, and p(n,x) = x*p(n-1,x) + 2*(x^2)*p(n-1,x) + 1. See A192872.

A192879 is also generated as the coefficient sequence of x in the reduction x^2->x+1 of the polynomial v(n,x) defined by v(0,x) = 2, v(1,x) = x+1, and v(n,x) = x*v(n-1,x) + 2*(x^2)*v(n-1,x) + 1, for n>0, v(n,x) = F(n)*x^(n-1) + L(n)*x^n, where F(n) = A000045(n) (Fibonacci numbers) and L(n) = A000032(n) (Lucas numbers).

			The first six polynomials and reductions:
  p(0,x) = 3 -> 3
  p(1,x) = x -> x
  p(2,x) = 4*x^2 -> 4+4*x
  p(3,x) = 5*x^3 -> 5+10*x
  p(4,x) = 9*x^4 -> 18+27*x
  p(5,x) = 14*x^5 -> 42+27*x
In general, p(n,x) = (A104449(n))*x^n -> A192878(n) + A192879(n)*x.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A192872, A192878.

a:=[0,1,4];; for n in [4..40] do a[n]:=2*a[n-1]+2*a[n-2]-a[n-3]; od; a; # G. C. Greubel, Jan 07 2019

I:=[0,1,4]; [n le 3 select I[n] else 2*Self(n-1) +2*Self(n-2) -Self(n-3): n in [1..40]]; // G. C. Greubel, Jan 07 2019

with(combinat); seq( fibonacci(2*n) + fibonacci(n)*fibonacci(n-1), n=0..40); # G. C. Greubel, Feb 13 2020

(See A192878.)
LinearRecurrence[{2,2,-1}, {0,1,4}, 30] (* G. C. Greubel, Jan 07 2019 *)
a[n_] := a[n] = 2*a[n-1]+2*a[n - 2]-a[n-3]; a[0] = 0; a[1]=1; a[2]=4; Table[a[n], {n,0,40}] (* Rigoberto Florez, Feb 06 2020 *)
Table[Fibonacci[n]*Fibonacci[n-1]+Fibonacci[2n], {n,0,40}] (* Rigoberto Florez, Feb 06 2020 *)

a(n) = round((2^(-1-n)*((-1)^n*2^(1+n)+(3+sqrt(5))^n*(-1+3*sqrt(5))-(3-sqrt(5))^n*(1+3*sqrt(5))))/5) \\ Colin Barker, Sep 29 2016

concat(0, Vec(x*(1+2*x)/((1+x)*(1-3*x+x^2)) + O(x^40))) \\ Colin Barker, Sep 29 2016

(x*(1+2*x)/((1+x)*(1-3*x+x^2))).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Jan 07 2019

a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3), with a(0) = 0, a(1) = 1, a(2) = 4.
G.f.: x * (1+2*x) / ((1+x) * (1-3*x+x^2)). - Colin Barker, Jun 18 2012
a(n) = (2^(-1-n) * ((-1)^n*2^(1+n) + (3+sqrt(5))^n * (-1+3*sqrt(5)) - (3-sqrt(5))^n * (1+3*sqrt(5))))/5. - Colin Barker, Sep 29 2016
a(n) = F(n-1)*F(n) + F(2n), where F(n) is a Fibonacci number. - Rigoberto Florez, Feb 06 2020
E.g.f.: (exp(-x) + exp(3*x/2) * (3*sqrt(5)*sinh(sqrt(5)*x/2) - cosh(sqrt(5)*x/2)))/5. - Stefano Spezia, Feb 06 2020
a(n)*F(n) = the number of ways to tile a 3-arm starfish (with n-1 cells on each arm and one cell in the center) using squares and dominos. - Greg Dresden and Hasita Kanamarlapudi, Oct 02 2023

A341694 Square array T(n, k) read by antidiagonals upwards, n, k > 0: T(n, k) = A227736(n, k) for k = 1..A005811(n), and T(n, k) = T(n, k - A005811(n)) + ... + T(n, k-1) for k > A005811(n).

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 2, 2, 2, 1, 1, 1, 2, 3, 1, 1, 1, 3, 2, 5, 1, 3, 2, 1, 4, 2, 8, 1, 3, 3, 3, 3, 7, 2, 13, 1, 1, 1, 3, 5, 5, 11, 2, 21, 1, 1, 2, 4, 3, 8, 9, 18, 2, 34, 1, 2, 1, 1, 5, 3, 13, 17, 29, 2, 55, 1, 2, 1, 1, 4, 9, 3, 21, 31, 47, 2, 89, 1

Offset: 1

Rémy Sigrist, Feb 17 2021

This table contains all Fibonacci sequences of order m > 0 with positive terms:
- order 1 corresponds to constant sequences (n in A126646),
- order 2 corresponds to Fibonacci-like sequences (n in A043569),
- order 3 corresponds to tribonacci-like sequences (n in A043570),
- order 4 corresponds to tetranacci-like sequences (n in A043571).
For any n > 0, the row A341746(n) corresponds to the n-th row from which the first term has been removed.

			Array T(n, k) begins:
  n\k|  1  2  3  4  5   6   7   8   9   10   11   12   13    14
  ---+---------------------------------------------------------
    1|  1  1  1  1  1   1   1   1   1    1    1    1    1     1 --> A000012
    2|  1  1  2  3  5   8  13  21  34   55   89  144  233   377 --> A000045
    3|  2  2  2  2  2   2   2   2   2    2    2    2    2     2 --> A007395
    4|  2  1  3  4  7  11  18  29  47   76  123  199  322   521 --> A000032
    5|  1  1  1  3  5   9  17  31  57  105  193  355  653  1201 --> A000213
    6|  1  2  3  5  8  13  21  34  55   89  144  233  377   610 --> A000045
    7|  3  3  3  3  3   3   3   3   3    3    3    3    3     3 --> A010701
    8|  3  1  4  5  9  14  23  37  60   97  157  254  411   665 --> A104449
    9|  1  2  1  4  7  12  23  42  77  142  261  480  883  1624 --> A275778
   10|  1  1  1  1  4   7  13  25  49   94  181  349  673  1297 --> A000288

Rémy Sigrist, Table of n, a(n) for n = 1..11325
Rémy Sigrist, Colored representation of the table for n, k = 1..1024
Rémy Sigrist, PARI program for A341694
Wikipedia, Fibonacci numbers of higher order
Index entries for sequences related to Fibonacci numbers

Cf. A005811, A043569, A043570, A043571, A126646, A136480, A227736, A341699, A341746.

Cf. A000012, A000032, A000045, A000213, A000288, A007395, A010701, A104449, A275778.

PARI
```
See Links section.
```

T(A341746(n), k) = T(n, k+1).

T(n, 1) = A136480(n).

A353595 Array read by ascending antidiagonals. Generalized Fibonacci numbers F(n, k) = (psi^(k - 1)(phi + n) - phi^(k - 1)(psi + n)) / (psi - phi) where phi = (1+sqrt(5))/2 and psi = (1-sqrt(5))/2. F(n, k) for n >= 0 and k >= 0.

Original entry on oeis.org

0, 1, 1, 2, 1, 1, 3, 1, 2, 2, 4, 1, 3, 3, 3, 5, 1, 4, 4, 5, 5, 6, 1, 5, 5, 7, 8, 8, 7, 1, 6, 6, 9, 11, 13, 13, 8, 1, 7, 7, 11, 14, 18, 21, 21, 9, 1, 8, 8, 13, 17, 23, 29, 34, 34, 10, 1, 9, 9, 15, 20, 28, 37, 47, 55, 55, 11, 1, 10, 10, 17, 23, 33, 45, 60, 76, 89, 89

Offset: 0

Peter Luschny, May 09 2022

The definition declares the Fibonacci numbers for all integers n and k. It gives the classical Fibonacci numbers as F(0, n) = A000045(n). A different enumeration is given in A352744.

			Array starts:
n\k 0, 1,  2,  3,  4,  5,  6,   7,   8,   9, ...
--------------------------------------------------------
[0]  0, 1,  1,  2,  3,  5,  8, 13,  21,  34, ... A000045
[1]  1, 1,  2,  3,  5,  8, 13, 21,  34,  55, ... A000045 (shifted once)
[2]  2, 1,  3,  4,  7, 11, 18, 29,  47,  76, ... A000032
[3]  3, 1,  4,  5,  9, 14, 23, 37,  60,  97, ... A104449
[4]  4, 1,  5,  6, 11, 17, 28, 45,  73, 118, ... [4] + A022095
[5]  5, 1,  6,  7, 13, 20, 33, 53,  86, 139, ... [5] + A022096
[6]  6, 1,  7,  8, 15, 23, 38, 61,  99, 160, ... [6] + A022097
[7]  7, 1,  8,  9, 17, 26, 43, 69, 112, 181, ... [7] + A022098
[8]  8, 1,  9, 10, 19, 29, 48, 77, 125, 202, ... [8] + A022099
[9]  9, 1, 10, 11, 21, 32, 53, 85, 138, 223, ... [9] + A022100

Peter Luschny, The Fibonacci Function.
Wikipedia, Cassini and Catalan identities.

Cf. A000045, A000032, A104449, A094588 (main diagonal).

Cf. A352744, A354265 (generalized Lucas numbers).

function fibrec(n::Int)
    n == 0 && return (BigInt(0), BigInt(1))
    a, b = fibrec(div(n, 2))
    c = a * (b * 2 - a)
    d = a * a + b * b
    iseven(n) ? (c, d) : (d, c + d)
end
function Fibonacci(n::Int, k::Int)
    k == 0 && return BigInt(n)
    k == 1 && return BigInt(1)
    k  < 0 && return (-1)^(k-1)*Fibonacci(-n - 1, 2 - k)
    a, b = fibrec(k - 1)
    a*n + b
end
for n in -6:6
    println([n], [Fibonacci(n, k) for k in -6:6])
end

f := n -> combinat:-fibonacci(n): F := (n, k) -> n*f(k - 1) + f(k):
seq(seq(F(n - k, k), k = 0..n), n = 0..11);
# The next implementation is for illustration only but is not recommended
# as it relies on floating point arithmetic. Illustrates the case n,k < 0.
phi := (1 + sqrt(5))/2: psi := (1 - sqrt(5))/2:
F := (n, k) -> (psi^(k-1)*(psi + n) - phi^(k-1)*(phi + n)) / (psi - phi):
for n from -6 to 6 do lprint(seq(simplify(F(n, k)), k = -6..6)) od;

(* Works also for n < 0 and k < 0. Uses a remark from Bill Gosper. *)
c := I*ArcSinh[1/2] - Pi/2;
F[n_, k_] := (n Sin[(k - 1) c] - I Sin[k c]) / (I^k Sqrt[5/4]);
Table[Simplify[F[n, k]], {n, 0, 6}, {k, 0, 6}] // TableForm

Functional equation extends Cassini's theorem:
F(n, k) = (-1)^(k - 1)*F(-n - 1, 2 - k).
F(n, k) = ((1 - phi)^(k - 1)*(1 - phi + n) - phi^(k - 1)*(phi + n))/(1 - 2*phi).
F(n, k) = n*fib(k - 1) + fib(k), where fib(n) are the classical Fibonacci numbers A000045 extended in the usual way for negative n.
F(n, k) - F(n-1, k) = fib(k-1).
F(n, k) = F(n, k-1) + F(n, k-2).
F(n, k) = (n*sin((k - 1)*c) - i*sin(k*c))/(i^k*sqrt(5/4)) where c = i*arcsinh(1/2) - Pi/2, for all n, k in Z. Based on a remark of Bill Gosper.

cp's OEIS Frontend

A160244 A104449(n+1)+prime(n), sum of a Lucas and the prime sequence.

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A000285 a(0) = 1, a(1) = 4, and a(n) = a(n-1) + a(n-2) for n >= 2.

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A013655 a(n) = F(n+1) + L(n), where F(n) and L(n) are Fibonacci and Lucas numbers, respectively.

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A001060 a(n) = a(n-1) + a(n-2) with a(0)=2, a(1)=5. Sometimes called the Evangelist Sequence.

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A356988 a(n) = n - a^[2](n - a^[3](n-1)) with a(1) = 1, where a^[2](n) = a(a(n)) and a^[3](n) = a(a(a(n))).

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A054486 Expansion of (1+2*x)/(1-3*x+x^2).

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A054486 Expansion of (1+2x)/(1-3x+x^2).

A353595 Array read by ascending antidiagonals. Generalized Fibonacci numbers F(n, k) = (psi^(k - 1)(phi + n) - phi^(k - 1)(psi + n)) / (psi - phi) where phi = (1+sqrt(5))/2 and psi = (1-sqrt(5))/2. F(n, k) for n >= 0 and k >= 0.