cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A000204 Lucas numbers (beginning with 1): L(n) = L(n-1) + L(n-2) with L(1) = 1, L(2) = 3.

Original entry on oeis.org

1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843, 1364, 2207, 3571, 5778, 9349, 15127, 24476, 39603, 64079, 103682, 167761, 271443, 439204, 710647, 1149851, 1860498, 3010349, 4870847, 7881196, 12752043, 20633239, 33385282, 54018521, 87403803, 141422324
Offset: 1

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Author

Keywords

Comments

See A000032 for the version beginning 2, 1, 3, 4, 7, ...
Also called Schoute's accessory series (see Jean, 1984). - N. J. A. Sloane, Jun 08 2011
L(n) is the number of matchings in a cycle on n vertices: L(4)=7 because the matchings in a square with edges a, b, c, d (labeled consecutively) are the empty set, a, b, c, d, ac and bd. - Emeric Deutsch, Jun 18 2001
This comment covers a family of sequences which satisfy a recurrence of the form a(n) = a(n-1) + a(n-m), with a(n) = 1 for n = 1..m - 1, a(m) = m + 1. The generating function is (x + m*x^m)/(1 - x - x^m). Also a(n) = 1 + n*Sum_{i=1..n/m} binomial(n - 1 - (m - 1)*i, i - 1)/i. This gives the number of ways to cover (without overlapping) a ring lattice (or necklace) of n sites with molecules that are m sites wide. Special cases: m = 2: A000204, m = 3: A001609, m = 4: A014097, m = 5: A058368, m = 6: A058367, m = 7: A058366, m = 8: A058365, m = 9: A058364.
L(n) is the number of points of period n in the golden mean shift. The number of orbits of length n in the golden mean shift is given by the n-th term of the sequence A006206. - Thomas Ward, Mar 13 2001
Row sums of A029635 are 1, 1, 3, 4, 7, ... - Paul Barry, Jan 30 2005
a(n) counts circular n-bit strings with no repeated 1's. E.g., for a(5): 00000 00001 00010 00100 00101 01000 01001 01010 10000 10010 10100. Note #{0...} = fib(n+1), #{1...} = fib(n-1), #{000..., 001..., 100...} = a(n-1), #{010..., 101...} = a(n-2). - Len Smiley, Oct 14 2001
Row sums of the triangle in A182579. - Reinhard Zumkeller, May 07 2012
If p is prime then L(p) == 1 (mod p). L(2^k) == -1 (mod 2^(k+1)) for k = 0,1,2,... - Thomas Ordowski, Sep 25 2013
Satisfies Benford's law [Brown-Duncan, 1970; Berger-Hill, 2017]. - N. J. A. Sloane, Feb 08 2017

Examples

			G.f. = x + 3*x^2 + 4*x^3 + 7*x^4 + 11*x^5 + 18*x^6 + 29*x^7 + 47*x^8 + ...
		

References

  • P. Bachmann, Niedere Zahlentheorie (1902, 1910), reprinted Chelsea, NY, 1968, vol. 2, p. 69.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 46.
  • Leonhard Euler, Introductio in analysin infinitorum (1748), sections 216 and 229.
  • G. Everest, A. van der Poorten, I. Shparlinski and T. Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. p. 255.
  • G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, p. 148.
  • Silvia Heubach and Toufik Mansour, Combinatorics of Compositions and Words, CRC Press, 2010.
  • V. E. Hoggatt, Jr., Fibonacci and Lucas Numbers. Houghton, Boston, MA, 1969.
  • R. V. Jean, Mathematical Approach to Pattern and Form in Plant Growth, Wiley, 1984. See p. 5. - N. J. A. Sloane, Jun 08 2011
  • Thomas Koshy, "Fibonacci and Lucas Numbers with Applications", John Wiley and Sons, 2001.
  • 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).
  • S. Vajda, Fibonacci and Lucas numbers and the Golden Section, Ellis Horwood Ltd., Chichester, 1989.

Crossrefs

Programs

  • Haskell
    a000204 n = a000204_list !! n
    a000204_list = 1 : 3 : zipWith (+) a000204_list (tail a000204_list)
    -- Reinhard Zumkeller, Dec 18 2011
    
  • Magma
    [Lucas(n): n in [1..30]]; // G. C. Greubel, Dec 17 2017
    
  • Maple
    A000204 := proc(n) option remember; if n <=2 then 2*n-1; else procname(n-1)+procname(n-2); fi; end;
    with(combinat): A000204 := n->fibonacci(n+1)+fibonacci(n-1);
    # alternative Maple program:
    L:= n-> (<<1|1>, <1|0>>^n. <<2, -1>>)[1, 1]:
    seq(L(n), n=1..50);  # Alois P. Heinz, Jul 25 2008
    # Alternative:
    a := n -> `if`(n=1, 1, `if`(n=2, 3, hypergeom([(1-n)/2, -n/2], [1-n], -4))):
    seq(simplify(a(n)), n=1..39); # Peter Luschny, Sep 03 2019
  • Mathematica
    c = (1 + Sqrt[5])/2; Table[Expand[c^n + (1 - c)^n], {n, 30}] (* Artur Jasinski, Oct 05 2008 *)
    Table[LucasL[n, 1], {n, 36}] (* Zerinvary Lajos, Jul 09 2009 *)
    LinearRecurrence[{1, 1},{1, 3}, 50] (* Sture Sjöstedt, Nov 28 2011 *)
    lukeNum[n_] := If[n < 1, 0, LucasL[n]]; (* Michael Somos, May 18 2015 *)
    lukeNum[n_] := SeriesCoefficient[x D[Log[1 / (1 - x - x^2)], x], {x, 0, n}]; (* Michael Somos, May 18 2015 *)
  • PARI
    A000204(n)=fibonacci(n+1)+fibonacci(n-1) \\ Michael B. Porter, Nov 05 2009
    
  • Python
    from functools import cache
    @cache
    def a(n): return [1, 3][n-1] if n < 3 else a(n-1) + a(n-2)
    print([a(n) for n in range(1, 41)]) # Michael S. Branicky, Nov 13 2022
    
  • Python
    [(i:=-1)+(j:=2)] + [(j:=i+j)+(i:=j-i) for  in range(100)] # _Jwalin Bhatt, Apr 02 2025
  • Sage
    def A000204():
        x, y = 1, 2
        while true:
           yield x
           x, y = x + y, x
    a = A000204(); print([next(a) for i in range(39)])  # Peter Luschny, Dec 17 2015
    
  • Scala
    def lucas(n: BigInt): BigInt = {
      val zero = BigInt(0)
      def fibTail(n: BigInt, a: BigInt, b: BigInt): BigInt = n match {
        case `zero` => a
        case _ => fibTail(n - 1, b, a + b)
      }
      fibTail(n, 2, 1)
    }
    (1 to 50).map(lucas()) // _Alonso del Arte, Oct 20 2019
    

Formula

Expansion of x(1 + 2x)/(1 - x - x^2). - Simon Plouffe, dissertation 1992; multiplied by x. - R. J. Mathar, Nov 14 2007
a(n) = A000045(2n)/A000045(n). - Benoit Cloitre, Jan 05 2003
For n > 1, L(n) = F(n + 2) - F(n - 2), where F(n) is the n-th Fibonacci number (A000045). - Gerald McGarvey, Jul 10 2004
a(n+1) = 4*A054886(n+3) - A022388(n) - 2*A022120(n+1) (a conjecture; note that the above sequences have different offsets). - Creighton Dement, Nov 27 2004
a(n) = Sum_{k=0..floor((n+1)/2)} (n+1)*binomial(n - k + 1, k)/(n - k + 1). - Paul Barry, Jan 30 2005
L(n) = A000045(n+3) - 2*A000045(n). - Creighton Dement, Oct 07 2005
L(n) = A000045(n+1) + A000045(n-1). - John Blythe Dobson, Sep 29 2007
a(n) = 2*Fibonacci(n-1) + Fibonacci(n), n >= 1. - Zerinvary Lajos, Oct 05 2007
L(n) is the term (1, 1) in the 1 X 2 matrix [2, -1].[1, 1; 1, 0]^n. - Alois P. Heinz, Jul 25 2008
a(n) = phi^n + (1 - phi)^n = phi^n + (-phi)^(-n) = ((1 + sqrt(5))^n + (1 - sqrt(5))^n)/(2^n) where phi is the golden ratio (A001622). - Artur Jasinski, Oct 05 2008
a(n) = A014217(n+1) - A014217(n-1). See A153263. - Paul Curtz, Dec 22 2008
a(n) = ((1 + sqrt(5))^n - (1 - sqrt(5))^n)/(2^n*sqrt(5)) + ((1 + sqrt(5))^(n - 1) - (1 - sqrt(5))^(n - 1))/(2^(n - 2)*sqrt(5)). - Al Hakanson (hawkuu(AT)gmail.com), Jan 12 2009, Jan 14 2009
From Hieronymus Fischer, Oct 20 2010 (Start)
Continued fraction for n odd: [L(n); L(n), L(n), ...] = L(n) + fract(Fib(n) * phi).
Continued fraction for n even: [L(n); -L(n), L(n), -L(n), L(n), ...] = L(n) - 1 + fract(Fib(n)*phi). Also: [L(n) - 2; 1, L(n) - 2, 1, L(n) - 2, ...] = L(n) - 2 + fract(Fib(n)*phi). (End)
INVERT transform of (1, 2, -1, -2, 1, 2, ...). - Gary W. Adamson, Mar 07 2012
L(2n - 1) = floor(phi^(2n - 1)); L(2n) = ceiling(phi^(2n)). - Thomas Ordowski, Jun 15 2012
a(n) = hypergeom([(1 - n)/2, -n/2], [1 - n], -4) for n >= 3. - Peter Luschny, Sep 03 2019
E.g.f.: 2*(exp(x/2)*cosh(sqrt(5)*x/2) - 1). - Stefano Spezia, Jul 26 2022

Extensions

Additional comments from Yong Kong (ykong(AT)curagen.com), Dec 16 2000
Plouffe Maple line edited by N. J. A. Sloane, May 13 2008

A002878 Bisection of Lucas sequence: a(n) = L(2*n+1).

Original entry on oeis.org

1, 4, 11, 29, 76, 199, 521, 1364, 3571, 9349, 24476, 64079, 167761, 439204, 1149851, 3010349, 7881196, 20633239, 54018521, 141422324, 370248451, 969323029, 2537720636, 6643838879, 17393796001, 45537549124, 119218851371, 312119004989, 817138163596, 2139295485799
Offset: 0

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Author

Keywords

Comments

In any generalized Fibonacci sequence {f(i)}, Sum_{i=0..4n+1} f(i) = a(n)*f(2n+2). - Lekraj Beedassy, Dec 31 2002
The continued fraction expansion for F((2n+1)*(k+1))/F((2n+1)*k), k>=1 is [a(n),a(n),...,a(n)] where there are exactly k elements (F(n) denotes the n-th Fibonacci number). E.g., continued fraction for F(12)/F(9) is [4, 4,4]. - Benoit Cloitre, Apr 10 2003
See A135064 for a possible connection with Galois groups of quintics.
Sequence of all positive integers k such that continued fraction [k,k,k,k,k,k,...] belongs to Q(sqrt(5)). - Thomas Baruchel, Sep 15 2003
All positive integer solutions of Pell equation a(n)^2 - 5*b(n)^2 = -4 together with b(n)=A001519(n), n>=0.
a(n) = L(n,-3)*(-1)^n, where L is defined as in A108299; see also A001519 for L(n,+3).
Inverse binomial transform of A030191. - Philippe Deléham, Oct 04 2005
General recurrence is a(n) = (a(1)-1)*a(n-1) - a(n-2), a(1) >= 4, lim_{n->infinity} a(n) = x*(k*x+1)^n, k =(a(1)-3), x=(1+sqrt((a(1)+1)/(a(1)-3)))/2. Examples in OEIS: a(1)=4 gives A002878. a(1)=5 gives A001834. a(1)=6 gives A030221. a(1)=7 gives A002315. a(1)=8 gives A033890. a(1)=9 gives A057080. a(1)=10 gives A057081. - Ctibor O. Zizka, Sep 02 2008
Let r = (2n+1), then a(n), n>0 = Product_{k=1..floor((r-1)/2)} (1 + sin^2 k*Pi/r); e.g., a(3) = 29 = (3.4450418679...)*(4.801937735...)*(1.753020396...). - Gary W. Adamson, Nov 26 2008
a(n+1) is the Hankel transform of A001700(n)+A001700(n+1). - Paul Barry, Apr 21 2009
a(n) is equal to the permanent of the (2n) X (2n) tridiagonal matrix with sqrt(5)'s along the main diagonal, i's along the superdiagonal and the subdiagonal (i is the imaginary unit), and 0's everywhere else. - John M. Campbell, Jun 09 2011
Conjecture: for n > 0, a(n) = sqrt(Fibonacci(4*n+3) + Sum_{k=2..2*n} Fibonacci(2*k)). - Alex Ratushnyak, May 06 2012
Pisano period lengths: 1, 3, 4, 3, 2, 12, 8, 6, 12, 6, 5, 12, 14, 24, 4, 12, 18, 12, 9, 6, ... . - R. J. Mathar, Aug 10 2012
The continued fraction [a(n); a(n), a(n), ...] = phi^(2n+1), where phi is the golden ratio, A001622. - Thomas Ordowski, Jun 05 2013
Solutions (x, y) = (a(n), a(n+1)) satisfying x^2 + y^2 = 3xy + 5. - Michel Lagneau, Feb 01 2014
Conjecture: except for the number 3, a(n) are the numbers such that a(n)^2+2 are Lucas numbers. - Michel Lagneau, Jul 22 2014
Comment on the preceding conjecture: It is clear that all a(n) satisfy a(n)^2 + 2 = L(2*(2*n+1)) due to the identity (17 c) of Vajda, p. 177: L(2*n) + 2*(-1)^n = L(n)^2 (take n -> 2*n+1). - Wolfdieter Lang, Oct 10 2014
Limit_{n->oo} a(n+1)/a(n) = phi^2 = phi + 1 = (3+sqrt(5))/2. - Derek Orr, Jun 18 2015
If d[k] denotes the sequence of k-th differences of this sequence, then d[0](0), d[1](1), d[2](2), d[3](3), ... = A048876, cf. message to SeqFan list by P. Curtz on March 2, 2016. - M. F. Hasler, Mar 03 2016
a(n-1) and a(n) are the least phi-antipalindromic numbers (A178482) with 2*n and 2*n+1 digits in base phi, respectively. - Amiram Eldar, Jul 07 2021
Triangulate (hyperbolic) 2-space such that around every vertex exactly 7 triangles touch. Call any 7 triangles having a common vertex the first layer and let the (n+1)-st layer be all triangles that do not appear in any of the first n layers and have a common vertex with the n-th layer. Then the n-th layer contains 7*a(n-1) triangles. E.g., the first layer (by definition) contains 7 triangles, the second layer (the "ring" of triangles around the first layer) consists of 28 triangles, the third layer (the next "ring") consists of 77 triangles, and so on. - Nicolas Nagel, Aug 13 2022

Examples

			G.f. = 1 + 4*x + 11*x^2 + 29*x^3 + 76*x^4 + 199*x^5 + 521*x^6 + ... - _Michael Somos_, Jan 13 2019
		

References

  • J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 91.
  • 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).
  • Steven Vajda, Fibonacci and Lucas numbers and the Golden Section, Ellis Horwood Ltd., Chichester, 1989.

Crossrefs

Cf. A000204. a(n) = A060923(n, 0), a(n)^2 = A081071(n).
Cf. A005248 [L(2n) = bisection (even n) of Lucas sequence].
Cf. A001906 [F(2n) = bisection (even n) of Fibonacci sequence], A000045, A002315, A004146, A029907, A113224, A153387, A153416, A178482, A192425, A285992 (prime subsequence).
Cf. similar sequences of the type k*F(n)*F(n+1)+(-1)^n listed in A264080.

Programs

  • GAP
    List([0..40], n-> Lucas(1,-1,2*n+1)[2] ); # G. C. Greubel, Jul 15 2019
    
  • Haskell
    a002878 n = a002878_list !! n
    a002878_list = zipWith (+) (tail a001906_list) a001906_list
    -- Reinhard Zumkeller, Jan 11 2012
    
  • Magma
    [Lucas(2*n+1): n in [0..40]]; // Vincenzo Librandi, Apr 16 2011
    
  • Maple
    A002878 := proc(n)
        option remember;
        if n <= 1 then
            op(n+1,[1,4]);
        else
            3*procname(n-1)-procname(n-2) ;
        end if;
    end proc: # R. J. Mathar, Apr 30 2017
  • Mathematica
    a[n_]:= FullSimplify[GoldenRatio^n - GoldenRatio^-n]; Table[a[n], {n, 1, 40, 2}]
    a[1]=1; a[2]=4; a[n_]:=a[n]= 3a[n-1] -a[n-2]; Array[a, 40]
    LinearRecurrence[{3, -1}, {1, 4}, 41] (* Jean-François Alcover, Sep 23 2017 *)
    Table[Sum[(-1)^Floor[k/2] Binomial[n -Floor[(k+1)/2], Floor[k/2]] 3^(n - k), {k, 0, n}], {n, 0, 40}] (* L. Edson Jeffery, Feb 26 2018 *)
    a[ n_] := Fibonacci[2n] + Fibonacci[2n+2]; (* Michael Somos, Jul 31 2018 *)
    a[ n_]:= LucasL[2n+1]; (* Michael Somos, Jan 13 2019 *)
  • PARI
    a(n)=fibonacci(2*n)+fibonacci(2*n+2) \\ Charles R Greathouse IV, Jun 16 2011
    
  • PARI
    for(n=1,40,q=((1+sqrt(5))/2)^(2*n-1);print1(contfrac(q)[1],", ")) \\ Derek Orr, Jun 18 2015
    
  • PARI
    Vec((1+x)/(1-3*x+x^2) + O(x^40)) \\ Altug Alkan, Oct 26 2015
    
  • Python
    a002878 = [1, 4]
    for n in range(30): a002878.append(3*a002878[-1] - a002878[-2])
    print(a002878) # Gennady Eremin, Feb 05 2022
  • Sage
    [lucas_number2(2*n+1,1,-1) for n in (0..40)] # G. C. Greubel, Jul 15 2019
    

Formula

a(n+1) = 3*a(n) - a(n-1).
G.f.: (1+x)/(1-3*x+x^2). - Simon Plouffe in his 1992 dissertation
a(n) = S(2*n, sqrt(5)) = S(n, 3) + S(n-1, 3); S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(n, 3) = A001906(n+1) (even-indexed Fibonacci numbers).
a(n) ~ phi^(2*n+1). - Joe Keane (jgk(AT)jgk.org), May 15 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i); then (-1)^n*q(n, -1) = a(n). - Benoit Cloitre, Nov 10 2002
a(n) = A005248(n+1) - A005248(n) = -1 + Sum_{k=0..n} A005248(k). - Lekraj Beedassy, Dec 31 2002
a(n) = 2^(-n)*A082762(n) = 4^(-n)*Sum_{k>=0} binomial(2*n+1, 2*k)*5^k; see A091042. - Philippe Deléham, Mar 01 2004
a(n) = (-1)^n*Sum_{k=0..n} (-5)^k*binomial(n+k, n-k). - Benoit Cloitre, May 09 2004
From Paul Barry, May 27 2004: (Start)
Both bisection and binomial transform of A000204.
a(n) = Fibonacci(2n) + Fibonacci(2n+2). (End)
Sequence lists the numerators of sinh((2*n-1)*psi) where the denominators are 2; psi=log((1+sqrt(5))/2). Offset 1. a(3)=11. - Al Hakanson (hawkuu(AT)gmail.com), Mar 25 2009
a(n) = A001906(n) + A001906(n+1). - Reinhard Zumkeller, Jan 11 2012
a(n) = floor(phi^(2n+1)), where phi is the golden ratio, A001622. - Thomas Ordowski, Jun 10 2012
a(n) = A014217(2*n+1) = A014217(2*n+2) - A014217(2*n). - Paul Curtz, Jun 11 2013
Sum_{n >= 0} 1/(a(n) + 5/a(n)) = 1/2. Compare with A005248, A001906, A075796. - Peter Bala, Nov 29 2013
a(n) = lim_{m->infinity} Fibonacci(m)^(4n+1)*Fibonacci(m+2*n+1)/ Sum_{k=0..m} Fibonacci(k)^(4n+2). - Yalcin Aktar, Sep 02 2014
From Peter Bala, Mar 22 2015: (Start)
The aerated sequence (b(n))n>=1 = [1, 0, 4, 0, 11, 0, 29, 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 = -1, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy.
b(n) = (1/2)*((-1)^n - 1)*F(n) + (1 + (-1)^(n-1))*F(n+1), where F(n) is a Fibonacci number. The o.g.f. is x*(1 + x^2)/(1 - 3*x^2 + x^4).
Exp( Sum_{n >= 1} 2*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 2*F(n)*x^n.
Exp( Sum_{n >= 1} (-2)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 2*F(n)*(-x)^n.
Exp( Sum_{n >= 1} 4*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 4*A029907(n)*x^n.
Exp( Sum_{n >= 1} (-4)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 4*A029907(n)*(-x)^n. Cf. A002315, A004146, A113224 and A192425. (End)
a(n) = sqrt(5*F(2*n+1)^2-4), where F(n) = A000045(n). - Derek Orr, Jun 18 2015
For n > 1, a(n) = 5*F(2*n-1) + L(2*n-3) with F(n) = A000045(n). - J. M. Bergot, Oct 25 2015
For n > 0, a(n) = L(n-1)*L(n+2) + 4*(-1)^n. - J. M. Bergot, Oct 25 2015
For n > 2, a(n) = a(n-2) + F(n+2)^2 + F(n-3)^2 = L(2*n-3) + F(n+2)^2 + F(n-3)^2. - J. M. Bergot, Feb 05 2016 and Feb 07 2016
E.g.f.: ((sqrt(5) - 5)*exp((3-sqrt(5))*x/2) + (5 + sqrt(5))*exp((3+sqrt(5))*x/2))/(2*sqrt(5)). - Ilya Gutkovskiy, Apr 24 2016
a(n) = Sum_{k=0..n} (-1)^floor(k/2)*binomial(n-floor((k+1)/2), floor(k/2))*3^(n-k). - L. Edson Jeffery, Feb 26 2018
a(n)*F(m+2n-1) = F(m+4n-2)-F(m), with Fibonacci number F(m), empirical observation. - Dan Weisz, Jul 30 2018
a(n) = -a(-1-n) for all n in Z. - Michael Somos, Jul 31 2018
Sum_{n>=0} 1/a(n) = A153416. - Amiram Eldar, Nov 11 2020
a(n) = Product_{k=1..n} (1 + 4*sin(2*k*Pi/(2*n+1))^2). - Seiichi Manyama, Apr 30 2021
Sum_{n>=0} (-1)^n/a(n) = (1/sqrt(5)) * A153387 (Carlitz, 1967). - Amiram Eldar, Feb 05 2022
The continued fraction [a(n);a(n),a(n),...] = phi^(2*n+1), with phi = A001622. - A.H.M. Smeets, Feb 25 2022
a(n) = 2*sinh((2*n + 1)*arccsch(2)). - Peter Luschny, May 25 2022
This gives the sequence with 2 1's prepended: b(1)=b(2)=1 and, for k >= 3, b(k) = Sum_{j=1..k-2} (2^(k-j-1) - 1)*b(j). - Neal Gersh Tolunsky, Oct 28 2022 (formula due to Jon E. Schoenfield)
For n > 0, a(n) = 1 + 1/(Sum_{k>=1} F(k)/phi^(2*n*k + k)). - Diego Rattaggi, Nov 08 2023
From Peter Bala, Apr 16 2025: (Start)
a(3*n+1) = a(n)^3 + 3*a(n).
a(5*n+2) = a(n)^5 + 5*a(n)^3 + 5*a(n).
a(7*n+3) = a(n)^7 + 7*a(n)^5 + 14*a(n)^3 + 7*a(n).
For the coefficients see A034807.
The general result is: for k >= 0, a(k*n + (k-1)/2) = 2 * T(k, a(n)/2), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind and a(n) = ((1 + sqrt(5))/2)^(2*n+1) + ((1 - sqrt(5))/2)^(2*n+1).
Sum_{n >= 0} (-1)^n/a(n) = (1/4)* (theta_3(phi) - theta_3(phi^2)) = 0.815947983588122..., where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122) and phi = (sqrt(5) - 1)/2. See Borwein and Borwein, Exercise 3 a, p. 94 and Carlitz, 1967. (End)
From Peter Bala, May 15 2025: (Start)
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/5 (telescoping series: 5/(a(n) - 1/a(n)) = 1/A001906(n+1) + 1/A001906(n) ).
More generally, for k >= 1, Sum_{n >= 1} (-1)^(n+1)/(a(k*n) - s(k)/a(k*n)) = 1/(1 + a(k)) where s(k) = a(0) + a(1) + ... + a(k-1) = Lucas(2*k) - 2.
For k >= 1, Sum_{n >= 1} (-1)^(n+1)/(a(n) + L(2*k)^2/a(n)) = (1/5) * A064170(k+2).
Sum_{n >= 1} 1/(a(n) + 9/a(n)) = 3/10 (follows from 1/(a(n) + 9/a(n)) = L(2*n)/A081076(n) - L(2*n+2)/A081076(n+1) ).
More generally, it appears that for k >= 1, Sum_{n >= 1} 1/(a(n) + L(2*k)^2/a(n)) is rational.
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(5) [telescoping product: Product_{k = 1..n} ((a(k) + 1)/(a(k) - 1))^2 = 5*(1 - 4/A240926(n+1)) ]. (End)

Extensions

Chebyshev and Pell comments from Wolfdieter Lang, Aug 31 2004

A289780 p-INVERT of the positive integers (A000027), where p(S) = 1 - S - S^2.

Original entry on oeis.org

1, 4, 14, 47, 156, 517, 1714, 5684, 18851, 62520, 207349, 687676, 2280686, 7563923, 25085844, 83197513, 275925586, 915110636, 3034975799, 10065534960, 33382471801, 110713382644, 367182309614, 1217764693607, 4038731742156, 13394504020957, 44423039068114
Offset: 0

Views

Author

Clark Kimberling, Aug 10 2017

Keywords

Comments

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x).
Taking p(S) = 1 - S gives the INVERT transform of s, so that p-INVERT is a generalization of the INVERT transform (e.g., A033453).
Guide to p-INVERT sequences using p(S) = 1 - S - S^2:
t(A000012) = t(1,1,1,1,1,1,1,...) = A001906
t(A000290) = t(1,4,9,16,25,36,...) = A289779
t(A000027) = t(1,2,3,4,5,6,7,8,...) = A289780
t(A000045) = t(1,2,3,5,8,13,21,...) = A289781
t(A000032) = t(2,1,3,4,7,11,14,...) = A289782
t(A000244) = t(1,3,9,27,81,243,...) = A289783
t(A000302) = t(1,4,16,64,256,...) = A289784
t(A000351) = t(1,5,25,125,625,...) = A289785
t(A005408) = t(1,3,5,7,9,11,13,...) = A289786
t(A005843) = t(2,4,6,8,10,12,14,...) = A289787
t(A016777) = t(1,4,7,10,13,16,...) = A289789
t(A016789) = t(2,5,8,11,14,17,...) = A289790
t(A008585) = t(3,6,9,12,15,18,...) = A289795
t(A000217) = t(1,3,6,10,15,21,...) = A289797
t(A000225) = t(1,3,7,15,31,63,...) = A289798
t(A000578) = t(1,8,27,64,625,...) = A289799
t(A000984) = t(1,2,6,20,70,252,...) = A289800
t(A000292) = t(1,4,10,20,35,56,...) = A289801
t(A002620) = t(1,2,4,6,9,12,16,...) = A289802
t(A001906) = t(1,3,8,21,55,144,...) = A289803
t(A001519) = t(1,1,2,5,13,34,...) = A289804
t(A103889) = t(2,1,4,3,6,5,8,7,,...) = A289805
t(A008619) = t(1,1,2,2,3,3,4,4,...) = A289806
t(A080513) = t(1,2,2,3,3,4,4,5,...) = A289807
t(A133622) = t(1,2,1,3,1,4,1,5,...) = A289809
t(A000108) = t(1,1,2,5,14,42,...) = A081696
t(A081696) = t(1,1,3,9,29,97,...) = A289810
t(A027656) = t(1,0,2,0,3,0,4,0,5...) = A289843
t(A175676) = t(1,0,0,2,0,0,3,0,...) = A289844
t(A079977) = t(1,0,1,0,2,0,3,...) = A289845
t(A059841) = t(1,0,1,0,1,0,1,...) = A289846
t(A000040) = t(2,3,5,7,11,13,...) = A289847
t(A008578) = t(1,2,3,5,7,11,13,...) = A289828
t(A000142) = t(1!, 2!, 3!, 4!, ...) = A289924
t(A000201) = t(1,3,4,6,8,9,11,...) = A289925
t(A001950) = t(2,5,7,10,13,15,...) = A289926
t(A014217) = t(1,2,4,6,11,17,29,...) = A289927
t(A000045*) = t(0,1,1,2,3,5,...) = A289975 (* indicates prepended 0's)
t(A000045*) = t(0,0,1,1,2,3,5,...) = A289976
t(A000045*) = t(0,0,0,1,1,2,3,5,...) = A289977
t(A290990*) = t(0,1,2,3,4,5,...) = A290990
t(A290990*) = t(0,0,1,2,3,4,5,...) = A290991
t(A290990*) = t(0,0,01,2,3,4,5,...) = A290992

Examples

			Example 1:  s = (1,2,3,4,5,6,...) = A000027 and p(S) = 1 - S.
S(x) = x + 2x^2 + 3x^3 + 4x^4 + ...
p(S(x)) = 1 - (x + 2x^2 + 3x^3 + 4x^4 + ... )
- p(0) + 1/p(S(x)) = -1 + 1 + x + 3x^2 + 8x^3 + 21x^4 + ...
T(x) = 1 + 3x + 8x^2 + 21x^3 + ...
t(s) = (1,3,8,21,...) = A001906.
***
Example 2:  s = (1,2,3,4,5,6,...) = A000027 and p(S) = 1 - S - S^2.
S(x) =  x + 2x^2 + 3x^3 + 4x^4 + ...
p(S(x)) = 1 - ( x + 2x^2 + 3x^3 + 4x^4 + ...) - ( x + 2x^2 + 3x^3 + 4x^4 + ...)^2
- p(0) + 1/p(S(x)) = -1 + 1 + x + 4x^2 + 14x^3 + 47x^4 + ...
T(x) = 1 + 4x + 14x^2 + 47x^3 + ...
t(s) = (1,4,14,47,...) = A289780.
		

Crossrefs

Cf. A000027.

Programs

  • GAP
    P:=[1,4,14,47];; for n in [5..10^2] do P[n]:=5*P[n-1]-7*P[n-2]+5*P[n-3]-P[n-4]; od; P; # Muniru A Asiru, Sep 03 2017
  • Mathematica
    z = 60; s = x/(1 - x)^2; p = 1 - s - s^2;
    Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
    Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289780 *)
  • PARI
    x='x+O('x^99); Vec((1-x+x^2)/(1-5*x+7*x^2-5*x^3+x^4)) \\ Altug Alkan, Aug 13 2017
    

Formula

G.f.: (1 - x + x^2)/(1 - 5 x + 7 x^2 - 5 x^3 + x^4).
a(n) = 5*a(n-1) - 7*a(n-2) + 5*a(n-3) - a(n-4).

A001610 a(n) = a(n-1) + a(n-2) + 1, with a(0) = 0 and a(1) = 2.

Original entry on oeis.org

0, 2, 3, 6, 10, 17, 28, 46, 75, 122, 198, 321, 520, 842, 1363, 2206, 3570, 5777, 9348, 15126, 24475, 39602, 64078, 103681, 167760, 271442, 439203, 710646, 1149850, 1860497, 3010348, 4870846, 7881195, 12752042, 20633238, 33385281, 54018520
Offset: 0

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Author

Keywords

Comments

For prime p, p divides a(p-1). - T. D. Noe, Apr 11 2009 [This result follows immediately from the fact that A032190(n) = (1/n)*Sum_{d|n} a(d-1)*phi(n/d). - Petros Hadjicostas, Sep 11 2017]
Generalization. If a(0,x)=0, a(1,x)=2 and, for n>=2, a(n,x)=a(n-1,x)+x*a(n-2,x)+1, then we obtain a sequence of polynomials Q_n(x)=a(n,x) of degree floor((n-1)/2), such that p is prime iff all coefficients of Q_(p-1)(x) are multiple of p (sf. A174625). Thus a(n) is the sum of coefficients of Q_(n-1)(x). - Vladimir Shevelev, Apr 23 2010
Odd composite numbers n such that n divides a(n-1) are in A005845. - Zak Seidov, May 04 2010; comment edited by N. J. A. Sloane, Aug 10 2010
a(n) is the number of ways to modify a circular arrangement of n objects by swapping one or more adjacent pairs. E.g., for 1234, new arrangements are 2134, 2143, 1324, 4321, 1243, 4231 (taking 4 and 1 to be adjacent) and a(4) = 6. - Toby Gottfried, Aug 21 2011
For n>2, a(n) equals the number of Markov equivalence classes with skeleton the cycle on n+1 nodes. See Theorem 2.1 in the article by A. Radhakrishnan et al. below. - Liam Solus, Aug 23 2018
From Gus Wiseman, Feb 12 2019: (Start)
For n > 0, also the number of nonempty subsets of {1, ..., n + 1} containing no two cyclically successive elements (cyclically successive means 1 succeeds n + 1). For example, the a(5) = 17 stable subsets are:
{1}, {2}, {3}, {4}, {5}, {6},
{1,3}, {1,4}, {1,5}, {2,4}, {2,5}, {2,6}, {3,5}, {3,6}, {4,6},
{1,3,5}, {2,4,6}.
(End)
Also the rank of the n-Lucas cube graph. - Eric W. Weisstein, Aug 01 2023

References

  • 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).

Crossrefs

Programs

  • GAP
    List([0..40], n-> Lucas(1,-1,n+1)[2] -1); # G. C. Greubel, Jul 12 2019
  • Haskell
    a001610 n = a001610_list !! n
    a001610_list =
       0 : 2 : map (+ 1) (zipWith (+) a001610_list (tail a001610_list))
    -- Reinhard Zumkeller, Aug 21 2011
    
  • Magma
    I:=[0,2]; [n le 2 select I[n] else Self(n-1)+Self(n-2)+1: n in [1..40]]; // Vincenzo Librandi, Mar 20 2015
    
  • Magma
    [Lucas(n+1) -1: n in [0..40]]; // G. C. Greubel, Jul 12 2019
    
  • Mathematica
    t = {0, 2}; Do[AppendTo[t, t[[-1]] + t[[-2]] + 1], {n, 2, 40}]; t
    RecurrenceTable[{a[n] == a[n - 1] +a[n - 2] +1, a[0] == 0, a[1] == 2}, a, {n, 0, 40}] (* Robert G. Wilson v, Apr 13 2013 *)
    CoefficientList[Series[x (2 - x)/((1 - x - x^2) (1 - x)), {x, 0, 40}], x] (* Vincenzo Librandi, Mar 20 2015 *)
    Table[Fibonacci[n] + Fibonacci[n + 2] - 1, {n, 0, 40}] (* Eric W. Weisstein, Feb 13 2018 *)
    LinearRecurrence[{2, 0, -1}, {2, 3, 6}, 20] (* Eric W. Weisstein, Feb 13 2018 *)
    Table[LucasL[n] - 1, {n, 20}] (* Eric W. Weisstein, Aug 01 2023 *)
    LucasL[Range[20]] - 1 (* Eric W. Weisstein, Aug 01 2023 *)
  • PARI
    a(n)=([0,1,0; 0,0,1; -1,0,2]^n*[0;2;3])[1,1] \\ Charles R Greathouse IV, Sep 08 2016
    
  • PARI
    vector(40, n, f=fibonacci; f(n+1)+f(n-1)-1) \\ G. C. Greubel, Jul 12 2019
    
  • Sage
    [lucas_number2(n+1,1,-1) -1 for n in (0..40)] # G. C. Greubel, Jul 12 2019
    

Formula

a(n) = A000204(n)-1 = A000032(n+1)-1 = A000071(n+1) + A000045(n).
G.f.: x*(2-x)/((1-x-x^2)*(1-x)) = (2*x-x^2)/(1-2*x+x^3). [Simon Plouffe in his 1992 dissertation]
a(n) = F(n) + F(n+2) - 1 where F(n) is the n-th Fibonacci number. - Zerinvary Lajos, Jan 31 2008
a(n) = A014217(n+1) - A000035(n+1). - Paul Curtz, Sep 21 2008
a(n) = Sum_{i=1..floor((n+1)/2)} ((n+1)/i)*C(n-i,i-1). In more general case of polynomials Q_n(x)=a(n,x) (see our comment) we have Q_n(x) = Sum_{i=1..floor((n+1)/2)}((n+1)/i)*C(n-i,i-1)*x^(i-1). - Vladimir Shevelev, Apr 23 2010
a(n) = Sum_{k=0..n-1} Lucas(k), where Lucas(n) = A000032(n). - Gary Detlefs, Dec 07 2010
a(0)=0, a(1)=2, a(2)=3; for n>=3, a(n) = 2*a(n-1) - a(n-3). - George F. Johnson, Jan 28 2013
For n > 1, a(n) = A048162(n+1) + 3. - Toby Gottfried, Apr 13 2013
For n > 0, a(n) = A169985(n + 1) - 1. - Gus Wiseman, Feb 12 2019

A052952 a(n) = Fibonacci(n+2) - (1-(-1)^n)/2.

Original entry on oeis.org

1, 1, 3, 4, 8, 12, 21, 33, 55, 88, 144, 232, 377, 609, 987, 1596, 2584, 4180, 6765, 10945, 17711, 28656, 46368, 75024, 121393, 196417, 317811, 514228, 832040, 1346268, 2178309, 3524577, 5702887, 9227464, 14930352, 24157816, 39088169, 63245985, 102334155
Offset: 0

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Author

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

Keywords

Comments

Equals row sums of triangle A173284. - Gary W. Adamson, Feb 14 2010
The Kn21 sums (see A180662 for definition) of the 'Races with Ties' triangle A035317 produce this sequence. - Johannes W. Meijer, Jul 20 2011
a(n-1), for n >= 1, gives the number of compositions of n with relative prime parts, and parts not exceeding 2. See the row sums of triangle A030528 where for even n the leading 1 is missing. - Wolfdieter Lang, Jul 27 2023

Examples

			G.f. = 1 + x + 3*x^2 + 4*x^3 + 8*x^4 + 12*x^5 + 21*x^6 + 33*x^7 + ...
		

Crossrefs

Partial sums of A008346, first differences of A129696.
Cf. also A000032, A000045, A030528.

Programs

  • GAP
    List([0..40], n-> Fibonacci(n+2) -(1-(-1)^n)/2); # G. C. Greubel, Jul 10 2019
  • Haskell
    a052952 n = a052952_list !! n
    a052952_list = 1 : 1 : zipWith (+)
       a059841_list (zipWith (+) a052952_list $ tail a052952_list)
    -- Reinhard Zumkeller, Jan 06 2012
    
  • Magma
    [Fibonacci(n+2)-(1-(-1)^n)/2: n in [0..40]]; // Vincenzo Librandi, Dec 02 2016
    
  • Maple
    A052952 :=proc(n)
        option remember;
        local t1;
        if n <= 1 then
            return 1 ;
        fi:
        if n mod 2 = 1 then
            t1:=0
        else
            t1:=1;
        fi:
        procname(n-1)+procname(n-2)+t1;
    end proc;
    seq(A052952(n), n=0..40) ; # N. J. A. Sloane, May 25 2008
  • Mathematica
    Table[Fibonacci[n+2] -(1-(-1)^n)/2, {n, 0, 40}] (* Vincenzo Librandi, Dec 02 2016 *)
    Sum[(-1)^k*Fibonacci[Range[2,41], 1-k], {k,0,1}] (* G. C. Greubel, Oct 21 2019 *)
    CoefficientList[Series[1/((1-x-x^2)*(1-x^2)),{x,0,40}],x] (* Harvey P. Dale, Sep 12 2020 *)
  • PARI
    {a(n) = fibonacci(n+2) - n%2};
    
  • Sage
    [fibonacci(n+2) -(1-(-1)^n)/2 for n in (0..40)] # G. C. Greubel, Jul 10 2019
    

Formula

G.f.: 1/((1-x-x^2)*(1-x^2)).
a(n) = A074331(n+1).
a(n) = A054450(n+1, 1) (second column of triangle).
a(n) = 2*a(n-2) + a(n-3) + 1, with a(0)=1, a(1)=1, a(2)=3.
a(n) = Sum_{alpha=RootOf(-1+z+z^2)} (3+alpha)*alpha^(-1-n)/3 - Sum_{beta=RootOf(-1+z^2)} beta^(-1-n)/2.
a(2*k) = Sum_{j=0..k} F(2*j+1) = F(2*(k+1)) for k >= 0; a(2*k-1) = Sum_{j=0..k} F(2*j) = F(2*k+1)-1 for k >= 1 (F = A000045, Fibonacci numbers).
a(n) = a(n-1) + a(n-2) + (1+(-1)^n)/2.
a(n) = Sum_{k=0..floor(n/2)} binomial(n-k+1, k). - Paul Barry, Oct 23 2004
a(n) = floor(phi^(n+2) / sqrt(5)), where phi is the golden ratio: phi = (1+sqrt(5))/2. - Reinhard Zumkeller, Apr 19 2005
a(n) = Fibonacci(n+1) + a(n-2) with n>1, a(0)=a(1)=1. - Zerinvary Lajos, Mar 17 2008
a(n) = floor(Fibonacci(n+3)^2/Fibonacci(n+4)). - Gary Detlefs, Nov 29 2010
a(n) = (A001595(n+3) - A066983(n+4))/2. - Gary Detlefs, Dec 19 2010
a(4*n) = F(4*n+2); a(4*n+1) = F(4*n+3) - 1; a(4*n+2) = F(4*n+4); a(4*n+3) = F(4*n+5) - 1. - Johannes W. Meijer, Jul 20 2011
a(n+1) = a(n) + a(n-1) + A059841(n+1). - Reinhard Zumkeller, Jan 06 2012
a(n) = floor(|F((1+i)*(n+2))|), n >= 0, with the complex Fibonacci function F: C -> C, z -> F(z) with F(z) := (exp(log(phi)*z) - exp(i*Pi*z)*exp(-log(phi)*z))/(2*phi-1) with the modulus |z|, the imaginary unit i and the golden section phi:=(1+sqrt(5))/2. A Conjecture: For F(z) see, e.g., the T. Koshy reference. ch. 45, p. 523, where F is called f, given in A000045. - Wolfdieter Lang, Jul 24 2012
5*a(n) = (L(n+3)-1)*(L(n+4)+3) -14 -Sum_{k=0..n} L(k+1)*L(k+5) = (L(n+3)-1)*(L(n+4)+3) -L(2*n+7) +A168309(n), where L=A000032. - J. M. Bergot, Jun 13 2014
a(n) = floor(phi*Fibonacci(n+1)), where phi is the golden section. - Michel Dekking, Dec 02 2016
a(n) = -(-1)^n * a(-4-n) for all n in Z. - Michael Somos, Dec 03 2016
a(n) = Sum_{k=0..n} Sum_{i=0..n} C(n-k-1,k-i). - Wesley Ivan Hurt, Sep 21 2017
a(n) = floor(1/(Sum_{k>=n+4} 1/Fibonacci(k))) [Ohtsuka and Nakamura]. - Michel Marcus, Aug 09 2018
a(n) = floor(abs(chebyshevU(n/2, 3/2))). - Federico Provvedi, Feb 23 2022
E.g.f.: exp(x/2)*(5*cosh(sqrt(5)*x/2) + 3*sqrt(5)*sinh(sqrt(5)*x/2))/5 - sinh(x). - Stefano Spezia, Mar 09 2024

Extensions

Additional formulas and more terms from Wolfdieter Lang, May 02 2000
Better description from Olivier Gérard, Jun 05 2001

A001350 Associated Mersenne numbers.

Original entry on oeis.org

0, 1, 1, 4, 5, 11, 16, 29, 45, 76, 121, 199, 320, 521, 841, 1364, 2205, 3571, 5776, 9349, 15125, 24476, 39601, 64079, 103680, 167761, 271441, 439204, 710645, 1149851, 1860496, 3010349, 4870845, 7881196, 12752041, 20633239, 33385280, 54018521, 87403801, 141422324
Offset: 0

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Author

Keywords

Comments

a(n) is last term in the period of the continued fraction expansion of phi^n (phi being the golden number). E.g.: n=10, phi^10=[122,1,121,1,121,1,121,...] (and the period may only have 1 or 2 terms). Also, a(n) = floor(phi^n)-((n+1) mod 2), or a(n) = A014217(n)-((n+1) mod 2). - Thomas Baruchel, Nov 05 2002 [continued fraction value corrected by Jon E. Schoenfield, Jan 20 2019]
a(n) is the resultant of the polynomials x^2-x-1 and x^(n+1)-x^n-1 for n >= 1. - Richard Choulet, Aug 05 2007
This is a divisibility sequence; that is, if n divides m, then a(n) divides a(m). - Michael Somos, Feb 12 2012
Gives the number of arrangements of black and white beads on a necklace with a total of n beads satisfying (1) there is at least one black bead (2) between any two black beads the number of white beads is even and (3) rotations and flippings of a necklace are considered distinct (see Butler). - Peter Bala, Mar 06 2014
This is the case P1 = 1, P2 = 0, Q = -1 of the 3-parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Mar 31 2014
The resultant of the (s_2, s_2+n) pair, where s_n(X) is X^n-X-1, is -a(n). See Rush link. - Michel Marcus, Sep 30 2019

Examples

			G.f. = x + x^2 + 4*x^3 + 5*x^4 + 11*x^5 + 16*x^6 + 29*x^7 + 45*x^8 + 76*x^9 + ...
n=1: a(9)/a(3) = 76/4 = 19; a(18)/a(6) = 5776/16 = 361 = 19^2. - _Bob Selcoe_, Jun 01 2014
		

References

  • 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).

Crossrefs

Programs

  • Magma
    [Floor(-(1 - ((1 + Sqrt(5))/2)^n - (-(1 + Sqrt(5))/2)^(-n) + (-1)^n)): n in [0..40]]; // Vincenzo Librandi, Aug 15 2011
    
  • Maple
    A001350 := n -> add(binomial(k-1, 2*k-n)*n/(n-k), k=0..n-1);
    seq(A001350(n), n=0..39); # Peter Luschny, Sep 26 2014
  • Mathematica
    Clear[f, n]; f[n_] = -(1 - ((1 + Sqrt[5])/2)^n - (-(1 + Sqrt[5])/2)^(-n) + (-1)^n); Table[FullSimplify[ExpandAll[f[n]]], {n, 0, 30}] (* Roger L. Bagula and Gary W. Adamson, Nov 26 2008 *)
    a[ n_] := LucasL[n] - 1 - (-1)^n; (* Michael Somos, May 18 2015 *)
    a[ n_] := SeriesCoefficient[ x D[ Log[ 1 + x / (1 - x - x^2)], x], {x, 0, n}]; (* Michael Somos, May 18 2015 *)
    LinearRecurrence[{1, 2, -1, -1}, {0, 1, 1, 4}, 40] (* Jean-François Alcover, Jan 07 2019 *)
  • PARI
    {a(n) = fibonacci(n+1) + fibonacci(n-1) - 1 - (-1)^n};
    
  • PARI
    {a(n) = my(w = quadgen(5)); simplify( -(w^n - 1) * ((-1/w)^n - 1))}; /* Michael Somos, Feb 12 2012 */
    
  • Python
    from sympy import lucas
    def A001350(n): return lucas(n)-((n&1^1)<<1) # Chai Wah Wu, Sep 23 2023

Formula

G.f.: x*(1+x^2)/((1-x^2)*(1-x-x^2)). - Simon Plouffe in his 1992 dissertation
a(n) = a(n-1) + a(n-2) + 1 -(-1)^n. a(-n) = (-1)^n * a(n).
a(n) = A050140(Fibonacci(n)). - Thomas Baruchel, Nov 05 2002
Convolution of F(n) and {1, 0, 2, 0, 2, ...}. a(n) = Sum_{k=0..n} ((1+(-1)^k)-0^k)*F(n-k) = Sum_{k=0..n} F(k)*((1+(-1)^(n-k))-0^(n-k)). - Paul Barry, Jul 19 2004
a(n) = 2*A074331(n) - A000045(n). - Paul Barry, Jul 19 2004
a(n) = Lucas_number(n) - 1 - (-1)^n = A000032(n) - 1 - (-1)^n. - Hieronymus Fischer, Feb 18 2006
a(n) = -(1 - ((1 + sqrt(5))/2)^n - (-(1 + sqrt(5))/2)^(-n) + (-1)^n). - Roger L. Bagula and Gary W. Adamson, Nov 26 2008
a(n) = n * Sum_{k=1..n} (Sum_{i=ceiling((n-k)/2)..(n-k)} (binomial(i,n-k-i)*binomial(k+i-1,k-1))/k*(-1)^(k+1)), n>0. - Vladimir Kruchinin, Sep 03 2010
a(n) = a(n-1) + 2*a(n-2) - a(n-3) - a(n-4). - Colin Barker, Apr 11 2014
a(n) = sqrt(A152152(n)). - Colin Barker, Apr 11 2014
a(n) = a(2*n)/A000032(n) when n is odd; a(n) = a(2*n)/(A000032(n+2)) when n is even. - Bob Selcoe, Jun 01 2014
a(12n+6)/a(4n+2) = (a(6n+3)/a(2n+1))^2. - Bob Selcoe, Jun 01 2014
a(n) = Sum_{k=0..n-1} binomial(k-1, 2*k-n)*n/(n-k). - Peter Luschny, Sep 26 2014
From Peter Bala, Mar 19 2015: (Start)
a(n) = -(alpha^n - 1)*(beta^n - 1), where alpha = 1/2*(1 + sqrt(5)) and beta = (1/2)*(1 - sqrt(5)).
a(n) = -det(I - M^n) where I is the 2 X 2 identity matrix and M = [ 1, 1; 1, 0 ]. Cf. A129744.
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + Sum_{n >= 1} Fibonacci(n)*x^n. Cf. A004146. (End)
a(n) = A052952(n-1) + A052952(n-3). - R. J. Mathar, Jul 02 2018
a(n) = (L(2*n+1) - L(n+1)) mod (L(n+1)-1) for n > 0 where L(k)=A000032(k). - Art Baker, Jan 17 2019
a(n) = Sum_{j=n..2*n-1} L(j) mod Sum_{j=0..n-1} L(j) where L(j)=A000032(j). - Art Baker, Jan 20 2019
Convolution of (1, 0, 3, 0, 5, 0, 7, ...) and (1, 1, 1, 2, 3, 5, 8, 13, ...). - Gary W. Adamson, Jul 08 2019
a(n) = Sum_{d|n} d*A060280(d) = Sum_{d|n} A031367(d). [Baake, Roberts, Weiss, eq(2)]. - R. J. Mathar, Oct 19 2021

Extensions

Additional comments from Michael Somos, Aug 01 2002

A169985 Round phi^n to the nearest integer.

Original entry on oeis.org

1, 2, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843, 1364, 2207, 3571, 5778, 9349, 15127, 24476, 39603, 64079, 103682, 167761, 271443, 439204, 710647, 1149851, 1860498, 3010349, 4870847, 7881196, 12752043, 20633239, 33385282, 54018521, 87403803
Offset: 0

Views

Author

N. J. A. Sloane, Sep 26 2010

Keywords

Comments

Phi = (1+sqrt(5))/2, see A001622.
a(n) is the number of subsets of {1,2,...,n} with no two consecutive elements where n and 1 are considered to be consecutive. - Geoffrey Critzer, Sep 23 2013
Equals the Lucas sequence beginning at 1 (A000204) with 2 inserted between 1 and 3.
The Lucas sequence beginning at 2 (A000032) can be written as L(n) = phi^n + (-1/phi)^n. Since |(-1/phi)^n|<1/2 for n>1, this sequence is {L(n)} (with the first two terms switched). As a consequence, for n>1: a(n) is obtained by rounding phi^n up for even n and down for odd n; a(n) is also the nearest integer to 1/|phi^n - a(n)|. - Danny Rorabaugh, Apr 15 2015

Examples

			a(4) = 7 because we have: {}, {1}, {2}, {3}, {4}, {1,3}, {2,4}. - _Geoffrey Critzer_, Sep 23 2013
		

Crossrefs

Programs

  • GAP
    Concatenation([1,2], List([2..40], n-> Lucas(1,-1,n)[2] )); # G. C. Greubel, Jul 09 2019
    
  • Magma
    [Round(Sqrt(Fibonacci(2*n) + 2*Fibonacci(2*n-1))): n in [0..40]]; // Vincenzo Librandi, Apr 16 2015
    
  • Mathematica
    nn=34; CoefficientList[Series[(1+x-x^3)/(1-x-x^2),{x,0,nn}],x] (* Geoffrey Critzer, Sep 23 2013 *)
    Round[GoldenRatio^Range[0,40]] (* Harvey P. Dale, Jul 13 2014 *)
    Table[If[n<=1, n+1, LucasL[n]], {n, 0, 40}] (* G. C. Greubel, Jul 09 2019 *)
  • PARI
    my(x='x+O('x^40)); Vec((1+x-x^3)/(1-x-x^2)) \\ G. C. Greubel, Feb 13 2019
    
  • Python
    from gmpy2 import isqrt, fib2
    def A169985(n): return int((m:=isqrt(k:=(lambda x:(x[1]<<1)+x[0])(fib2(n<<1))))+(k-m*(m+1)>=1)) # Chai Wah Wu, Jun 19 2024
  • Sage
    [round(golden_ratio^n) for n in range(40)] # Danny Rorabaugh, Apr 16 2015
    

Formula

O.g.f.: (1 + x - x^3)/(1 - x - x^2). - Geoffrey Critzer, Sep 23 2013
a(n) = round(sqrt(F(2n) + 2*F(2n-1))), for n >= 0, allowing F(-1) = 1. Also phi^n -> sqrt(F(2n) + 2*F(2n-1)), within < 0.02% by n = 4, therefore converging rapidly. - Richard R. Forberg, Jun 23 2014
For k > 0, a(2k) = A169986(2k) and a(2k+1) = A014217(2k+1). - Danny Rorabaugh, Apr 15 2015
For n > 1, a(n) = A001610(n - 1) + 1. - Gus Wiseman, Feb 12 2019
a(n) = A000032(n) for n>=2. - G. C. Greubel, Jul 09 2019

A128588 Expansion of g.f. x*(1+x+x^2)/(1-x-x^2).

Original entry on oeis.org

1, 2, 4, 6, 10, 16, 26, 42, 68, 110, 178, 288, 466, 754, 1220, 1974, 3194, 5168, 8362, 13530, 21892, 35422, 57314, 92736, 150050, 242786, 392836, 635622, 1028458, 1664080, 2692538, 4356618, 7049156, 11405774, 18454930, 29860704, 48315634, 78176338
Offset: 1

Views

Author

Gary W. Adamson, Mar 11 2007

Keywords

Comments

Previous name was: A007318 * A128587.
a(n)/a(n-1) tends to phi, 1.618... = A001622.
Regardless of initial two terms, any linearly recurring sequence with signature (1,1) will yield an a(n)/a(n+1) ratio tending to phi (the Golden Ratio). - Harvey P. Dale, Mar 29 2017
Apart from the initial term, double the Fibonacci numbers. O.g.f.: x*(1+x+x^2)/(1-x-x^2). a(n) gives the number of binary strings of length n-1 avoiding the substrings 000 and 111. a(n) also gives the number of binary strings of length n-1 avoiding the substrings 010 and 101. - Peter Bala, Jan 22 2008
Row lengths of triangle A232642. - Reinhard Zumkeller, May 14 2015
a(n) is the number of binary strings of length n-1 avoiding the substrings 000 and 111. - Allan C. Wechsler, Feb 13 2025

Crossrefs

Programs

  • GAP
    Concatenation([1], List([2..40], n-> 2*Fibonacci(n))); # G. C. Greubel, Jul 10 2019
  • Haskell
    a128588 n = a128588_list !! (n-1)
    a128588_list = 1 : cows where
                       cows = 2 : 4 : zipWith (+) cows (tail cows)
    -- Reinhard Zumkeller, May 14 2015
    
  • Magma
    [1] cat [2*Fibonacci(n): n in [2..40]]; // G. C. Greubel, Jul 10 2019
    
  • Maple
    a:= n-> `if`(n<2, n, 2*(<<0|1>, <1|1>>^n)[1,2]):
    seq(a(n), n=1..50);  # Alois P. Heinz, Apr 28 2018
  • Mathematica
    nn=40; a=(1-x^3)/(1-x); b=x*(1-x^2)/(1-x); CoefficientList[Series[a^2 /(1-b^2), {x,0,nn}], x]  (* Geoffrey Critzer, Sep 01 2012 *)
    LinearRecurrence[{1,1}, {1,2,4}, 40] (* Harvey P. Dale, Mar 29 2017 *)
    Join[{1}, 2*Fibonacci[Range[2,40]]] (* G. C. Greubel, Jul 10 2019 *)
  • PARI
    {a(n) = if( n<2, n==1, 2 * fibonacci(n))}; /* Michael Somos, Jul 18 2015 */
    
  • Sage
    [1]+[2*fibonacci(n) for n in (2..40)] # G. C. Greubel, Jul 10 2019
    

Formula

G.f.: x*(1+x+x^2)/(1-x-x^2).
Binomial transform of A128587; a(n+2) = a(n+1) + a(n), n>3.
a(n) = A068922(n-1), n>2. - R. J. Mathar, Jun 14 2008
For n > 1: a(n+1) = a(n) + if a(n) odd then max{a(n),a(n-1)} else min{a(n),a(n-1)}, see also A038754. - Reinhard Zumkeller, Oct 19 2015
E.g.f.: 4*exp(x/2)*sinh(sqrt(5)*x/2)/sqrt(5) - x. - Stefano Spezia, Feb 19 2023

Extensions

New name from Joerg Arndt, Feb 16 2024

A022088 Fibonacci sequence beginning 0, 5.

Original entry on oeis.org

0, 5, 5, 10, 15, 25, 40, 65, 105, 170, 275, 445, 720, 1165, 1885, 3050, 4935, 7985, 12920, 20905, 33825, 54730, 88555, 143285, 231840, 375125, 606965, 982090, 1589055, 2571145, 4160200, 6731345, 10891545, 17622890, 28514435, 46137325, 74651760, 120789085
Offset: 0

Views

Author

Keywords

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, pp. 15, 34, 52.

Crossrefs

Cf. sequences of the form m*Fibonacci listed in A022086.

Programs

Formula

a(n) = round( (2*phi-1)*phi^n ) for n>3. - Thomas Baruchel, Sep 08 2004
a(n) = 5*Fibonacci(n).
a(n) = A119457(n+3,n-1) for n>1. - Reinhard Zumkeller, May 20 2006
G.f.: 5*x/(1-x-x^2). - Philippe Deléham, Nov 20 2008
a(n+2) = A014217(n+4) - A014217(n). - Paul Curtz, Dec 22 2008
a(n) = sqrt(5*(A000032(n)^2 - 4*(-1)^n)). - Alexander Samokrutov, Sep 02 2015
From Tom Copeland, Jan 25 2016: (Start)
The o.g.f. for the shifted series b(0)=0 and b(n) = a(n+1) is G(x) = 5*x*(1+x)/(1-x*(1+x)) = 5 L(-Cinv(-x)), where L(x) = x/(1-x) with inverse Linv(x) = x/(1+x) and Cinv(x) = x*(1-x), the inverse of the o.g.f. for the shifted Catalan numbers of A000108, C(x) = (1-sqrt(1-4*x))/2. Then Ginv(x) = -C(-Linv(x/5)) = (-1 + sqrt(1+4*x/(5+x)))/2.
a(n+1) = 5*Sum_{k=0..n} binomial(n-k,k) = 5 * A000045(n+1), from A267633, with the convention for zeros of the binomial assumed there. (End)
For n > 0, 1/a(n) = Sum_{k>=1} F(n*k)/(L(n+1)^(k+1)), where F(n) = A000045(n) and L(n) = A000032(n). - Diego Rattaggi, Oct 26 2022

A098600 a(n) = Fibonacci(n-1) + Fibonacci(n+1) - (-1)^n.

Original entry on oeis.org

1, 2, 2, 5, 6, 12, 17, 30, 46, 77, 122, 200, 321, 522, 842, 1365, 2206, 3572, 5777, 9350, 15126, 24477, 39602, 64080, 103681, 167762, 271442, 439205, 710646, 1149852, 1860497, 3010350, 4870846, 7881197, 12752042, 20633240, 33385281, 54018522
Offset: 0

Views

Author

Paul Barry, Sep 17 2004

Keywords

Comments

Row sums of A098599.

Crossrefs

Programs

  • Magma
    [Fibonacci(n-1) + Fibonacci(n+1) - (-1)^n: n in [0..50]]; // Vincenzo Librandi, Aug 31 2014
    
  • Mathematica
    Table[-(-1)^n + LucasL[n], {n, 0, 39}] (* Alonso del Arte, Aug 30 2014 *)
    Table[Fibonacci[n - 1] + Fibonacci[n + 1] - (-1)^n, {n, 0, 40}] (* Vincenzo Librandi, Aug 31 2014 *)
    CoefficientList[ Series[-(1 + 2x)/(-1 + 2x^2 + x^3), {x, 0, 40}], x] (* or *)
    LinearRecurrence[{0, 2, 1}, {1, 2, 2}, 40] (* Robert G. Wilson v, Mar 09 2018 *)
  • PARI
    a(n)=fibonacci(n-1) + fibonacci(n+1) - (-1)^n; \\ Joerg Arndt, Oct 18 2014
    
  • PARI
    Vec((1+2*x)/((1+x)*(1-x-x^2)) + O(x^30)) \\ Colin Barker, Jun 03 2016
    
  • SageMath
    [lucas_number2(n,1,-1) -(-1)^n for n in range(51)] # G. C. Greubel, Mar 26 2024

Formula

G.f.: (1+2*x) / ((1+x)*(1-x-x^2)).
a(n) = Sum_{k = 0..n} binomial(k, n-k) + binomial(k-1, n-k-1).
a(n) = A020878(n) - 1 = A001350(n) + 1.
a(n) = Lucas(n) - (-1)^n. - Paul Barry, Dec 01 2004
a(n) = A181716(n+1). - Richard R. Forberg, Aug 30 2014
a(n) = [x^n] ( (1 + x + sqrt(1 + 6*x + 5*x^2))/2 )^n. exp( Sum_{n >= 1} a(n)*x^n/n ) = Sum_{n >= 0} Fibonacci(n+2)*x^n. Cf. A182143. - Peter Bala, Jun 29 2015
From Colin Barker, Jun 03 2016: (Start)
a(n) = (-(-1)^n + ((1/2)*(1-sqrt(5)))^n + ((1/2)*(1+sqrt(5)))^n).
a(n) = 2*a(n-2) + a(n-3) for n > 2. (End)
E.g.f.: (2*exp(3*x/2)*cosh(sqrt(5)*x/2) - 1)*exp(-x). - Ilya Gutkovskiy, Jun 03 2016
a(n) = A014217(n) + A000035(n). - Paul Curtz, Jul 27 2023
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