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.

Showing 1-5 of 5 results.

A107328 Duplicate of A065034.

Original entry on oeis.org

3, 4, 8, 19, 48, 124, 323, 844, 2208, 5779, 15128, 39604, 103683, 271444, 710648, 1860499, 4870848, 12752044, 33385283, 87403804, 228826128, 599074579, 1568397608, 4106118244, 10749957123, 28143753124
Offset: 0

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Keywords

A005248 Bisection of Lucas numbers: a(n) = L(2*n) = A000032(2*n).

Original entry on oeis.org

2, 3, 7, 18, 47, 123, 322, 843, 2207, 5778, 15127, 39603, 103682, 271443, 710647, 1860498, 4870847, 12752043, 33385282, 87403803, 228826127, 599074578, 1568397607, 4106118243, 10749957122, 28143753123, 73681302247, 192900153618, 505019158607, 1322157322203
Offset: 0

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Author

Keywords

Comments

Drop initial 2; then iterates of A050411 do not diverge for these starting values. - David W. Wilson
All nonnegative integer solutions of Pell equation a(n)^2 - 5*b(n)^2 = +4 together with b(n)=A001906(n), n>=0. - Wolfdieter Lang, Aug 31 2004
a(n+1) = B^(n)AB(1), n>=0, with compositions of Wythoff's complementary A(n):=A000201(n) and B(n)=A001950(n) sequences. See the W. Lang link under A135817 for the Wythoff representation of numbers (with A as 1 and B as 0 and the argument 1 omitted). E.g., 3=`10`, 7=`010`, 18=`0010`, 47=`00010`, ..., in Wythoff code. a(0) = 2 = B(1) in Wythoff code.
Output of Tesler's formula (as well as that of Lu and Wu) for the number of perfect matchings of an m X n Möbius band where m and n are both even specializes to this sequence for m=2. - Sarah-Marie Belcastro, Jul 04 2009
Numbers having two 1's in their base-phi representation. - Robert G. Wilson v, Sep 13 2010
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
From Wolfdieter Lang, Feb 18 2013: (Start)
a(n) is also one half of the total number of round trips, each of length 2*n, on the graph P_4 (o-o-o-o) (the simple path with 4 points (vertices) and 3 lines (or edges)). See the array and triangle A198632 for the general case of the graph P_N (there N is n and the length is l=2*k).
O.g.f. for w(4,l) (with zeros for odd l): y*(d/dy)S(4,y)/S(4,y) with y=1/x and Chebyshev S-polynomials (coefficients A049310). See also A198632 for a rewritten form. One half of this o.g.f. for x -> sqrt(x) produces the g.f. (2-3x)/(1-3x+x^2) given below. (End)
Solutions (x, y) = (a(n), a(n+1)) satisfying x^2 + y^2 = 3xy - 5. - Michel Lagneau, Feb 01 2014
Except for the first term, positive values of x (or y) satisfying x^2 - 7xy + y^2 + 45 = 0. - Colin Barker, Feb 16 2014
Except for the first term, positive values of x (or y) satisfying x^2 - 18xy + y^2 + 320 = 0. - Colin Barker, Feb 16 2014
a(n) are the numbers such that a(n)^2-2 are Lucas numbers. - Michel Lagneau, Jul 22 2014
All sequences of this form, b(n+1) = 3*b(n) - b(n-1), regardless of initial values, which includes this sequence, yield this sequence as follows: a(n) = (b(j+n) + b(j-n))/b(j), for any j, except where b(j) = 0. Also note formula below relating this a(n) to all sequences of the form G(n+1) = G(n) + G(n-1). - Richard R. Forberg, Nov 18 2014
A non-simple continued fraction expansion for F(2n*(k+1))/F(2nk) k>=1 is a(n) + (-1)/(a(n) + (-1)/(a(n) + ... + (-1)/a(n))) where a(n) appears exactly k times (F(n) denotes the n-th Fibonacci number). E.g., F(16)/F(12) equals 7 + (-1)/(7 + (-1)/7). Furthermore, these a(n) are exactly the positive integers k such that the non-simple infinite continued fraction k + (-1)/(k + (-1)/(k + (-1)/(k + ...))) belongs to Q(sqrt(5)). Compare to Benoit Cloitre and Thomas Baruchel's comments at A002878. - Greg Dresden, Aug 13 2019
For n >= 1, a(n) is the number of cyclic up-down words of length 2*n over an alphabet of size 3. - Sela Fried, Apr 08 2025

Examples

			G.f. = 2 + 3*x + 7*x^2 + 18*x^3 + 47*x^4 + 123*x^5 + 322*x^6 + 843*x^7 + ... - _Michael Somos_, Aug 11 2009
		

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).
  • Richard P. Stanley, Enumerative combinatorics, Vol. 2. Volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999.

Crossrefs

Cf. A000032, A002878 (odd-indexed Lucas numbers), A001906 (Chebyshev S(n-1, 3)), a(n) = sqrt(4+5*A001906(n)^2), A228842.
a(n) = A005592(n)+1 = A004146(n)+2 = A065034(n)-1.
First differences of A002878. Pairwise sums of A001519. First row of array A103997.
Cf. A153415, A201157. Also Lucas(k*n): A000032 (k = 1), A014448 (k = 3), A056854 (k = 4), A001946 (k = 5), A087215 (k = 6), A087281 (k = 7), A087265 (k = 8), A087287 (k = 9), A065705 (k = 10), A089772 (k = 11), A089775 (k = 12).

Programs

  • Haskell
    a005248 n = a005248_list !! n
    a005248_list = zipWith (+) (tail a001519_list) a001519_list
    -- Reinhard Zumkeller, Jan 11 2012
  • Magma
    [Lucas(2*n) : n in [0..100]]; // Vincenzo Librandi, Apr 14 2011
    
  • Maple
    a:= n-> (<<2|3>>. <<3|1>, <-1|0>>^n)[1$2]: seq(a(n), n=0..30); # Alois P. Heinz, Jul 31 2008
    with(combinat): seq(5*fibonacci(n)^2+2*(-1)^n, n= 0..26);
  • Mathematica
    a[0] = 2; a[1] = 3; a[n_] := 3a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 27}] (* Robert G. Wilson v, Jan 30 2004 *)
    Fibonacci[1 + 2n] + 1/2 (-Fibonacci[2n] + LucasL[2n]) (* Tesler. Sarah-Marie Belcastro, Jul 04 2009 *)
    LinearRecurrence[{3, -1}, {2, 3}, 50] (* Sture Sjöstedt, Nov 27 2011 *)
    LucasL[Range[0,60,2]] (* Harvey P. Dale, Sep 30 2014 *)
  • PARI
    {a(n) = fibonacci(2*n + 1) + fibonacci(2*n - 1)}; /* Michael Somos, Jun 23 2002 */
    
  • PARI
    {a(n) = 2 * subst( poltchebi(n), x, 3/2)}; /* Michael Somos, Jun 28 2003 */
    
  • Sage
    [lucas_number2(n,3,1) for n in range(37)] # Zerinvary Lajos, Jun 25 2008
    

Formula

a(n) = Fibonacci(2*n-1) + Fibonacci(2*n+1).
G.f.: (2-3*x)/(1-3*x+x^2). - Simon Plouffe in his 1992 dissertation.
a(n) = S(n, 3) - S(n-2, 3) = 2*T(n, 3/2) with S(n-1, 3) = A001906(n) and S(-2, x) = -1. U(n, x)=S(n, 2*x) and T(n, x) are Chebyshev's U- and T-polynomials.
a(n) = a(k)*a(n - k) - a(n - 2k) for all k, i.e., a(n) = 2*a(n) - a(n) = 3*a(n - 1) - a(n - 2) = 7*a(n - 2) - a(n - 4) = 18*a(n - 3) - a(n - 6) = 47*a(n - 4) - a(n - 8) etc., a(2n) = a(n)^2 - 2. - Henry Bottomley, May 08 2001
a(n) = A060924(n-1, 0) = 3*A001906(n) - 2*A001906(n-1), n >= 1. - Wolfdieter Lang, Apr 26 2001
a(n) ~ phi^(2*n) where phi=(1+sqrt(5))/2. - Joe Keane (jgk(AT)jgk.org), May 15 2002
a(0)=2, a(1)=3, a(n) = 3*a(n-1) - a(n-2) = a(-n). - Michael Somos, Jun 28 2003
a(n) = phi^(2*n) + phi^(-2*n) where phi=(sqrt(5)+1)/2, the golden ratio. E.g., a(4)=47 because phi^(8) + phi^(-8) = 47. - Dennis P. Walsh, Jul 24 2003
With interpolated zeros, trace(A^n)/4, where A is the adjacency matrix of path graph P_4. Binomial transform is then A049680. - Paul Barry, Apr 24 2004
a(n) = (floor((3+sqrt(5))^n) + 1)/2^n. - Lekraj Beedassy, Oct 22 2004
a(n) = ((3-sqrt(5))^n + (3+sqrt(5))^n)/2^n (Note: substituting the number 1 for 3 in the last equation gives A000204, substituting 5 for 3 gives A020876). - Creighton Dement, Apr 19 2005
a(n) = (1/(n+1/2))*Sum_{k=0..n} B(2k)*L(2n+1-2k)*binomial(2n+1, 2k) where B(2k) is the (2k)-th Bernoulli number. - Benoit Cloitre, Nov 02 2005
a(n) = term (1,1) in the 1 X 2 matrix [2,3] . [3,1; -1,0]^n. - Alois P. Heinz, Jul 31 2008
a(n) = 2*cosh(2*n*psi), where psi=log((1+sqrt(5))/2). - Al Hakanson, Mar 21 2009
From Sarah-Marie Belcastro, Jul 04 2009: (Start)
a(n) - (a(n) - F(2n))/2 - F(2n+1) = 0. (Tesler)
Product_{r=1..n} (1 + 4*(sin((4r-1)*Pi/(4n)))^2). (Lu/Wu) (End)
a(n) = Fibonacci(2n+6) mod Fibonacci(2n+2), n > 1. - Gary Detlefs, Nov 22 2010
a(n) = 5*Fibonacci(n)^2 + 2*(-1)^n. - Gary Detlefs, Nov 22 2010
a(n) = A033888(n)/A001906(n), n > 0. - Gary Detlefs, Dec 26 2010
a(n) = 2^(2*n) * Sum_{k=1..2} (cos(k*Pi/5))^(2*n). - L. Edson Jeffery, Jan 21 2012
From Peter Bala, Jan 04 2013: (Start)
Let F(x) = Product_{n>=0} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let alpha = 1/2*(3 - sqrt(5)). This sequence gives the simple continued fraction expansion of 1 + F(alpha) = 2.31829 56058 81914 31334 ... = 2 + 1/(3 + 1/(7 + 1/(18 + ...))).
Also F(-alpha) = 0.64985 97768 07374 32950 has the continued fraction representation 1 - 1/(3 - 1/(7 - 1/(18 - ...))) and the simple continued fraction expansion 1/(1 + 1/((3-2) + 1/(1 + 1/((7-2) + 1/(1 + 1/((18-2) + 1/(1 + ...))))))).
F(alpha)*F(-alpha) has the simple continued fraction expansion 1/(1 + 1/((3^2-4) + 1/(1 + 1/((7^2-4) + 1/(1 + 1/((18^2-4) + 1/(1 + ...))))))).
Added Oct 13 2019: 1/2 + 1/2*F(alpha)/F(-alpha) = 1.5142923542... has the simple continued fraction expansion 1 + 1/((3 - 2) + 1/(1 + 1/((18 - 2) + 1/(1 + 1/(123 - 2) + 1/(1 + ...))))). (End)
G.f.: (W(0)+6)/(5*x), where W(k) = 5*x*k + x - 6 + 6*x*(5*k-9)/W(k+1) (continued fraction). - Sergei N. Gladkovskii, Aug 19 2013
Sum_{n >= 1} 1/( a(n) - 5/a(n) ) = 1. Compare with A001906, A002878 and A023039. - Peter Bala, Nov 29 2013
0 = a(n) * a(n+2) - a(n+1)^2 - 5 for all n in Z. - Michael Somos, Aug 24 2014
a(n) = (G(j+2n) + G(j-2n))/G(j), for n >= 0 and any j, positive or negative, except where G(j) = 0, and for any sequence of the form G(n+1) = G(n) + G(n-1) with any initial values for G(0), G(1), including non-integer values. G(n) includes Lucas, Fibonacci. Compare with A081067 for odd number offsets from j. - Richard R. Forberg, Nov 16 2014
a(n) = [x^n] ( (1 + 3*x + sqrt(1 + 6*x + 5*x^2))/2 )^n for n >= 1. - Peter Bala, Jun 23 2015
From J. M. Bergot, Oct 28 2015: (Start)
For n>0, a(n) = F(n-1) * L(n) + F(2*n+1) - (-1)^n with F(k) = A000045(k).
For n>1, a(n) = F(n+1) * L(n) + F(2*n-1) - (-1)^n.
For n>2, a(n) = 5*F(2*n-3) + 2*L(n-3) * L(n) + 8*(-1)^n. (End)
For n>1, a(n) = L(n-2)*L(n+2) -7*(-1)^n. - J. M. Bergot, Feb 10 2016
a(n) = 6*F(n-1)*L(n-1) - F(2*n-6) with F(n)=A000045(n) and L(n)=A000032(n). - J. M. Bergot, Apr 21 2017
a(n) = F(2*n) + 2*F(n-1)*L(n) with F(n)=A000045(n) and L(n)=A000032(n). - J. M. Bergot, May 01 2017
E.g.f.: exp(4*x/(1+sqrt(5))^2) + exp((1/4)*(1+sqrt(5))^2*x). - Stefano Spezia, Aug 13 2019
From Peter Bala, Oct 14 2019: (Start)
a(n) = F(2*n+2) - F(2*n-2) = A001906(n+1) - A001906(n-1).
a(n) = trace(M^n), where M is the 2 X 2 matrix [0, 1; 1, 1]^2 = [1, 1; 1, 2].
Consequently the Gauss congruences hold: a(n*p^k) = a(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k. See Zarelua and also Stanley (Ch. 5, Ex. 5.2(a) and its solution).
Sum_{n >= 1} (-1)^(n+1)/( a(n) + 1/a(n) ) = 1/5.
Sum_{n >= 1} (-1)^(n+1)/( a(n) + 3/(a(n) + 2/(a(n))) ) = 1/6.
Sum_{n >= 1} (-1)^(n+1)/( a(n) + 9/(a(n) + 4/(a(n) + 1/(a(n)))) ) = 1/9.
x*exp(Sum_{n >= 1} a(n)*x^/n) = x + 3*x^2 + 8*x^3 + 21*x^4 + ... is the o.g.f. for A001906. (End)
a(n) = n + 2 + Sum_{k=1..n-1} k*a(n-k). - Yu Xiao, May 30 2020
Sum_{n>=1} 1/a(n) = A153415. - Amiram Eldar, Nov 11 2020
Sum_{n>=0} 1/(a(n) + 3) = (2*sqrt(5) + 1)/10 (André-Jeannin, 1991). - Amiram Eldar, Jan 23 2022
a(n) = 2*cosh(2*n*arccsch(2)) = 2*cosh(2*n*asinh(1/2)). - Peter Luschny, May 25 2022
a(n) = (5/2)*(Sum_{k=-n..n} binomial(2*n, n+5*k)) - (1/2)*4^n. - Greg Dresden, Jan 05 2023
a(n) = Sum_{k>=0} Lucas(2*n*k)/(Lucas(2*n)^(k+1)). - Diego Rattaggi, Jan 12 2025

Extensions

Additional comments from Michael Somos, Jun 23 2001

A005592 a(n) = F(2n+1) + F(2n-1) - 1.

Original entry on oeis.org

1, 2, 6, 17, 46, 122, 321, 842, 2206, 5777, 15126, 39602, 103681, 271442, 710646, 1860497, 4870846, 12752042, 33385281, 87403802, 228826126, 599074577, 1568397606, 4106118242, 10749957121, 28143753122, 73681302246, 192900153617, 505019158606, 1322157322202
Offset: 0

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Author

Keywords

Comments

For any m, the maximum element in the continued fraction of F(2n+m)/F(m) is a(n). - Benoit Cloitre, Jan 10 2006
The continued fraction [a(n);1,a(n)-1,1,a(n)-1,...] = phi^(2n), where phi = 1.618... is the golden ratio, A001622. - Thomas Ordowski, Jun 07 2013
a(n) is the number of labeled subgraphs of the n-cycle C_n. For example, a(3)=17. There are 7 subgraphs of the triangle C_3 with 0 edges, 6 with 1 edge, 3 with 2 edges, and 1 with 3 edges (C_3 itself); here 7+6+3+1 = 17. - John P. McSorley, Oct 31 2016
a(n) equals the sum of the n-th row of triangle A277919. - John P. McSorley, Nov 25 2016

Examples

			G.f. = 1 + 2*x + 6*x^2 + 17*x^3 + 46*x^4 + 122*x^5 + 321*x^6 + 842*x^7 + ...
		

References

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

Crossrefs

Equals A004146+1 and A005248+1.
Bisection of A014217; the other bisection is A002878, which also bisects A000032.

Programs

  • Haskell
    a005592 n = a005592_list !! (n-1)
    a005592_list = map (subtract 1) $
                       tail $ zipWith (+) a001519_list $ tail a001519_list
    -- Reinhard Zumkeller, Aug 09 2013
  • Magma
    [Fibonacci(2*n+1)+Fibonacci(2*n-1)-1: n in [1..30]]; // Vincenzo Librandi, Aug 23 2011
    
  • Maple
    A005592:=-(2-2*z+z**2)/(z-1)/(z**2-3*z+1); # conjectured by Simon Plouffe in his 1992 dissertation
    # second Maple program:
    F:= n-> (<<0|1>, <1|1>>^n)[1,2]:
    a:= n-> F(2*n+1)+F(2*n-1)-1:
    seq(a(n), n=0..30);  # Alois P. Heinz, Nov 04 2016
  • Mathematica
    Table[Fibonacci[2n+1]+Fibonacci[2n-1]-1,{n,30}] (* Harvey P. Dale, Aug 22 2011 *)
    a[n_] := LucasL[2n]-1; Array[a, 30] (* Jean-François Alcover, Dec 09 2015 *)
  • PARI
    a(n)=fibonacci(2*n+1)+fibonacci(2*n-1)-1 \\ Charles R Greathouse IV, Aug 23 2011
    
  • Sage
    [lucas_number2(n,3,1)-1 for n in range(1,29)] # Zerinvary Lajos, Jul 06 2008
    

Formula

a(n) = Lucas(2*n)-1, with Lucas(n)=A000032(n).
a(n) = floor(r^(2*n)), where r = golden ratio = (1+sqrt(5))/2.
a(n) = floor(Fibonacci(5*n)/Fibonacci(3*n)). - Gary Detlefs, Mar 11 2011
a(n) = +4*a(n-1) -4*a(n-2) +1*a(n-3). - Joerg Arndt, Mar 11 2011
a(n) = A001519(2*n-1) + A001519(2*n+1) - 1. - Reinhard Zumkeller, Aug 09 2013
a(n) = 3*a(n) - a(n-1) + 1; a(n) = A004146(n) + 1, n>0. - Richard R. Forberg, Sep 04 2013
a(n) = 2*cosh(2*n*arcsinh(1/2)) - 1. - Ilya Gutkovskiy, Oct 31 2016
a(n) = floor(sqrt(5)*Fibonacci(2*n)), for n > 0 (Seamons, 1966). - Amiram Eldar, Feb 05 2022

Extensions

Formulae and comments by Clark Kimberling, Nov 24 2010
a(0)=1 prepended by Alois P. Heinz, Nov 04 2016

A104626 Numbers having three 1's in their base-phi representation.

Original entry on oeis.org

4, 5, 6, 8, 19, 48, 124, 323, 844, 2208, 5779, 15128, 39604, 103683, 271444, 710648, 1860499, 4870848, 12752044, 33385283, 87403804, 228826128, 599074579, 1568397608, 4106118244, 10749957123, 28143753124, 73681302248
Offset: 1

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Author

Eric W. Weisstein, Mar 17 2005

Keywords

Crossrefs

Programs

  • Magma
    [4,5,6] cat [1 + Fibonacci(2*n-3) + Fibonacci(2*n-5): n in [4..50]]; // G. C. Greubel, Aug 13 2018
  • Mathematica
    Join[{4, 5, 6}, Table[LucasL[2*n-4] + 1, {n, 4, 50}]] (* G. C. Greubel, Aug 13 2018 *)
  • PARI
    for(n=1,50, print1(if(n==1,4, if(n==2, 5, if(n==3, 6, 1 + fibonacci(2*n-3) + fibonacci(2*n-5)))), ", ")) \\ G. C. Greubel, Aug 13 2018
    

Formula

{n: A055778(n) = 3}. - R. J. Mathar, Sep 05 2010
a(n) = Lucas(2*n-4) + 1, for n>3. - Ralf Stephan, Nov 13 2010

Extensions

Terms 5 and 6 added by Jaroslav Krizek, May 25 2010
Edited by R. J. Mathar, Sep 05 2010

A092387 a(n) = Fibonacci(2*n+1) + Fibonacci(2*n-1) + 2.

Original entry on oeis.org

5, 9, 20, 49, 125, 324, 845, 2209, 5780, 15129, 39605, 103684, 271445, 710649, 1860500, 4870849, 12752045, 33385284, 87403805, 228826129, 599074580, 1568397609, 4106118245, 10749957124, 28143753125, 73681302249, 192900153620
Offset: 1

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Author

Benoit Cloitre, Mar 20 2004

Keywords

Comments

Let b(k)=sum(i=1,k,F(2*n*i)*binomial(k,i)) where F(k) denotes the k-th Fibonacci number. The (b(k)) sequence satisfies the recursion: b(k)=a(n)*(b(k-1)-b(k-2)).
Same as the number of Kekulé structures in polyphenanthrene in terms of the number of hexagons. - Parthasarathy Nambi, Aug 22 2006

Crossrefs

Equals A065034(n)+1.

Programs

  • Magma
    I:=[5, 9, 20]; [n le 3 select I[n] else 4*Self(n-1)-4*Self(n-2)+Self(n-3): n in [1..30]]; // Vincenzo Librandi, May 06 2012
  • Mathematica
    CoefficientList[Series[-(5-11*x+4*x^2)/((x-1)(x^2-3*x+1)),  {x,0,30}],x] (* Vincenzo Librandi, May 06 2012 *)

Formula

a(1)=5, a(2)=9, a(3)=20, a(n) = 4*a(n-1) - 4*a(n-2) + a(n-3).
a(n) = 3 + floor((1+phi)^n) where phi = (1+sqrt(5))/2.
a(n) = A005248(n) + 2.
From R. J. Mathar, Mar 18 2009: (Start)
G.f.: -x*(5 - 11*x + 4*x^2)/((x-1)(x^2 - 3*x + 1)).
a(n+1) - a(n) = A002878(n). (End)

Extensions

Better definition from Mohammed Bouayoun (bouyao(AT)wanadoo.fr), Mar 20 2004
Showing 1-5 of 5 results.