A006483 -id:A006483 - cp's OEIS frontend

A063727 a(n) = 2a(n-1) + 4a(n-2), a(0)=1, a(1)=2.

1, 2, 8, 24, 80, 256, 832, 2688, 8704, 28160, 91136, 294912, 954368, 3088384, 9994240, 32342016, 104660992, 338690048, 1096024064, 3546808320, 11477712896, 37142659072, 120196169728, 388962975744, 1258710630400

Offset: 0

Klaus E. Kastberg (kastberg(AT)hotkey.net.au), Aug 12 2001

Essentially the same as A085449.
Convergents to 2*golden ratio = (1+sqrt(5)).
Number of ways to tile an n-board with two types of colored squares and four types of colored dominoes.
The same sequence may be obtained by the following process. Starting a priori with the fraction 1/1, the numerators of fractions built according to the rule: add top and bottom to get the new bottom, add top and 5 times the bottom to get the new top. The limit of the sequence of fractions is sqrt(5). - Cino Hilliard, Sep 25 2005
a(n) is also the quasi-diagonal element A(i-1,i)=A(1,i-1) of matrix A(i,j) whose elements in first row A(1,k) and first column A(k,1) equal k-th Fibonacci Fib(k) and the generic element is the sum of adjacent (previous) in row and column minus the absolute value of their difference. - Carmine Suriano, May 13 2010
Equals INVERT transform of A006131: (1, 1, 5, 9, 29, 65, 181, ...). - Gary W. Adamson, Aug 12 2010
For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 2's along the three central diagonals. - John M. Campbell, Jul 19 2011
The numbers composing the denominators of the fractional limit to A134972. - Seiichi Kirikami, Mar 06 2012
Pisano period lengths:  1, 1, 8, 1, 5, 8, 48, 1, 24, 5, 10, 8, 42, 48, 40, 1, 72, 24, 18, 5, ... - R. J. Mathar, Aug 10 2012

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 235.
John Derbyshire, Prime Obsession, Joseph Henry Press, April 2004, see p. 16.

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..200 from Harry J. Smith)
J. Borowska and L. Lacinska, Recurrence form of determinant of a heptadiagonal symmetric Toeplitz matrix, J. Appl. Math. Comp. Mech. 13 (2014) 19-16, remark 2 for permanent of tridiagonal Toeplitz matrices a=2, b=2.
Index entries for linear recurrences with constant coefficients, signature (2,4).
Index entries for sequences related to Chebyshev polynomials.

Cf. A006483, A103435, A006131, A269991.
Second row of A234357. Row sums of triangle A016095.
The following sequences (and others) belong to the same family: A001333, A000129, A026150, A002605, A046717, A015518, A084057, A063727, A002533, A002532, A083098, A083099, A083100, A015519.

List([0..25],n->2^n*Fibonacci(n+1)); # Muniru A Asiru, Nov 24 2018

[n le 2 select n else 2*Self(n-1) + 4*Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 07 2018

a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=2*a[n-1]+4*a[n-2]od: seq(a[n], n=1..33); # Zerinvary Lajos, Dec 15 2008

a[n_]:=(MatrixPower[{{1,5},{1,1}},n].{{1},{1}})[[2,1]]; Table[Abs[a[n]],{n,-1,40}] (* Vladimir Joseph Stephan Orlovsky, Feb 19 2010 *)
CoefficientList[Series[1/(1 - 2 x - 4 x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Oct 31 2014 *)
LinearRecurrence[{2, 4}, {1, 2}, 50] (* G. C. Greubel, Jan 07 2018 *)

s(n)=if(n<2,n+1,(s(n-1)+(s(n-2)*2))*2); for(n=0,32,print(s(n)))

{ for (n=0, 200, if (n>1, a=2*a1 + 4*a2; a2=a1; a1=a, if (n, a=a1=2, a=a2=1)); write("b063727.txt", n, " ", a) ) } \\ Harry J. Smith, Aug 28 2009

[lucas_number1(n,2,-4) for n in range(1, 26)] # Zerinvary Lajos, Apr 22 2009

a(n) = 2 * A087206(n+1).
From Vladeta Jovovic, Aug 16 2001: (Start)
a(n) = sqrt(5)/10*((1+sqrt(5))^(n+1) - (1-sqrt(5))^(n+1)).
G.f.: 1/(1-2*x-4*x^2). (End)
From Mario Catalani (mario.catalani(AT)unito.it), Jun 13 2003: (Start)
a(2*n) = 4*a(n-1)^2 + a(n)^2.
A084057(n+1)/a(n) converges to sqrt(5). (End)
E.g.f.: exp(x)*(cosh(sqrt(5)*x)+sinh(sqrt(5)*x)/sqrt(5)). - Paul Barry, Sep 20 2003
a(n) = 2^n*Fibonacci(n+1). - Vladeta Jovovic, Oct 25 2003
a(n) = Sum_{k=0..floor(n/2)} C(n, 2*k+1)*5^k. - Paul Barry, Nov 15 2003
a(n) = U(n, i/2)*(-i*2)^n, i^2=-1. - Paul Barry, Nov 17 2003
Simplified formula: ((1+sqrt(5))^n-(1-sqrt(5))^n)/sqrt(20). Offset 1. a(3)=8. - Al Hakanson (hawkuu(AT)gmail.com), Jan 03 2009
First binomial transform of 1,1,5,5,25,25. - Al Hakanson (hawkuu(AT)gmail.com), Jul 20 2009
a(n) = A(n-1,n) = A(n,n-1); A(i,j) = A(i-1,j) + A(i,j-1) - abs(A(i-1,j) - A(i,j-1)). - Carmine Suriano, May 13 2010
G.f.: G(0) where G(k) = 1 + 2*x*(1+2*x)/(1 - 2*x*(1+2*x)/(2*x*(1+2*x) + 1/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jan 31 2013
G.f.: G(0)/(2*(1-x)), where G(k) = 1 + 1/(1 - x*(5*k-1)/(x*(5*k+4) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013
G.f.: Q(0)/2 , where Q(k) = 1 + 1/(1 - x*(4*k+2 + 4*x )/( x*(4*k+4 + 4*x ) + 1/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, Sep 21 2013
Sum_{n>=0} 1/a(n) = A269991. - Amiram Eldar, Feb 01 2021

Better description from Jason Earls and Vladeta Jovovic, Aug 16 2001

Incorrect comment removed by Greg Dresden, Jun 02 2020

A103435 a(n) = 2^n * Fibonacci(n).

Original entry on oeis.org

0, 2, 4, 16, 48, 160, 512, 1664, 5376, 17408, 56320, 182272, 589824, 1908736, 6176768, 19988480, 64684032, 209321984, 677380096, 2192048128, 7093616640, 22955425792, 74285318144, 240392339456, 777925951488, 2517421260800

Offset: 0

Ralf Stephan, Feb 08 2005

Cardinality of set of bracelets of size at most n that are tiled with two types of colored squares and four types of colored dominoes.
a(n) is also the diagonal element of the matrix A(i,j) whose first row (i=1) and first column (j=1) are the Fibonacci numbers: A(1,k)=A(k,1)=fib(k) and whose generic element is the sum of element in adjacent (preceding) row and column minus the absolute value of their difference. So a(n) = A(n,n) = A(i-1,j)+A(i,j-1)-abs(A(i-1,j)-A(i,j-1)). - Carmine Suriano, May 13 2010
a(n) is the coefficient of x in the reduction by x^2->x+1 of the polynomial p(n,x) given for d=sqrt(x+1) by p(n,x)=((x+d)^n-(x-d)^n)/(2d), for n>=1.  The constant terms under this reduction are the absolute values of terms of A086344.  See A192232 for a discussion of reduction. - Clark Kimberling, Jun 29 2011
The exponential convolution of A000032 and A000045. - Vladimir Reshetnikov, Oct 06 2016

			a(5)=160=A(5,5)=A(4,5)+A(5,4)-abs[A(4,5)+A(5,4)]=80+80-0. - _Carmine Suriano_, May 13 2010
G.f. = 2*x + 4*x^2 + 16*x^3 + 48*x^4 + 160*x^5 + 512*x^6 + 1664*x^7 + ...

Arthur T. Benjamin and Jennifer J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A., 2003, identity 236, p. 131.

Tom Edgar, Extending Some Fibonacci-Lucas Relations, The Fibonacci Quarterly, Vol. 54, No. 1 (2016), p. 79.
Harris Kwong, An Alternate Proof of Sury's Fibonacci-Lucas Relation, The American Mathematical Monthly, Vol. 121, No. 6 (2014), p. 514.
Diego Marques, A new Fibonacci-Lucas relation, Amer. Math. Monthly, Vol. 122, No. 7 (2015), p. 683.
Ivica Martinjak and Helmut Prodinger, Complementary Families of the Fibonacci-Lucas Relations, arXiv:1508.04949 [math.CO], 2015-2016.
B. Sury, A polynomial parent to a Fibonacci-Lucas relations, Amer. Math. Monthly, Vol. 121, No. 3 (2014), p. 236.
Charles R. Wall, Problem B-607, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 25, No. 4 (1987), p. 370; Product of Exponential Generating Functions, Solution to Problem B-607 by Bob Prielipp, ibid., Vol. 26, No. 4 (1988), pp. 374-375.
Index entries for linear recurrences with constant coefficients, signature (2,4).

First differences of A014334.
Partial sums of A087131.
Cf. A000032, A000045, A006483, A063727, A085449, A269991.

[2^n *Fibonacci(n): n in [0..50]]; // Vincenzo Librandi, Apr 04 2011

Expand[Table[((1 + Sqrt[5])^n - (1 - Sqrt[5])^n)5/(5 Sqrt[5]), {n, 0, 25}]] (* Zerinvary Lajos, Mar 22 2007 *)
Table[2^n Fibonacci[n],{n,0,40}] (* or *) LinearRecurrence[{2,4},{0,2},40] (* Harvey P. Dale, Oct 14 2020 *)

a(n)=fibonacci(n)<Charles R Greathouse IV, Feb 03 2014

concat(0, Vec(2*x/(1-2*x-4*x^2) + O(x^99))) \\ Altug Alkan, May 11 2016

a(n) = A006483(n) + 1 = 2*A085449(n) = 2*A063727(n-1), n>0.
G.f.: 2*x / (1 - 2*x - 4*x^2).
a(n) = Sum_{i=0..n-1}( 2^i * Lucas(i) ).
a(n) = 2*a(n-1) + 4*a(n-2). - Carmine Suriano, May 13 2010
a(n) = a(-n) * -(-4)^n for all n in Z. - Michael Somos, Sep 20 2014
E.g.f.: 2*sinh(sqrt(5)*x)*exp(x)/sqrt(5). - Ilya Gutkovskiy, May 10 2016
Sum_{n>=1} 1/a(n) = (1/2) * A269991. - Amiram Eldar, Nov 17 2020
a(n) == 2*n (mod 10). - Amiram Eldar, Jan 15 2022
a(n) = Sum_{k=0..n} binomial(n,k) * Fibonacci(k) * Lucas(n-k) (Wall, 1987). - Amiram Eldar, Jan 27 2022

A087131 a(n) = 2^n*Lucas(n), where Lucas = A000032.

Original entry on oeis.org

2, 2, 12, 32, 112, 352, 1152, 3712, 12032, 38912, 125952, 407552, 1318912, 4268032, 13811712, 44695552, 144637952, 468058112, 1514668032, 4901568512, 15861809152, 51329892352, 166107021312, 537533612032, 1739495309312

Offset: 0

Paul Barry, Aug 16 2003

Number of ways to tile an n-bracelet with two types of colored squares and four types of colored dominoes.
Inverse binomial transform of even Lucas numbers (A014448).
From L. Edson Jeffery, Apr 25 2011: (Start)
Let A be the unit-primitive matrix (see [Jeffery])
A=A_(10,4)=
(0 0 0 0 1)
(0 0 0 2 0)
(0 0 2 0 1)
(0 2 0 2 0)
(2 0 2 0 1).
Then a(n)=(Trace(A^n)-1)/2. Also a(n)=Trace((2*A_(5,1))^n), where A_(5,1)=[(0,1); (1,1)] is also a unit-primitive matrix. (End)
Also the number of connected dominating sets in the n-sun graph for n >= 3. - Eric W. Weisstein, May 02 2017
Also the number of total dominating sets in the n-sun graph for n >= 3. - Eric W. Weisstein, Apr 27 2018

Arthur T. Benjamin and Jennifer J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A., 2003, identity 237, p. 132.

L. Edson Jeffery, Unit-primitive matrices
Eric Weisstein's World of Mathematics, Connected Dominating Set.
Eric Weisstein's World of Mathematics, Sun Graph.
Eric Weisstein's World of Mathematics, Total Dominating Set.
Index entries for linear recurrences with constant coefficients, signature (2,4).

First differences of A006483 and A103435.

Cf. A000032, A014334, A014335, A084057, A269992.

[2] cat [2^n*Lucas(n): n in [1..30]]; // G. C. Greubel, Dec 18 2017

Table[Tr[MatrixPower[{{2, 2}, {2, 0}}, x]], {x, 1, 20}] (* Artur Jasinski, Jan 09 2007 *)
Join[{2}, Table[2^n LucasL[n], {n, 20}]] (* Eric W. Weisstein, May 02 2017 *)
Join[{2}, 2^# LucasL[#] & [Range[20]]] (* Eric W. Weisstein, May 02 2017 *)
LinearRecurrence[{2, 4}, {2, 12}, {0, 20}] (* Eric W. Weisstein, Apr 27 2018 *)
CoefficientList[Series[(2 (-1 + x))/(-1 + 2 x + 4 x^2), {x, 0, 20}], x] (* Eric W. Weisstein, Apr 27 2018 *)

for(n=0,30, print1(if(n==0, 2, 2^n*(fibonacci(n+1) + fibonacci(n-1))), ", ")) \\ G. C. Greubel, Dec 18 2017

first(n) = Vec(2*(1-x)/(1-2*x-4*x^2) + O(x^n)) \\ Iain Fox, Dec 19 2017

[lucas_number2(n,2,-4) for n in range(0, 25)] # Zerinvary Lajos, Apr 30 2009

a(n) = 2*A084057(n).
Recurrence: a(n) = 2a(n-1) + 4a(n-2), a(0)=2, a(1)=2.
G.f.: 2*(1-x)/(1-2*x-4*x^2).
a(n) = (1+sqrt(5))^n + (1-sqrt(5))^n.
For n>=2, a(n) = Trace of matrix [({2,2},{2,0})^n]. - Artur Jasinski, Jan 09 2007
a(n) = 2*[A063727(n)-A063727(n-1)]. - R. J. Mathar, Nov 16 2007
a(n) = (5*A052899(n)-1)/2. - L. Edson Jeffery, Apr 25 2011
a(n) = [x^n] ( 1 + x + sqrt(1 + 2*x + 5*x^2) )^n for n >= 1. - Peter Bala, Jun 23 2015
Sum_{n>=1} 1/a(n) = (1/2) * A269992. - Amiram Eldar, Nov 17 2020
From Amiram Eldar, Jan 15 2022: (Start)
a(n) == 2 (mod 10).
a(n) = 5 * A014334(n) + 2.
a(n) = 10 * A014335(n) + 2. (End)

Edited by Ralf Stephan, Feb 08 2005

A087206 a(n) = 2a(n-1) + 4a(n-2); with a(0)=1, a(1)=4.

Original entry on oeis.org

1, 4, 12, 40, 128, 416, 1344, 4352, 14080, 45568, 147456, 477184, 1544192, 4997120, 16171008, 52330496, 169345024, 548012032, 1773404160, 5738856448, 18571329536, 60098084864, 194481487872, 629355315200, 2036636581888

Offset: 0

Paul Barry, Aug 25 2003

Binomial transform of A056487. Unsigned version of A152174.
Number of words of length n over the alphabet {1,2,3,4} such that no odd letter is followed by an odd letter. - Armend Shabani, Feb 18 2017
From Sean A. Irvine, Jun 06 2025: (Start)
Also, the number of walks of length n starting at 0 in the following graph:
  1---2
  |\ /|
  | 0 |
  |/ \|
  4---3. (End)

Jens Christian Claussen, Time-evolution of the Rule 150 cellular automaton activity from a Fibonacci iteration, arXiv:math/0410429 [math.CO], 2004. See Table II, p. 4.
Sean A. Irvine, Walks on Graphs.
N. J. A. Sloane, On the Number of ON Cells in Cellular Automata, arXiv:1503.01168 [math.CO], 2015.
Index entries for linear recurrences with constant coefficients, signature (2,4).

Cf. A060925, A063782, A253064.

Equals (1/2) * A063727(n-1). Cf. A006483.

LinearRecurrence[{2, 4}, {1, 4}, 25] (* Jean-François Alcover, Sep 21 2017 *)

G.f.: (1+2x)/(1-2x-4x^2).
a(n) = (1-sqrt(5))^n*(1/2-3*sqrt(5)/10)+(1+sqrt(5))^n*(1/2+3*sqrt(5)/10).
a(n) = 2^n*Fibonacci(n+2). - Paul Barry, Mar 22 2004
a(n) = ((1+sqrt(5))^n-(1-sqrt(5))^n)/sqrt(80). Offset 2. a(4)=12. - Al Hakanson (hawkuu(AT)gmail.com), Apr 11 2009
G.f.: 1/(-2x-1/(-2x-1)). - Paul Barry, Mar 24 2010

Comment corrected by Philippe Deléham, Nov 27 2008

cp's OEIS Frontend

A059670 Numbers k such that F(k)*2^k + 1 (A006483) is prime, where F(k) is the k-th Fibonacci number.

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A063727 a(n) = 2*a(n-1) + 4*a(n-2), a(0)=1, a(1)=2.

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A103435 a(n) = 2^n * Fibonacci(n).

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A087131 a(n) = 2^n*Lucas(n), where Lucas = A000032.

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A087206 a(n) = 2*a(n-1) + 4*a(n-2); with a(0)=1, a(1)=4.

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A063727 a(n) = 2a(n-1) + 4a(n-2), a(0)=1, a(1)=2.

A087206 a(n) = 2a(n-1) + 4a(n-2); with a(0)=1, a(1)=4.