A269991 -id:A269991 - 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

A085449 Horadam sequence (0,1,4,2).

Original entry on oeis.org

0, 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

Ross La Haye, Aug 18 2003

a(n) / a(n-1) converges to sqrt(5) + 1 as n approaches infinity. sqrt(5) + 1 can also be written as Phi^3 - 1, 2 * Phi, Phi^2 + Phi - 1 and (L(n) / F(n)) + 1, where L(n) is the n-th Lucas number and F(n) is the n-th Fibonacci number as n approaches infinity.

Binomial transform is A001076. - Paul Barry, Aug 25 2003

			a(4) = 24 because a(3) = 8, a(2) = 2, s = 2, r = 4 and (2 * 8) + (4 * 2) = 24.
G.f. = x + 2*x^2 + 8*x^3 + 24*x^4 + 80*x^5 + 256*x^6 + 832*x^7 + ... - _Michael Somos_, Mar 07 2021

Karl V. Keller, Jr., Table of n, a(n) for n = 0..1000
F. P. Muga II, Extending the Golden Ratio and the Binet-de Moivre Formula, Preprint on ResearchGate, March 2014.
Eric Weisstein, Horadam Sequence.
Eric Weisstein, Fibonacci Number.
Eric Weisstein, Pell Number.
Eric Weisstein, Lucas Number.
Eric Weisstein, Lucas Sequence.
Index entries for linear recurrences with constant coefficients, signature (2, 4).

Cf. A024318, A000032, A000129, A001076, A085939, A269991.

Essentially the same as A063727.

a:=[0,1];; for n in [3..30] do a[n]:=2*a[n-1]+4*a[n-2]; od; a; # Muniru A Asiru, Oct 09 2018

[2^(n-1)*Fibonacci(n): n in [0..50]]; // G. C. Greubel, Oct 08 2018

a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=2*(a[n-1]+2*a[n-2]) od: seq(a[n], n=0..26); # Zerinvary Lajos, Mar 17 2008

Table[2^(n-1)*Fibonacci[n], {n,0,50}] (* G. C. Greubel, Oct 08 2018 *)

vector(50, n, n--; 2^(n-1)*fibonacci(n)) \\ G. C. Greubel, Oct 08 2018

a(n) = s*a(n-1) + r*a(n-2); for n > 1, where a(0) = 0, a(1) = 1, s = 2, r = 4.
From Paul Barry, Aug 25 2003: (Start)
G.f.: x/(1-2*x-4*x^2).
a(n) = sqrt(5)*((1+sqrt(5))^n - (1-sqrt(5))^n)/10.
a(n) = Sum_{k=0..floor(n/2)} C(n, 2*k+1)5^k . (End)
The signed version 0, 1, -2, ... has a(n)=sqrt(5)((sqrt(5)-1)^n-(-sqrt(5)-1)^n)/10. It is the second inverse binomial transform of A085449. - Paul Barry, Aug 25 2003
a(n) = 2^(n-1)*Fib(n). - Paul Barry, Mar 22 2004
Sum_{n>=1} 1/a(n) = A269991. - Amiram Eldar, Feb 01 2021
a(n) = -(-4)^n*a(-n) for all integer n. - Michael Somos, Mar 07 2021

A209084 a(n) = 2a(n-1) + 4a(n-2) with n>1, a(0)=0, a(1)=4.

Original entry on oeis.org

0, 4, 8, 32, 96, 320, 1024, 3328, 10752, 34816, 112640, 364544, 1179648, 3817472, 12353536, 39976960, 129368064, 418643968, 1354760192, 4384096256, 14187233280, 45910851584, 148570636288, 480784678912, 1555851902976, 5034842521600, 16293092655104

Offset: 0

Seiichi Kirikami, Mar 06 2012

a(n)/A063727(n) are convergents for A134972.

Abs(Sum_{i=0..n} C(n,n-i)*a(i)-(sqrt(5)-1)* A033887(n))->0. - Seiichi Kirikami, Jan 20 2016

E. W. Cheney, Introduction to Approximation Theory, McGraw-Hill, Inc., 1966.

Bruno Berselli, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (2,4).

Cf. A063727, A033887, A085449, A103435, A134972, A269991.

Cf. A086344 (this sequence with signs).

I:=[0,4]; [n le 2 select I[n] else 2*Self(n-1)+4*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jan 16 2016

RecurrenceTable[{a[n]==2*a[n-1]+4*a[n-2], a[0]==0, a[1]==4}, a, {n, 30}]
LinearRecurrence[{2, 4}, {0, 4}, 40] (* Vincenzo Librandi, Jan 16 2016 *)

concat(0, Vec(4*x/(1-2*x-4*x^2) + O(x^40))) \\ Michel Marcus, Jan 16 2016

a(n) = (2/sqrt(5))*((1+sqrt(5))^n-(1-sqrt(5))^n).
G.f.: 4*x/(1-2*x-4*x^2). - Bruno Berselli, Mar 08 2012
a(n) = 4*A085449(n) = 2*A103435(n). - Bruno Berselli, Mar 08 2012
Sum_{n>=1} 1/a(n) = (1/4) * A269991. - Amiram Eldar, Feb 01 2021

A269992 Decimal expansion of Sum_{n>=1} 2^(1-n)/L(n), where L = A000032 (Lucas numbers).

Original entry on oeis.org

1, 2, 5, 5, 2, 2, 1, 1, 3, 4, 3, 2, 9, 8, 4, 8, 6, 0, 3, 1, 4, 0, 2, 6, 6, 7, 2, 7, 4, 4, 0, 3, 3, 6, 0, 1, 5, 6, 0, 5, 4, 3, 5, 7, 0, 4, 4, 4, 4, 3, 0, 0, 3, 8, 3, 6, 8, 8, 7, 0, 6, 2, 4, 1, 4, 9, 3, 0, 9, 6, 6, 8, 6, 0, 2, 5, 3, 8, 6, 3, 0, 8, 6, 8, 9, 0

Offset: 1

Clark Kimberling, Mar 12 2016

			1.2552211343298486031402667274403360...

Cf. A000032, A084057, A087131, A269991.

x = N[Sum[2^(1 - n)/LucasL[n], {n, 1, 500}], 100]
RealDigits[x][[1]]

L(n) = real((2 + quadgen(5)) * quadgen(5)^n); \\ A000032
suminf(n=1, 2^(1-n)/L(n)) \\ Michel Marcus, Nov 17 2020

Equals Sum_{n>=1} 1/A084057(n) = 2 * Sum_{n>=1} 1/A087131(n). - Amiram Eldar, Feb 01 2021

cp's OEIS Frontend

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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A085449 Horadam sequence (0,1,4,2).

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A209084 a(n) = 2*a(n-1) + 4*a(n-2) with n>1, a(0)=0, a(1)=4.

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A269992 Decimal expansion of Sum_{n>=1} 2^(1-n)/L(n), where L = A000032 (Lucas numbers).

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

A209084 a(n) = 2a(n-1) + 4a(n-2) with n>1, a(0)=0, a(1)=4.