|
| |
|
|
A001076
|
|
Denominators of continued fraction convergents to sqrt(5).
(Formerly M3538 N1434)
|
|
107
|
|
|
|
0, 1, 4, 17, 72, 305, 1292, 5473, 23184, 98209, 416020, 1762289, 7465176, 31622993, 133957148, 567451585, 2403763488, 10182505537, 43133785636, 182717648081, 774004377960, 3278735159921, 13888945017644, 58834515230497, 249227005939632, 1055742538989025
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,3
|
|
|
COMMENTS
|
a(2*n+1) with b(2*n+1) := A001077(2*n+1), n >= 0, give all (positive integer) solutions to Pell equation b^2 - 5*a^2 = -1, a(2*n) with b(2*n) := A001077(2*n), n >= 1, give all (positive integer) solutions to Pell equation b^2 - 5*a^2 = +1 (cf. Emerson reference).
Bisection: a(2*n+1) = T(2*n+1, sqrt(5))/sqrt(5) = A007805(n), n >= 0 and a(2*n) = 4*S(n-1,18), n >= 0, with T(n,x), resp. S(n,x), Chebyshev's polynomials of the first, resp. second kind. S(-1,x)=0. See A053120, resp. A049310. S(n,18)=A049660(n+1). - Wolfdieter Lang, Jan 10 2003
Apart from initial terms, this is the Pisot sequence E(4,17), a(n) = floor(a(n-1)^2/a(n-2) + 1/2).
This is also the Horadam sequence (0,1,1,4), having the recurrence relation a(n) = s*a(n-1) + r*a(n-2); for n > 1, where a(0) = 0, a(1) = 1, s = 4, r = 1. a(n) / a(n-1) converges to 5^1/2 + 2 as n approaches infinity. 5^(1/2) + 2 can also be written as (2 * Phi) + 1 and Phi^2 + Phi. - Ross La Haye, Aug 18 2003
Numerators of continued fraction [4, 4, 4, ...], where the convergents to [4, 4, 4, ...] = (4/1, 17/4, 72/17, ...). Let X = the 2 X 2 matrix [0, 1; 1, 4]; then X^n = [a(n-1), a(n); a(n), a(n+1)]; e.g., X^3 = [4, 17; 17, 72]. Let C = the limit of a(n)/a(n-1) = 2 + sqrt(5) = 4.236067977...; then C^n = a(n+1) + (1/C)*a(n), where (1/C) = 0.236067977... . Example: C^3 = 76.01315556..., = 72 + 17*(0.2360679...). - Gary W. Adamson, Dec 15 2007, corrected by Greg Dresden, Sep 16 2019, corrected by _Alex Mark_, Jul 21 2020
Sqrt(5) = 4/2 + 4/17 + 4/(17*305) + 4/(305*5473) + 4/(5473*98209) + ... . - Gary W. Adamson, Dec 15 2007
a(p) == 20^((p-1)/2)) mod p, for odd primes p. - Gary W. Adamson, Feb 22 2009
For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 4's along the main diagonal and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
Moreover, a(n) is the second binomial transform of (0,1,0,5,0,25,...) (see also A033887). This fact can be proved similarly like the proof of Paul Barry's remark in A033887 by using the following scaling identity for delta-Fibonacci numbers: y^n b(n;x/y) = Sum_{k=0..n} binomial(n,k) (y-1)^(n-k) b(k;x) and the fact that b(n;2) = (1-(-1)^n) 5^floor(n/2). - Roman Witula, Jul 12 2012
Binomial transform of 0, 1, 2, 8, 24, 80, 256, ... (A063727 with offset 1). - R. J. Mathar, Feb 05 2014
For n >= 1, a(n) equals the number of words of length n-1 on alphabet {0,1,...,4} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
With offset 1 is the INVERT transform of A006190: (1, 3, 10, 33, 109, 360, ...). - Gary W. Adamson, Jul 24 2015
This is a divisibility sequence (i.e., if n|m then a(n)|a(m)).
gcd(a(n),a(n+k)) = a(gcd(n, k)) for all positive integers n and k. (End)
The initial 0 of this sequence is in contradiction with the fact that 0 is no valid denominator and according to all standard references, the first convergent of a continued fraction is p(0)/q(0) = b(0)/1 where b(0) is the first term of the continued fraction, given by the integer part of the number. One may artificially define q(-1) = 0 to have a recurrent relation q(n) = b(n)*q(n-1) + q(n-2), n >= 1, but then its index should be -1. - M. F. Hasler, Nov 01 2019
Number of 4-compositions of n restricted to odd parts (and allowed zeros); see Hopkins & Ouvry reference. - Brian Hopkins, Aug 17 2020
Also called the 4-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence.
a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are 4 kinds of squares available. (End)
|
|
|
REFERENCES
|
A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 23.
S. Koshkin, Non-classical linear divisibility sequences ..., Fib. Q., 57 (No. 1, 2019), 68-80. See Table 1.
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).
V. Thébault, Les Récréations Mathématiques. Gauthier-Villars, Paris, 1952, p. 282.
|
|
|
LINKS
|
Yixing Fu, E. J. König, J. H. Wilson, Yang-Zhi Chou, and J. H. Pixley, Magic-angle semimetals, arXiv:1809.04604 [cond-mat.str-el], 2018.
|
|
|
FORMULA
|
a(n) = 4*a(n-1) + a(n-2), n > 1. a(0)=0, a(1)=1.
G.f.: x/(1 - 4*x - x^2).
a(n) = ((2+sqrt(5))^n - (2-sqrt(5))^n)/(2*sqrt(5)).
a(n) = ((-i)^(n-1))*S(n-1, 4*i), with i^2 =-1 and S(n, x) := U(n, x/2) Chebyshev's polynomials of the second kind. See A049310. S(-1, x)=0.
a(n) = Sum_{i=0..n} Sum_{j=0..n} Fibonacci(i+j)*n!/(i!j!(n-i-j)!)/2. - Paul Barry, Feb 06 2004
a(n) = F(1) + F(4) + F(7) + ... + F(3n-2), for n > 0.
a(n) = Sum_{k=0..n} Sum_{j=0..n} C(n, j)*C(j, k)F(j)/2. - Paul Barry, Feb 14 2005
Let M = {{0, 1}, {1, 4}}, v[1] = {0, 1}, v[n] = M.v[n - 1]; then a(n) = v[n][[1]]. - Roger L. Bagula, May 29 2005
a(n) = F(n, 4), the n-th Fibonacci polynomial evaluated at x=4. - T. D. Noe, Jan 19 2006
a(n) = (Sum_{k=0..n} Fibonacci(3*k-2)) + 1. - Gary Detlefs, Dec 26 2010
a(n) = (3*(-1)^n*F(n) + 5*F(n)^3)/2, n >= 0. See the general D. Jennings formula given in a comment on triangle A111125, where also the reference is given. Here the second (k=1) row [3,1] applies. - Wolfdieter Lang, Sep 01 2012
Sum_{k>=1} (-1)^(k-1)/(a(k)*a(k+1)) = (Sum_{k>=1} (-1)^(k-1)/(F_k*F_(k+1)))^3 = phi^(-3), where F_n is the n-th Fibonacci numbers (A000045) and phi is golden ratio (A001622). - Vladimir Shevelev, Feb 23 2013
G.f.: Q(0)*x/(2-4*x), where Q(k) = 1 + 1/(1 - x*(5*k-4)/(x*(5*k+1) - 2/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Oct 11 2013
The o.g.f. A(x) = x/(1 - 4*x - x^2) satisfies A(x) + A(-x) + 8*A(x)*A(-x) = 0 or equivalently (1 + 8*A(x))*(1 + 8*A(-x)) = 1. The o.g.f. for A049660 equals -A(sqrt(x))*A(-sqrt(x)). - Peter Bala, Apr 02 2015
Some properties:
(1) a(n)*a(n+1) = 4*Sum_{k=1..n} a(k)^2;
(2) a(n)^2 + a(n+1)^2 = a(2*n+1);
(3) a(n)^2 - a(n-2)^2 = 4*a(n-1)*(a(n) + a(n-2));
(4) a(m*(p+1)) = a(m*p)*a(m+1) + a(m*p-1)*a(m);
(5) a(n-k)*a(n+k) = a(n)^2 + (-1)^(n+k+1)*a(k)^2;
(6) a(n-1)*a(n+1) = a(n)^2 + (-1)^n (particular case of (5)!);
(7) a(2*n) = 2*a(n)*(2*a(n) + a(n-1));
(8) 3*Sum_{k=2..n+1} a(k)*a(k-1) is equal to a(n+1)^2 if n odd, and is equal to a(n+1)^2 - 1 if n is even;
(9) a(n) - a(n-2*k+1) = alpha(k)*a(n-2*k+1) + a(n-4*k+2), where alpha(k) = (2+sqrt(5))^(2*k-1) + (2-sqrt(5))^(2*k-1);
(10) 31|Sum_{k=n..n+9} a(k), for all positive n. (End)
O.g.f.: x*exp(Sum_{n >= 1} Lucas(3*n)*x^n/n) = x + 4*x^2 + 17*x^3 + .... - Peter Bala, Oct 11 2019
a(n) = Sum_{k=0..floor(n/2)} 4^(n-2*k-1)*C(n-k-1,n-2*k-1). - Vladimir Kruchinin, Oct 02 2022
|
|
|
EXAMPLE
|
-,-,-,--,---, ...
G.f. = x + 4*x^2 + 17*x^3 + 72*x^4 + 305*x^5 + 1292*x^6 + 5473*x^7 + 23184*x^8 + ...
|
|
|
MAPLE
|
K:=z/(1-4*z-z^2): Kser:=series(K, z=0, 30): seq(coeff(Kser, z, n), n= 0..21); # Zerinvary Lajos, Nov 08 2007
with(combinat): a:=n->fibonacci(n+2, 4)-4*fibonacci(n+1, 4): seq(a(n), n=0..25); # Zerinvary Lajos, Apr 04 2008
|
|
|
MATHEMATICA
|
Join[{0}, Denominator[Convergents[Sqrt[5], 30]]] (* Harvey P. Dale, Dec 10 2011 *)
a[ n_] := ((2 + Sqrt[5])^n - (2 - Sqrt[5])^n) /(2 Sqrt[5]) // Simplify; (* Michael Somos, Feb 23 2014 *)
|
|
|
PROG
|
(MuPAD) numlib::fibonacci(3*n)/2 $ n = 0..30; // Zerinvary Lajos, May 09 2008
(Sage) [lucas_number1(n, 4, -1) for n in range(23)] # Zerinvary Lajos, Apr 23 2009
(Sage) [fibonacci(3*n)/2 for n in range(23)] # Zerinvary Lajos, May 15 2009
(PARI) {a(n) = fibonacci(3*n) / 2}; /* Michael Somos, Aug 11 2009 */
(PARI) {a(n) = imag( (2 + quadgen(20))^n )}; /* Michael Somos, Feb 23 2014 */
(PARI) {a(n) = polchebyshev(n-1, 2, 2*I)/I^(n-1)}; /* Michael Somos, Nov 02 2021 */
(Magma) I:=[0, 1]; [n le 2 select I[n] else 4*Self(n-1) + Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 24 2018
(GAP) a:=[0, 1];; for n in [3..30] do a[n]:=4*a[n-1]+a[n-2]; od; a; # Muniru A Asiru, Mar 31 2018
(Maxima) a(n):=sum(4^(n-1-2*k)*binomial(n-k-1, n-2*k-1), k, 0, floor((n)/2)); /* Vladimir Kruchinin, Oct 02 2022 */
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,easy,cofr,nice
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
