|
| |
|
|
A079935
|
|
a(n) = 4*a(n-1) - a(n-2) with a(1) = 1, a(2) = 3.
|
|
22
|
|
|
|
1, 3, 11, 41, 153, 571, 2131, 7953, 29681, 110771, 413403, 1542841, 5757961, 21489003, 80198051, 299303201, 1117014753, 4168755811, 15558008491, 58063278153, 216695104121, 808717138331, 3018173449203, 11263976658481
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,2
|
|
|
COMMENTS
|
Greedy frac multiples of sqrt(3): a(1)=1, Sum_{n>0} frac(a(n)*x)) < 1 at x=sqrt(3).
The n-th greedy frac multiple of x is the smallest integer that does not cause Sum_{k=1..n} frac(a(k)*x) to exceed unity; an infinite number of terms appear as the denominators of the convergents to the continued fraction of x.
In general, Sum_{k=0..n} binomial(2n-k,k)*j^(n-k) = (-1)^n* U(2n, i*sqrt(j)/2), i=sqrt(-1). - Paul Barry, Mar 13 2005
The Hankel transform of this sequence is [1,2,0,0,0,0,0,0,0,0,0,...]. - Philippe Deléham, Nov 21 2007
This sequence is a particular case of the following situation:
a(0)=1, a(1)=a, a(2)=b with the recurrence relation a(n+3) = (a(n+2)*a(n+1)+q)/a(n)
where q is given in Z to have Q = (a*b^2 + q*b + a + q)/(a*b) itself in Z.
The g.f is f: f(z) = (1 + a*z + (b-Q)*z^2 + (a*b + q - a*Q)*z^3)/(1 - Q*z^2 + z^4);
so we have the linear recurrence: a(n+4) = Q*a(n+2) - a(n).
The general form of a(n) is given by:
a(2*m) = Sum_{p=0..floor(m/2)} (-1)^p*binomial(m-p,p)*Q^(m-2*p) + (b-Q)*Sum_{p=0..floor((m-1)/2)} (-1)^p*binomial(m-1-p,p)*Q^(m-1-2*p) and
a(2*m+1) = a*Sum_{p=0..floor(m/2)} (-1)^p*binomial(m-p,p)*Q^(m-2*p) + (a*b+q-a*Q)*Sum_{p=0..floor((m-1)/2)} (-1)^p*binomial(m-1-p,p)*Q^(m-1-2*p).
(End)
x-values in the solution to 3*x^2 - 2 = y^2. - Sture Sjöstedt, Nov 25 2011
[X(n) = S(n, 4) - S(n-1, 4), Y(n) = X(n-1)] gives all positive solutions of X^2 + Y^2 - 4*X*Y = -2, for n = -oo..+oo, where the Chebyshev S-polynomials are given in A049310, with S(-1, 0) = 0, and S(-|n|, x) = - S(|n|-2, x), for |n| >= 2.
This binary indefinite quadratic form has discriminant D = +12. There is only this family representing -2 properly with X and Y positive, and there are no improper solutions.
See the formula for a(n) = X(n-1), for n >= 1, in terms of S-polynomials below.
This comment is inspired by a paper by Robert K. Moniot (private communication). See his Oct 04 2020 comment in A027941 related to the case of x^2 + y^2 - 3*x*y = -1 (special Markov solutions). (End)
a(n) is also the output of Tesler's formula for the number of perfect matchings of an m x n Mobius band where m and n are both even, specialized to m=2. (The twist is on the length-n side.) - Sarah-Marie Belcastro, Feb 15 2022
|
|
|
LINKS
|
|
|
|
FORMULA
|
For n > 0, a(n) = ceiling( (2+sqrt(3))^n/(3+sqrt(3)) ).
G.f.: (1-x)/(1-4*x+x^2).
E.g.f.: exp(2*x)*(sinh(sqrt(3)*x)/sqrt(3) + cosh(sqrt(3)*x)).
a(n) = ( (3+sqrt(3))*(2+sqrt(3))^n + (3-sqrt(3))*(2-sqrt(3))^n )/6 (offset 0). (End)
a(n) = Sum_{k=0..n} binomial(2*n-k, k)*2^(n-k). - Paul Barry, Jan 22 2005 [offset 0]
a(n) = (-1)^n*U(2*n, i*sqrt(2)/2), U(n, x) Chebyshev polynomial of second kind, i=sqrt(-1). - Paul Barry, Mar 13 2005 [offset 0]
a(n) = Jacobi_P(n,-1/2,1/2,2)/Jacobi_P(n,-1/2,1/2,1). - Paul Barry, Feb 03 2006 [offset 0]
a(n) = sqrt(2+(2-sqrt(3))^(2*n-1) + (2+sqrt(3))^(2*n-1))/sqrt(6). - Gerry Martens, Jun 05 2015
a(n) = (1/2 + sqrt(3)/6)*(2-sqrt(3))^n + (1/2 - sqrt(3)/6)*(2+sqrt(3))^n. - Robert Israel, Jun 05 2015
a(n) = S(n-1,4) - S(n-2,4) = (-1)^(n-1)*S(2*(n-1), i*sqrt(2)), with Chebyshev S-polynomials (A049310), the imaginary unit i, S(-1, x) = 0, for n >= 1. See also the formula above by Paul Barry (with offset 0). - Wolfdieter Lang, Oct 12 2020
a(n) = sqrt(2/3)*cosh((-1 - 2*n) arccsch(sqrt(2))), where arccsch is the inverse hyperbolic cosecant function (with offset 0). - Peter Luschny, Oct 13 2020
|
|
|
EXAMPLE
|
a(4) = 41 since frac(1*x) + frac(3*x) + frac(11*x) + frac(41*x) < 1, while frac(1*x) + frac(3*x) + frac(11*x) + frac(k*x) > 1 for all k > 11 and k < 41.
|
|
|
MAPLE
|
f:= gfun:-rectoproc({a(n) = 4*a(n-1) - a(n-2), a(1)=1, a(2)=3}, a(n), remember):
|
|
|
MATHEMATICA
|
a[n_] := (MatrixPower[{{1, 2}, {1, 3}}, n].{{1}, {1}})[[1, 1]]; Table[ a[n], {n, 0, 23}]] (* Robert G. Wilson v, Jan 13 2005 *)
LinearRecurrence[{4, -1}, {1, 3}, 30] (* or *) CoefficientList[Series[ (1-x)/(1-4x+x^2), {x, 0, 30}], x] (* Harvey P. Dale, Apr 26 2011 *)
a[n_] := Sqrt[2/3] Cosh[(-1 - 2 n) ArcCsch[Sqrt[2]]];
Table[Simplify[a[n-1]], {n, 1, 12}] (* Peter Luschny, Oct 13 2020 *)
|
|
|
PROG
|
(Sage) [lucas_number1(n, 4, 1)-lucas_number1(n-1, 4, 1) for n in range(1, 25)] # Zerinvary Lajos, Apr 29 2009
(Haskell)
a079935 n = a079935_list !! (n-1)
a079935_list =
1 : 3 : zipWith (-) (map (4 *) $ tail a079935_list) a079935_list
(Magma) I:=[1, 3]; [n le 2 select I[n] else 4*Self(n-1)-Self(n-2): n in [1..40]]; // Vincenzo Librandi, Jun 06 2015
(PARI) my(x='x+O('x^30)); Vec((1-x)/(1-4*x+x^2)) \\ G. C. Greubel, Feb 25 2019
|
|
|
CROSSREFS
|
Cf. A001835 (same except for the first term).
Cf. similar sequences listed in A238379.
|
|
|
KEYWORD
|
nonn,easy
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
