OFFSET
0,2
COMMENTS
With different sign pattern, see A000748.
Conjecture: Let M be any endomorphism on any vector space, such that M^3 = 1 (identity). Then (1-M)^n = A057681(n) - A057682(n)*M + z(n)*M^2, where z(0) = z(1) = 0 and, apparently, z(n+2) = a(n). - Stanislav Sykora, Jun 10 2012
LINKS
Robert Israel, Table of n, a(n) for n = 0..4170
T. Alden Gassert, Discriminants of simplest 3^n-tic extensions, arXiv preprint arXiv:1409.7829 [math.NT], 2014.
A. F. Horadam, Special properties of the sequence W_n(a,b; p,q), Fib. Quart., 5.5 (1967), 424-434. Case n->n+1, a=0,b=1; p=3, q=-3.
Vladimir Kruchinin, Composition of ordinary generating functions, arXiv:1009.2565 [math.CO], 2010.
Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38 (2000) 408-419. Eqs. (38) and (45),lhs, m=3.
Vladimir Shevelev, Combinatorial identities generated by difference analogs of hyperbolic and trigonometric functions of order n, arXiv:1706.01454 [math.CO], 2017.
Index entries for linear recurrences with constant coefficients, signature (3,-3).
FORMULA
a(n) = S(n, sqrt(3))*(sqrt(3))^n with S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310.
G.f.: 1/(1-3*x+3*x^2).
Binomial transform of A057079. a(n) = Sum_{k=0..n} 2*binomial(n, k)*cos((k-1)Pi/3). - Paul Barry, Aug 19 2003
For n > 5, a(n) = -27*a(n-6) - Gerald McGarvey, Apr 21 2005
a(n) = Sum_{k=0..n} A109466(n,k)*3^k. - Philippe Deléham, Nov 12 2008
a(n) = Sum_{k=1..n} binomial(k,n-k) * 3^k *(-1)^(n-k) for n>0; a(0)=1. - Vladimir Kruchinin, Feb 07 2011
By the conjecture: Start with x(0)=1, y(0)=0, z(0)=0 and set x(n+1) = x(n) - z(n), y(n+1) = y(n) - x(n), z(n+1) = z(n) - y(n). Then a(n) = z(n+2). This recurrence indeed ends up in a repetitive cycle of length 6 and multiplicative factor -27, confirming G. McGarvey's observation. - Stanislav Sykora, Jun 10 2012
G.f.: Q(0) where Q(k) = 1 + k*(3*x+1) + 9*x - 3*x*(k+1)*(k+4)/Q(k+1) ; (continued fraction). - Sergei N. Gladkovskii, Mar 15 2013
G.f.: G(0)/(2-3*x), where G(k)= 1 + 1/(1 - x*(k+3)/(x*(k+4) + 2/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 16 2013
a(n) = Sum_{k = 0..floor(n/3)} (-1)^k*binomial(n+2,3*k+2). Sykora's conjecture in the Comments section follows easily from this. - Peter Bala, Nov 21 2016
From Vladimir Shevelev, Jul 30 2017: (Start)
a(n) = 2*3^(n/2)*cos(Pi*(n-2)/6);
a(n) = K_2(n+2) - K_1(n+2);
MAPLE
seq(3^(n/2)*orthopoly[U](n, sqrt(3)/2), n=0..100); # Robert Israel, Nov 21 2016
MATHEMATICA
Join[{a=1, b=3}, Table[c=3*b-3*a; a=b; b=c, {n, 100}]] (* Vladimir Joseph Stephan Orlovsky, Jan 17 2011 *)
CoefficientList[Series[1/(1 - 3 x + 3 x^2), {x, 0, 35}], x] (* Michael De Vlieger, Jul 30 2017 *)
PROG
(Sage) [lucas_number1(n, 3, 3) for n in range(1, 37)] # Zerinvary Lajos, Apr 23 2009
(PARI) a(n)=([0, 1; -3, 3]^n*[1; 3])[1, 1] \\ Charles R Greathouse IV, Apr 08 2016
(Magma) I:=[1, 3]; [n le 2 select I[n] else 3*Self(n-1) - 3*Self(n-2): n in [1..30]]; // G. C. Greubel, Oct 23 2018
CROSSREFS
KEYWORD
easy,sign
AUTHOR
Wolfdieter Lang, Aug 11 2000
STATUS
approved