login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

A057083
Scaled Chebyshev U-polynomials evaluated at sqrt(3)/2; expansion of 1/(1 - 3*x + 3*x^2).
43
1, 3, 6, 9, 9, 0, -27, -81, -162, -243, -243, 0, 729, 2187, 4374, 6561, 6561, 0, -19683, -59049, -118098, -177147, -177147, 0, 531441, 1594323, 3188646, 4782969, 4782969, 0, -14348907, -43046721, -86093442, -129140163, -129140163, 0
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
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.
FORMULA
a(n) = S(n, sqrt(3))*(sqrt(3))^n with S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310.
a(2*n) = A057078(n)*3^n; a(2*n+1)= A010892(n)*3^(n+1).
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);
For m,n>=1, a(n+m) = a(n-1)*K_1(m+1) + K_2(n+1)*K_2(m+1) + K_1(n+1)*a(m-1) where K_1 = A057681, K_2 = A057682. (End)
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
KEYWORD
easy,sign
AUTHOR
Wolfdieter Lang, Aug 11 2000
STATUS
approved