OFFSET
0,6
COMMENTS
Apply the Chebyshev transform (1/(1+x^2), x/(1+x^2)) followed by the binomial involution (1/(1-x),-x/(1-x)) (expressed as Riordan arrays) to -A001045(n). All elements are multiples of a power of 3. - Ralf Stephan, Jan 28 2005
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3).
FORMULA
G.f.: x(1-x)^2/(1-3x+3x^2); a(n)=-sum{j=0..n, (-1)^j*C(n, j)*sum{k=0..floor(j/2), (-1)^k*C(n-k, k)A001045(j-2k)}}.
Recurrence: a(n+2) = 3a(n-1) - 3a(n), starting with 0, 1, 1, 1. - Ralf Stephan, Jan 28 2005
MATHEMATICA
Join[{0, 1}, LinearRecurrence[{3, -3}, {1, 1}, 50]] (* Harvey P. Dale, Apr 12 2014 *)
CoefficientList[Series[x (1 - x)^2/(1 - 3 x + 3 x^2), {x, 0, 50}], x] (* Vincenzo Librandi, Apr 13 2014 *)
PROG
(PARI) {a(n) = if( n<0, 0, polcoeff( x * (1 - x)^2 / (1 - 3*x + 3*x^2) + x * O(x^n), n))} /* Michael Somos, Sep 29 2007 */
(PARI) {a(n) = if(n<2, n>0, 3^(n\2-1) * (-1)^((n+1)\6) * (1 + (-1)^((n-1)\3) * (n%3==1)))} /* Michael Somos, Sep 29 2007 */
CROSSREFS
KEYWORD
easy,sign
AUTHOR
Paul Barry, Jan 30 2005
STATUS
approved