A103312 A transform of the Jacobsthal numbers.

0, 1, 1, 1, 0, -3, -9, -18, -27, -27, 0, 81, 243, 486, 729, 729, 0, -2187, -6561, -13122, -19683, -19683, 0, 59049, 177147, 354294, 531441, 531441, 0, -1594323, -4782969, -9565938, -14348907, -14348907, 0, 43046721, 129140163, 258280326, 387420489, 387420489, 0, -1162261467, -3486784401

Offset: 0

Author

Paul Barry, Jan 30 2005

Keywords

• easy
• sign

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).

Crossrefs

Cf. A057681.

Programs

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 *)

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 */

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