OFFSET
0,2
COMMENTS
Binomial transform gives signed version of A093040.
The positive sequence has g.f. (1 - x)^2/((1 - x + x^2)(1 - 3*x + x^2)) and a(n) = Sum_{k=0..n} binomial(n+k+1, n-k)*(1+(-1)^k)/2. - Paul Barry, Jul 06 2009
Floretion Algebra Multiplication Program, FAMP Code: 2basei[C*F]; C = - .5'j + .5'k - .5j' + .5k' - 'ii' - .5'ij' - .5'ik' - .5'ji' - .5'ki'; F = + .5'i + .5'ii' + .5'ij' + .5'ik'
REFERENCES
C. Dement, Floretion Integer Sequences (work in progress).
LINKS
Muniru A Asiru, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (-4,-5,-4,-1).
FORMULA
a(n) + a(n+1) = (-1)^(n+1)*A109961(n+1).
a(n) + a(n+1) + a(n+2) = (-1)^n*A001906(n+2) = (-1)^n*F(2*n+4).
Lim_{n -> inf} a(n)/a(n-1) = -(1 + A001622). - A.H.M. Smeets, Sep 11 2018
a(n) = -4*a(n-1) - 5*a(n-2) - 4*a(n-3) - a(n-4). - Muniru A Asiru, Sep 11 2018
MAPLE
seq(coeff(series((1+x)^2/((1+x+x^2)*(1+3*x+x^2)), x, n+1), x, n), n = 0 .. 35); # Muniru A Asiru, Sep 11 2018
MATHEMATICA
LinearRecurrence[{-4, -5, -4, -1}, {1, -2, 4, -10}, 40] (* Vincenzo Librandi, Sep 12 2018 *)
CoefficientList[Series[(1 + x)^2/((1 + x + x^2)*(1 + 3 x + x^2)), {x, 0, 50}], x] (* Stefano Spezia, Sep 12 2018 *)
PROG
(PARI) x='x+O('x^99); Vec((1+x)^2/((1+x+x^2)*(1+3*x+x^2))) \\ Altug Alkan, Sep 11 2018
(GAP) a:=[1, -2, 4, -10];; for n in [5..35] do a[n]:=-4*a[n-1]-5*a[n-2]-4*a[n-3]-a[n-4]; od; a; # Muniru A Asiru, Sep 11 2018
(Magma) I:=[1, -2, 4, -10]; [n le 4 select I[n] else -4*Self(n-1)-5*Self(n-2)- 4*Self(n-3)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, Sep 12 2018
CROSSREFS
KEYWORD
sign,easy
AUTHOR
Creighton Dement, Oct 13 2005
STATUS
approved