OFFSET
0,2
COMMENTS
The main diagonal of the difference table is -A000079(n) = -2^n.
a(n) mod 9 is of period 6: repeat 8, 7, 5, 1, 2, 4.
a(n) + a(n+1) = -3, -6, -3, 3, 15, ...; all are multiples of 3.
LINKS
Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,1,-2).
FORMULA
a(2n+1) = A141725(n-1), a(2n+2) = 2*a(2n+1).
a(n+1) = 2*a(n) + (period 2: repeat 0, 9), n>0.
a(n) = -A157823(n) - (period 2: repeat 6, 3).
a(n+1) = a(n) - A156067(n).
a(n+2) = a(n) + 3*2^(n-1), n>0.
a(n+4) = a(n) + 15*2^(n-1), n>0.
a(n+6) = a(n) + 63*2^(n-1), n>0.
a(n) = (2^n - 3*(-1)^n - 9)/2 for n>0. - Colin Barker, Jan 30 2015
G.f.: (9*x^3+x^2-1) / ((x-1)*(x+1)*(2*x-1)). - Colin Barker, Jan 30 2015
MATHEMATICA
a[0] = -1; a[n_] := 2^(n-1) + 3*Mod[n, 2] - 6; Table[a[n], {n, 0, 33}] (* Jean-François Alcover, Feb 04 2015 *)
PROG
(PARI) Vec((9*x^3+x^2-1)/((x-1)*(x+1)*(2*x-1)) + O(x^100)) \\ Colin Barker, Jan 30 2015
CROSSREFS
KEYWORD
sign,easy
AUTHOR
Paul Curtz, Jan 29 2015
STATUS
approved