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A279634
Coefficients in the expansion of 1/([r] + [2r]x + [3r]x^2 + ...); [ ] = floor, r = 3/2.
9
1, -3, 5, -9, 18, -36, 72, -144, 288, -576, 1152, -2304, 4608, -9216, 18432, -36864, 73728, -147456, 294912, -589824, 1179648, -2359296, 4718592, -9437184, 18874368, -37748736, 75497472, -150994944, 301989888, -603979776, 1207959552, -2415919104, 4831838208
OFFSET
0,2
COMMENTS
After first 3 terms, agrees with A005010 except for signs; in particular 9 divides a(n) for n >= 3.
Suppose r = c/d is a rational number and (a(n)) is the coefficient series for 1/([r] + [2r]x + [3r]x^2 + ...). Let (s(k)) be the increasing sequence of indices n(k) for which a(n(k)) > = 0. In the table below, "yes" indicates that a check of the first 1000 terms indicates that (n(k)) is (eventually) periodic. Column 1 gives selected values of r, and column 2 gives the corresponding coefficient series.
3/2 A279634 yes
4/3 A279675 no
5/3 A279676 no
5/4 A279677 yes
7/4 A279678 yes
6/5 A279778 no
7/5 A279779 no
8/5 A279780 yes
9/5 A279781 no
FORMULA
G.f.: 1/(1 + 3x + 4x^2 + 6x^3 + ...).
G.f.: (1 - x) (1 - x^2)/(1 + 2x).
E.g.f.: - (1/8) - (3/4)*x + (1/4)*x^2 + (9/8)*exp(-2*x). - Alejandro J. Becerra Jr., Feb 16 2021
MATHEMATICA
z = 50; f[x_] := f[x] = Sum[Floor[(3/2)*(k + 1)] x^k, {k, 0, z}]; f[x]
CoefficientList[Series[1/f[x], {x, 0, z}], x]
LinearRecurrence[{-2}, {1, -3, 5, -9}, 40] (* Harvey P. Dale, Jul 28 2023 *)
CROSSREFS
Cf. A005010.
Sequence in context: A288229 A293332 A288135 * A028411 A018098 A108859
KEYWORD
sign,easy
AUTHOR
Clark Kimberling, Dec 18 2016
STATUS
approved