OFFSET
0,3
COMMENTS
Starting with g(0) = {0}, generate g(n) for n > 0 inductively using these rules:
(1) if x is in g(n-1), then x + 1 is in g(n); and
(2) if x is in g(n-1) and x < 3, then x/3 is in g(n).
The sum of numbers in g(n) is a(n)/3^(n-1).
FORMULA
Conjecture: a(n) = 4*a(n-1) + 9*a(n-2) + 18*a(n-3) - 81*a(n-4) - 162*a(n-5) - 243*a(n-6).
EXAMPLE
g(0) = {0}, sum = 0.
g(1) = {1}, sum = 1.
g(2) = {1/3,2/1}, sum = 7/3.
g(3) = {1/9,2/3,4/3,3/1}, sum = 46/9.
MATHEMATICA
z = 5; x = 1/3; g[0] = {0}; g[1] = {1};
g[n_] := g[n] = Union[1 + g[n - 1], (1/3) Select[g[n - 1], # < 3 &]]
Table[g[n], {n, 0, z}]
Table[Total[g[n]], {n, 0, z}]
u = Numerator[Table[Total[g[n]], {n, 0, z}] ]
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Nov 09 2015
STATUS
approved