A274142 Number of integers in n-th generation of tree T(1/2) defined in Comments.

1, 1, 1, 2, 2, 4, 5, 8, 11, 17, 25, 37, 54, 81, 119, 177, 261, 388, 574, 851, 1260, 1868, 2767, 4101, 6077, 9006, 13347, 19781, 29315, 43448, 64392, 95436, 141444, 209636, 310705, 460501, 682519, 1011581, 1499295, 2222155, 3293534, 4881472, 7235018, 10723311, 15893460, 23556367, 34913897, 51747400

Offset

0,4

Comments

Let T* be the infinite tree with root 0 generated by these rules: if p is in T*, then p+1 is in T* and x*p is in T*. Let g(n) be the set of nodes in the n-th generation, so that g(0) = {0}, g(1) = {1}, g(2) = {2,x}, g(3) = {3,2x,x+1,x^2}, etc. Let T(r) be the tree obtained by substituting r for x.

Guide to related sequences:

r sequence

1/2 A274142

1/3 A274143

1/4 A274144

2/3 A274145

3/4 A274146

-1/2 A274147

-1/3 A274148

-1/4 A274149

-2/3 A274150

-3/4 A274151

3/2 A274152

5/2 A274153

-3/2 A274154

-5/2 A274155

2^(1/2) A000045 (Fibonacci numbers)

2^(1/3) A000930

2^(1/4) A003269

2^(-1/2) A274156

3^(-1/2) A274157

2^(-1/3) A274158

3^(-1/3) A274159

i A274160

2i A206743

3i A274162

4i A274163

i/2 A274149

i/3 A274165

i+1 A274166

i-1 A274167

(-1+3i)/2 A274168

Links

Kenny Lau, Table of n, a(n) for n = 0..5847

Example

If r = 1/2, then g(3) = {3,2r,r+1, r^2}, in which the integers are 3 and 1, so that a(3) = 2.

Mathematica

z = 18; t = Join[{{0}}, Expand[NestList[DeleteDuplicates[Flatten[Map[{# + 1, x*#} &, #], 1]] &, {1}, z]]];
u = Table[t[[k]] /. x -> 1/2, {k, 1, z}];
Table[Count[Map[IntegerQ, u[[k]]], True], {k, 1, z}]
(* second program: *)
T[0] = {0}; T[n_] := T[n] = Complement[Join[T[n-1]+1, x*T[n-1]], T[n-1]]; Reap[For[n = 0, n <= 25, n++, cnt = Count[T[n] /. x -> 1/2, _Integer]; Print[n, " ", cnt]; Sow[cnt]]][[2, 1]] (* Jean-François Alcover, Jun 14 2016 *)

Crossrefs

Cf. A274143-A274160, A274162, A274163, A274165-A274168.

Sequence in context: A238478 A013979 A107458 * A006206 A060280 A095719

Adjacent sequences: A274139 A274140 A274141 * A274143 A274144 A274145

Keyword

nonn

Author

Clark Kimberling, Jun 11 2016

Extensions

More terms from Jean-François Alcover, Jun 14 2016

More terms from Kenny Lau, Jul 04 2016

Status

approved