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A274143 Number of integers in n-th generation of tree T(1/3) defined in Comments.

%I A274143 #16 Jul 04 2016 05:58:25
%S A274143 1,1,1,1,2,2,2,4,4,5,8,9,12,16,20,26,34,44,57,74,97,125,162,212,272,
%T A274143 356,462,597,780,1010,1311,1706,2210,2873,3732,4841,6294,8168,10608,
%U A274143 13781,17886,23237,30172,39177,50891,66072,85813,111446,144706,187947,244059,316937,411618,534503,694153,901461
%N A274143 Number of integers in n-th generation of tree T(1/3) defined in Comments.
%C A274143 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.
%C A274143 See A274142 for a guide to related sequences.
%H A274143 Kenny Lau, <a href="/A274143/b274143.txt">Table of n, a(n) for n = 0..8805</a>
%e A274143 For r = 1/3, we have g(3) = {3,2r,r+1, r^2}, in which only 3 is an integer, so that a(3) = 1.
%p A274143 A274143 := proc(r)
%p A274143     local gs,n,gs2,el,a ;
%p A274143     gs := [2,r] ;
%p A274143     for n from 3 do
%p A274143         gs2 := [] ;
%p A274143         for el in gs do
%p A274143             gs2 := [op(gs2),el+1,r*el] ;
%p A274143         end do:
%p A274143         gs := gs2 ;
%p A274143         a := 0 ;
%p A274143         for el in gs do
%p A274143             if type(el,'integer') then
%p A274143                  a := a+1 :
%p A274143             end if;
%p A274143         end do:
%p A274143         print(n,a) ;
%p A274143     end do:
%p A274143 end proc:
%p A274143 A274143(1/3) ; # _R. J. Mathar_, Jun 17 2016
%t A274143 z = 18; t = Join[{{0}}, Expand[NestList[DeleteDuplicates[Flatten[Map[{# + 1, x*#} &, #], 1]] &, {1}, z]]];
%t A274143 u = Table[t[[k]] /. x -> 1/3, {k, 1, z}];
%t A274143 Table[Count[Map[IntegerQ, u[[k]]], True], {k, 1, z}]
%Y A274143 Cf. A274142.
%K A274143 nonn
%O A274143 0,5
%A A274143 _Clark Kimberling_, Jun 11 2016
%E A274143 More terms from _Kenny Lau_, Jul 04 2016