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A299235 Number of 2's in the n-th {2,3}-power tower; see Comments.

%I A299235 #9 Aug 07 2024 11:56:38
%S A299235 1,0,2,1,1,3,0,2,2,2,1,1,4,3,1,0,3,2,3,2,3,2,2,1,2,1,5,4,4,3,2,1,1,0,
%T A299235 4,3,3,2,4,3,3,2,4,3,3,2,3,2,2,1,3,2,2,1,6,5,5,4,5,4,4,3,3,2,2,1,2,1,
%U A299235 1,0,5,4,4,3,4,3,3,2,5,4,4,3,4,3,3,2
%N A299235 Number of 2's in the n-th {2,3}-power tower; see Comments.
%C A299235 Suppose that S is a set of real numbers. An S-power-tower, t, is a number t = x(1)^x(2)^...^x(k), where k >= 1 and x(i) is in S for i = 1..k. We represent t by (x(1), x(2), ..., x(k)), which for k > 1 is defined as (x(1), (x(2), ..., x(k))); (2,3,2) means 2^9. The number k is the *height* of t. If every element of S exceeds 1 and all the power towers are ranked in increasing order, the position of each in the resulting sequence is its *rank*. See A299229 for a guide to related sequences.
%C A299235 Every nonnegative integer occurs infinitely many times in the sequence. In particular, a(n) = 0 when the tower consists exclusively of 3's. The position of the n-th 0 in the sequence is the rank of the n-th {3}-power tower, given by 9*2^(n-2)-2 for n > 1.
%H A299235 Clark Kimberling, <a href="/A299235/b299235.txt">Table of n, a(n) for n = 1..10000</a>
%e A299235 t(80) = (3,2,2,2,2,3), so that a(80) = 4.
%t A299235 t[1] = {2}; t[2] = {3}; t[3] = {2, 2}; t[4] = {2, 3}; t[5] = {3, 2};
%t A299235 t[6] = {2, 2, 2}; t[7] = {3, 3}; t[8] = {3, 2, 2}; t[9] = {2, 2, 3};
%t A299235 t[10] = {2, 3, 2}; t[11] = {3, 2, 3}; t[12] = {3, 3, 2};
%t A299235 z = 190; g[k_] := If[EvenQ[k], {2}, {3}]; f = 6;
%t A299235 While[f < 13, n = f; While[n < z, p = 1;
%t A299235   While[p < 12, m = 2 n + 1; v = t[n]; k = 0;
%t A299235     While[k < 2^p, t[m + k] = Join[g[k], t[n + Floor[k/2]]]; k = k + 1];
%t A299235    p = p + 1; n = m]]; f = f + 1]
%t A299235 Table[Count[t[n], 2], {n, 1, 100}];  (* A299235 *)
%t A299235 Table[Count[t[n], 3], {n, 1, 100}];  (* A299236 *)
%Y A299235 Cf. A299229, A299236 (complement).
%K A299235 nonn,easy
%O A299235 1,3
%A A299235 _Clark Kimberling_, Feb 06 2018