OFFSET
1,1
COMMENTS
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.
LINKS
Clark Kimberling, Table of n, a(n) for n = 1..1000
EXAMPLE
The first six terms are the ranks of these towers: t(2) = (3), t(7) = (3,3), t(11) = (3,2,3), t(12) = (3,3,2), t(15) = (2,3,3), t(16) = (3,3,3).
MATHEMATICA
t[1] = {2}; t[2] = {3}; t[3] = {2, 2}; t[4] = {2, 3}; t[5] = {3, 2};
t[6] = {2, 2, 2}; t[7] = {3, 3}; t[8] = {3, 2, 2}; t[9] = {2, 2, 3};
t[10] = {2, 3, 2}; t[11] = {3, 2, 3}; t[12] = {3, 3, 2};
z = 190; g[k_] := If[EvenQ[k], {2}, {3}]; f = 6;
While[f < 13, n = f; While[n < z, p = 1;
While[p < 12, m = 2 n + 1; v = t[n]; k = 0;
While[k < 2^p, t[m + k] = Join[g[k], t[n + Floor[k/2]]]; k = k + 1];
p = p + 1; n = m]]; f = f + 1]
Select[Range[1000], Count[t[#], 2] > Count[t[#], 3] &]; (* A299240 *)
Select[Range[1000], Count[t[#], 2] == Count[t[#], 3] &]; (* A299241 *)
Select[Range[1000], Count[t[#], 2] < Count[t[#], 3] &]; (* this sequence *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Feb 07 2018
STATUS
approved