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A299237 a(n) = the index m satisfying t(m) = reversal of t(n), where t(n) is the n-th {2,3}-power tower; see Comments.

%I A299237 #14 Aug 07 2024 13:12:29
%S A299237 1,2,3,5,4,6,7,9,8,10,11,15,13,19,12,16,21,31,14,20,17,23,22,32,25,33,
%T A299237 27,39,43,63,18,24,26,34,35,47,51,67,28,40,44,64,29,41,45,65,36,48,52,
%U A299237 68,37,49,53,69,55,79,87,127,71,95,103,135,30,42,46,66
%N A299237 a(n) = the index m satisfying t(m) = reversal of t(n), where t(n) is the n-th {2,3}-power tower; see Comments.
%C A299237 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 A299237 This sequence is a self-inverse permutation of the positive integers.
%H A299237 Clark Kimberling, <a href="/A299237/b299237.txt">Table of n, a(n) for n = 1..1000</a>
%e A299237 t(12) = (3,3,2) and t(15) = (2,3,3) = reversal of t(12); therefore a(12) = 15.
%t A299237 t[1] = {2}; t[2] = {3}; t[3] = {2, 2}; t[4] = {2, 3}; t[5] = {3, 2};
%t A299237 t[6] = {2, 2, 2}; t[7] = {3, 3}; t[8] = {3, 2, 2}; t[9] = {2, 2, 3};
%t A299237 t[10] = {2, 3, 2}; t[11] = {3, 2, 3}; t[12] = {3, 3, 2};
%t A299237 z = 190; g[k_] := If[EvenQ[k], {2}, {3}]; f = 6;
%t A299237 While[f < 13, n = f; While[n < z, p = 1;
%t A299237   While[p < 12, m = 2 n + 1; v = t[n]; k = 0;
%t A299237     While[k < 2^p, t[m + k] = Join[g[k], t[n + Floor[k/2]]]; k = k + 1];
%t A299237    p = p + 1; n = m]]; f = f + 1]
%t A299237 r[n_] := Reverse[t[n]]
%t A299237 Flatten[Table[Select[Range[2000], t[#] == r[n] &], {n, 1, 1500}]]; (* A299237 *)
%Y A299237 Cf. A299229, A299239 (fixed points of the permutation; palindromes).
%K A299237 nonn,easy
%O A299237 1,2
%A A299237 _Clark Kimberling_, Feb 07 2018