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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.
3

%I #14 Aug 07 2024 13:12:29

%S 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 27,39,43,63,18,24,26,34,35,47,51,67,28,40,44,64,29,41,45,65,36,48,52,

%U 68,37,49,53,69,55,79,87,127,71,95,103,135,30,42,46,66

%N 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 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 This sequence is a self-inverse permutation of the positive integers.

%H Clark Kimberling, <a href="/A299237/b299237.txt">Table of n, a(n) for n = 1..1000</a>

%e t(12) = (3,3,2) and t(15) = (2,3,3) = reversal of t(12); therefore a(12) = 15.

%t t[1] = {2}; t[2] = {3}; t[3] = {2, 2}; t[4] = {2, 3}; t[5] = {3, 2};

%t t[6] = {2, 2, 2}; t[7] = {3, 3}; t[8] = {3, 2, 2}; t[9] = {2, 2, 3};

%t t[10] = {2, 3, 2}; t[11] = {3, 2, 3}; t[12] = {3, 3, 2};

%t z = 190; g[k_] := If[EvenQ[k], {2}, {3}]; f = 6;

%t While[f < 13, n = f; While[n < z, p = 1;

%t While[p < 12, m = 2 n + 1; v = t[n]; k = 0;

%t While[k < 2^p, t[m + k] = Join[g[k], t[n + Floor[k/2]]]; k = k + 1];

%t p = p + 1; n = m]]; f = f + 1]

%t r[n_] := Reverse[t[n]]

%t Flatten[Table[Select[Range[2000], t[#] == r[n] &], {n, 1, 1500}]]; (* A299237 *)

%Y Cf. A299229, A299239 (fixed points of the permutation; palindromes).

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Feb 07 2018