A299238 a(n) = the index m satisfying t(m) = 5 - t(n), where t(n) is the n-th {2,3}-power tower; see Comments.

2, 1, 7, 5, 4, 16, 3, 15, 12, 11, 10, 9, 34, 33, 8, 6, 32, 31, 26, 25, 24, 23, 22, 21, 20, 19, 70, 69, 68, 67, 18, 17, 14, 13, 66, 65, 64, 63, 54, 53, 52, 51, 50, 49, 48, 47, 46, 45, 44, 43, 42, 41, 40, 39, 142, 141, 140, 139, 138, 137, 136, 135, 38, 37, 36

Offset: 1

Author

Clark Kimberling, Feb 07 2018

Keywords

• nonn
• easy

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.

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

Examples

			t(12) = (3,3,2) and t(9) = (2,2,3) = 5 - (3,3,2), so that a(12) = 9. (Note: 5 - (x(1), x(2), ..., x(k)) means (5-x(1), 5-x(2), ..., 5-x(k))).
		

Links

• Clark Kimberling, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A299229.

Programs

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]
Flatten[Table[Select[Range[1000], t[#] == 5 - t[n] &], {n, 1, 150}]] (* A299238 *)