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Rectangular array by antidiagonals: row n gives the ranks of {2,3}-power towers that start with n 2's, for n >= 1; see Comments.
2

%I #16 Aug 07 2024 15:12:11

%S 1,4,3,10,9,6,15,21,19,13,17,31,43,39,27,23,35,63,87,79,55,25,47,71,

%T 127,175,159,111,29,51,95,143,255,351,319,223,33,59,103,191,287,511,

%U 703,639,447,37,67,119,207,383,575,1023,1407,1279,895,41,75,135,239

%N Rectangular array by antidiagonals: row n gives the ranks of {2,3}-power towers that start with n 2's, for n >= 1; 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 As sequences, this one and A299326 partition the positive integers.

%e Northwest corner:

%e 1 4 10 15 17 23 25

%e 3 9 21 31 35 47 51

%e 6 19 43 63 71 95 103

%e 13 39 87 127 143 191 207

%e 27 79 175 255 287 383 415

%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 t[8] = {3, 2, 2}; t[9] = {2, 2, 3}; t[10] = {2, 3, 2};

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

%t z = 500; g[k_] := If[EvenQ[k], {2}, {3}];

%t f = 6; While[f < 13, n = f; While[n < z, p = 1;

%t While[p < 17, 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 s = Select[Range[60000], Count[First[Split[t[#]]], 3] == 0 & ];

%t r[n_] := Select[s, Length[First[Split[t[#]]]] == n &, 12]

%t TableForm[Table[r[n], {n, 1, 11}]] (* this array *)

%t w[n_, k_] := r[n][[k]];

%t Table[w[n - k + 1, k], {n, 11}, {k, n, 1, -1}] // Flatten (* this sequence *)

%Y Cf. A299229, A299326.

%K nonn,easy,tabl

%O 1,2

%A _Clark Kimberling_, Feb 08 2018