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A299236
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Number of 3's in the n-th {2,3}-power tower; see Comments.
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3
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0, 1, 0, 1, 1, 0, 2, 1, 1, 1, 2, 2, 0, 1, 2, 3, 1, 2, 1, 2, 1, 2, 2, 3, 2, 3, 0, 1, 1, 2, 2, 3, 3, 4, 1, 2, 2, 3, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 0, 1, 1, 2, 1, 2, 2, 3, 2, 3, 3, 4, 3, 4, 4, 5, 1, 2, 2, 3, 2, 3, 3, 4, 1, 2, 2, 3, 2, 3, 3, 4
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OFFSET
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1,7
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COMMENTS
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Suppose that S is a set of real numbers. As 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-1)); (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.
Every nonnegative integer occurs infinitely many times in the sequence. In particular, a(n) = 0 when the tower consists exclusively of 2's. The position of the n-th 0 in the sequence is the rank of the n-th {2}-power tower, given by 7*2^(n-3) - 1 for n > 2.
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LINKS
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EXAMPLE
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t(13) = (2,2,2,2), so that a(13) = 0.
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MATHEMATICA
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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]
Table[Count[t[n], 2], {n, 1, 100}]; (* A299235 *)
Table[Count[t[n], 3], {n, 1, 100}]; (* A299236 *)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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