

A299235


Number of 2's in the nth {2,3}power tower; see Comments.


3



1, 0, 2, 1, 1, 3, 0, 2, 2, 2, 1, 1, 4, 3, 1, 0, 3, 2, 3, 2, 3, 2, 2, 1, 2, 1, 5, 4, 4, 3, 2, 1, 1, 0, 4, 3, 3, 2, 4, 3, 3, 2, 4, 3, 3, 2, 3, 2, 2, 1, 3, 2, 2, 1, 6, 5, 5, 4, 5, 4, 4, 3, 3, 2, 2, 1, 2, 1, 1, 0, 5, 4, 4, 3, 4, 3, 3, 2, 5, 4, 4, 3, 4, 3, 3, 2
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OFFSET

1,3


COMMENTS

Suppose that S is a set of real numbers. As Spowertower, 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(k1)); (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 3's. The position of the nth 0 in the sequence is the rank of the nth {3}power tower, given by 9*2^(n2)2 for n > 1.


LINKS

Clark Kimberling, Table of n, a(n) for n = 1..10000


EXAMPLE

t(80) = (3,2,2,2,2,3), so that a(80) = 4.


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]
Table[Count[t[n], 2], {n, 1, 100}]; (* A299235 *)
Table[Count[t[n], 3], {n, 1, 100}]; (* A299236 *)


CROSSREFS

Cf. A299229, A299236 (complement).
Sequence in context: A230079 A105400 A194516 * A341259 A245840 A033774
Adjacent sequences: A299232 A299233 A299234 * A299236 A299237 A299238


KEYWORD

nonn,easy


AUTHOR

Clark Kimberling, Feb 06 2018


STATUS

approved



