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%I #14 Apr 04 2024 12:52:29
%S 1,0,1,1,1,0,0,0,1,0,1,0,0,0,0,0,1,1,0,0,1,1,1,1,1,1,0,1,0,0,0,1,0,1,
%T 1,0,1,0,1,0,0,0,0,1,0,0,1,0,1,0,0,0,0,0,0,1,1,0,1,0,1,0,1,0,0,1,0,0,
%U 1,1,1,1,1,0,1,1,0,1,1,0,0,1,1,0,1,0,1,0,1,1,0
%N Square array read by ascending antidiagonals: T(n,k) is the parity of the k-th iterate of the 5x+1 function started at n, with n >= 1 and k >= 0.
%C The 5x+1 function (A185452), denoted by T_5(x) in the literature, is defined as T_5(x) = (5x+1)/2 if x is odd, T_5(x) = x/2 if x is even.
%C As reported by Kontorovich and Lagarias (2009 and 2010), and analogously to A365495, the sequence of the first m terms in each row is periodic in n with period 2^m, with each of the 2^m possible binary vectors occurring exactly once (as the first m terms of a row) per period.
%C For example, for m = 3, the first 3 terms in rows 1..2^3 are respectively [1,1,0], [0,1,1], [1,0,0], [0,0,1], [1,1,1], [0,1,0], [1,0,1] and [0,0,0], and this pattern repeats from row 2^3 + 1 onwards.
%C As a consequence, Kontorovich and Lagarias note, each integer is uniquely determined by the sequence of the parity of its orbit, i.e., n is uniquely determined by the n-th row of the present array.
%H Paolo Xausa, <a href="/A365992/b365992.txt">Table of n, a(n) for n = 1..11325</a> (antidiagonals 1..150 of the array, flattened)
%H Alex V. Kontorovich and Jeffrey C. Lagarias, <a href="https://arxiv.org/abs/0910.1944">Stochastic Models for the 3x+1 and 5x+1 Problems</a>, arXiv:0910.1944 [math.NT], 2009, pp. 39-40, and in Jeffrey C. Lagarias, ed., <a href="http://www.ams.org/bookstore-getitem/item=mbk-78">The Ultimate Challenge: The 3x+1 Problem</a>, American Mathematical Society, 2010, pp. 164-165.
%H <a href="/index/3#3x1">Index entries for sequences related to 3x+1 (or Collatz) problem</a>
%F T(n,k) = A365991(n,k) mod 2.
%e The array begins:
%e n\k| 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 ...
%e -----------------------------------------------------
%e 1 | 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, ...
%e 2 | 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, ...
%e 3 | 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, ...
%e 4 | 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, ...
%e 5 | 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, ...
%e 6 | 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, ...
%e 7 | 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, ...
%e 8 | 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, ...
%e 9 | 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, ...
%e 10 | 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, ...
%e 11 | 1, 0, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, ...
%e 12 | 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, ...
%e 13 | 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, ...
%e 14 | 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, ...
%e 15 | 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, ...
%e ...
%t A365992list[dmax_]:=With[{a=Mod[Array[NestList[If[OddQ[#],(5#+1)/2,#/2]&,dmax-#,#]&,dmax,0],2]},Array[Diagonal[a,#]&,dmax,1-dmax]];A365992list[20] (* Generates 20 antidiagonals *)
%Y Cf. A185452, A347283, A365495, A365991, A371691 (main diagonal).
%K nonn,easy,tabl
%O 1
%A _Paolo Xausa_, Sep 25 2023
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