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A350548 Irregular triangle T(n,k) read by rows in which row n lists the iterates of the A350515 map from n to 0.

%I A350548 #10 Feb 20 2022 23:08:08
%S A350548 0,1,0,2,1,0,3,5,8,4,1,0,4,1,0,5,8,4,1,0,6,3,5,8,4,1,0,7,2,1,0,8,4,1,
%T A350548 0,9,14,7,2,1,0,10,3,5,8,4,1,0,11,17,26,13,4,1,0,12,6,3,5,8,4,1,0,13,
%U A350548 4,1,0,14,7,2,1,0,15,23,35,53,80,40,13,4,1,0
%N A350548 Irregular triangle T(n,k) read by rows in which row n lists the iterates of the A350515 map from n to 0.
%H A350548 Emre Yolcu, Scott Aaronson and Marijn J. H. Heule, <a href="https://arxiv.org/abs/2105.14697">An Automated Approach to the Collatz Conjecture</a>, arXiv:2105.14697 [cs.LO], 2021, pp. 21-25.
%H A350548 <a href="/index/3#3x1">Index entries for sequences related to 3x+1 (or Collatz) problem</a>
%F A350548 T(n,0) = n; T(n,k) = A350515(T(n,k-1)), where n >= 0 and k >= 1.
%F A350548 T(n,k) = (A350279(n+1,k+1)-1)/2, where n >= 0 and k >= 0.
%e A350548 Written as an irregular triangle, the sequence begins:
%e A350548   n\k   0   1   2   3   4   5   6
%e A350548   -------------------------------
%e A350548    0:   0
%e A350548    1:   1   0
%e A350548    2:   2   1   0
%e A350548    3:   3   5   8   4   1   0
%e A350548    4:   4   1   0
%e A350548    5:   5   8   4   1   0
%e A350548    6:   6   3   5   8   4   1   0
%e A350548    7:   7   2   1   0
%e A350548    8:   8   4   1   0
%e A350548    9:   9  14   7   2   1   0
%e A350548   10:  10   3   5   8   4   1   0
%e A350548   11:  11  17  26  13   4   1   0
%e A350548   ...
%t A350548 A350515[n_]:=If[Mod[n,3]==1,(n-1)/3,If[Mod[n,6]==0||Mod[n,6]==2,n/2,(3n+1)/2]];
%t A350548 nrows=20;Table[NestWhileList[A350515,n,#>0&],{n,0, nrows-1}]
%Y A350548 Cf. A070165, A349407, A350279, A350515.
%K A350548 nonn,easy,tabf
%O A350548 0,4
%A A350548 _Paolo Xausa_, Jan 04 2022