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 A224367 Triangle read by rows giving trajectory of -k/(2n+1) in Collatz problem, k = 1..2n. 0

%I #13 Nov 02 2021 13:30:30

%S 0,1,2,4,5,7,6,9,10,11,11,13,12,4,5,1,6,3,2,4,7,8,9,8,10,11,9,13,11,

%T 13,12,14,15,5,16,16,6,18,17,20,17,19,7,4,5,4,6,1,5,6,7,7,2,9,6,8,7,

%U 17,18,9,19,9,10,20,20,11,10,22,11,24,21,23,21,36,37

%N Triangle read by rows giving trajectory of -k/(2n+1) in Collatz problem, k = 1..2n.

%C Extension of A210483 with negative values, and subset of A224360.

%e The 2nd row [4, 5, 7, 6] gives the number of iterations of -k/5 (the first element is not counted):

%e k=1 => -1/5 ->2/5 -> 1/5 -> 8/5 -> 4/5 with 4 iterations;

%e k=2 => -2/5 -> -1/5 -> 2/5 -> 1/5 -> 8/5 -> 4/5 with 5 iterations;

%e k=3 => -3/5 -> -4/5 -> -2/5 -> -1/5 -> 2/5 -> 1/5 -> 8/5 -> 4/5 with 7 iterations;

%e k=4 => -4/5 -> -2/5 -> -1/5 -> 2/5 -> 1/5 -> 8/5 -> 4/5 with 6 iterations.

%e The array starts:

%e [0];

%e [1, 2];

%e [4, 5, 7, 6];

%e [9, 10, 11, 11, 13, 12];

%e [4, 5, 1, 6, 3, 2, 4, 7];

%e ...

%t Collatz[n_] := NestWhileList[If[EvenQ[Numerator[-#]], #/2, 3 # + 1] &, n, UnsameQ, All]; t = Join[{{0}}, Table[s = Collatz[-k/(2*n + 1)]; len = Length[s] - 2; If[s[[-1]] == 2, len = len - 1]; len, {n, 10}, {k, 2*n}]]; Flatten[t] (* program from _T. D. Noe_, adapted for this sequence - see A210483 *).

%Y Cf. A210483, A210468, A224299, A224360.

%K nonn,tabf

%O 0,3

%A _Michel Lagneau_, Apr 05 2013

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Last modified September 19 12:32 EDT 2024. Contains 376012 sequences. (Running on oeis4.)