A224360 Triangle read by rows: T(n,k) = -1 + length of the Collatz sequence of -(n-k)/(2k+1) for n >= 1 and k >= 0.
0, 1, 1, 4, 2, 4, 2, 0, 5, 9, 4, 3, 7, 10, 4, 5, 3, 6, 11, 5, 8, 4, 1, 0, 11, 1, 9, 14, 3, 6, 8, 13, 6, 8, 15, 4, 11, 4, 9, 12, 3, 10, 5, 5, 17, 4, 4, 7, 0, 2, 11, 16, 4, 18, 36, 6, 4, 14, 12, 4, 9, 16, 6, 9, 37, 13, 6, 5, 1, 16, 7, 13, 6, 1, 19, 16, 14, 7, 9
Offset: 1
Examples
The triangle of lengths begins 1; 2, 2; 5, 3, 5; 3, 1, 6, 10; 5, 4, 8, 11, 5; ... Individual numbers have the following Collatz sequences (the first term is not counted): [-1] => [1] because: -1 -> -1 with 0 iterations; [-2 -1/3] => [1, 1] because: -2 -> -1 => 1 iteration; -1/3 -> 0 => 1 iteration; [-3 -2/3 -1/5] => [4, 2, 4] because: -3 -> -8 -> -4 -> -2 -> -1 => 4 iterations; -2/3 -> -1/3 -> 0 => 2 iterations; -1/5 -> 2/5 -> 1/5 -> 8/5 -> 4/5 => 4 iterations.
Programs
-
Mathematica
Collatz2[n_] := Module[{lst = NestWhileList[If[EvenQ[Numerator[#]], #/2, 3 # + 1] &, n, Unequal, All]}, If[lst[[-1]] == -1, lst = Drop[lst, -2], If[lst[[-1]] == 2, lst = Drop[lst, -2], If[lst[[-1]] == 4, lst = Drop[lst, -1], If[MemberQ[Rest[lst], lst[[-1]]], lst = Drop[lst, -1]]]]]]; t = Table[s = Collatz2[-(n - k)/(2*k + 1)]; Length[s] - 1, {n, 13}, {k, 0, n - 1}]; Flatten[t] (* program from T. D. Noe, adapted for this sequence - see A210516 *)
Extensions
Better definition from Michel Marcus, Sep 14 2017
Comments