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A186008
Irregular triangle T(n,k) read by rows, in which row n has the pattern of conjectured dropping times in the Collatz iteration.
2
2, 4, 16, 12, 32, 8, 52, 128, 40, 56, 84, 136, 160, 180, 256, 60, 80, 136, 220, 288, 296, 448, 528, 636, 688, 712, 1024, 152, 232, 384, 648, 704, 788, 856, 1000, 1204, 1416, 1472, 1556, 1592, 1624, 1800, 1972, 2008, 2120, 2356, 2360, 2676, 2744, 2888, 2912, 3064, 3328, 3444, 3680, 3832, 4096
OFFSET
1,1
COMMENTS
Consider A126241, the sequence of dropping times in the Collatz iteration. Only zero and the numbers in A020914 can be dropping times. The dropping times in A126241 have a definite pattern. For example, 1 appears at positions n = 2 + 2*i, for i=0,1,2,3,... Similarly, 2 appears at positions n = 5 + 4*i; 4 appears at n = 3 + 16*i; 5 appears at n = 11 + {12,32}*i; and 7 appears at 7 + {8, 52, 128}*i. In general, if we let s=A020914(r) be the r-th possible stopping time, then A126241(n) = s for n = A122442(r) + T(r)*i, where T(r) is the r-th row of this triangle. The length of row n is A186009(n). The n-th row ends with 2^A020914(n).
The frequency of the r-th dropping time s=A020914(r) can be computed as A186009(r)/2^s. The first few frequencies are 1/2, 1/4, 1/16, 1/16, 3/128, 7/256, 3/256, 15/2048, and 85/8192.
The term "stopping time" is sometimes used instead of "dropping time", but the former usually refers to A006666.
This sequence is closely related to A177789.
REFERENCES
J. C. Lagarias, ed., The Ultimate Challenge: The 3x+1 Problem, Amer. Math. Soc., 2010. See pp. 33, 35ff.
EXAMPLE
The triangle begins
2
4
16
12, 32
8, 52, 128
40, 56, 84, 136, 160, 180, 256
60, 80, 136, 220, 288, 296, 448, 528, 636, 688, 712, 1024
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
T. D. Noe, Feb 09 2011
STATUS
approved