OFFSET
0,2
COMMENTS
We count only the 3x+1 steps of the usual Collatz iteration. We stop counting when the iteration produces a number less than the initial 2n+1. For a fixed dropping time k, let N(k)=A100982(k) and P(k)=2^(A020914(k)-1). There are exactly N(k) odd numbers less than P(k) with dropping time k. Moreover, the sequence is periodic: if d is one of the N(k) odd numbers, then k=a(d)=a(d+i*P(k)) for all i>=0. This periodicity makes it easy to compute the average dropping time of the reduced Collatz iteration: Sum_{k>0} k*N(k)/P(k) = 3.492651852186... (A122791).
REFERENCES
Victor Klee and Stan Wagon, Old and New Unsolved Problems in Plane Geometry and Number Theory, Mathematical Association of America (1991) pp. 225-229, 308-309. [called on p. 225 stopping time for 2n+1 and the function C(2*n+1) = A075677(n+1), n >= 0. - Wolfdieter Lang, Feb 20 2019]
LINKS
T. D. Noe, Table of n, a(n) for n = 0..10000
FORMULA
a(n) is the least k for which fr^[k](n) < 2*n + 1, for n >= 1 and k >= 1, where fr(n) = A075677(n+1) = A000265(3*n+2). No k satisfies this for n = 0: a(0) := 0 by convention. The dropping time a(n) is finite, for n >= 1, if the Collatz conjecture is true. - Wolfdieter Lang, Feb 20 2019
a(1+i*8) = 2, for i>=0, because A100982(2) = 1 is odd, and A020914(2) = 4 gives P(2) = 2^(4-1) = 8. - Ruud H.G. van Tol, Dec 19 2021
EXAMPLE
a(3)=4 because, starting with 7, the iteration produces 11,17,13,5 and the last term is less than 7.
n = 13: the fr trajectory for 2*13+1 = 27 is 41, 31, 47, 71, 107, 161, 121, 91, 137, 103, 155, 233, 175, 263, 395, 593, 445, 167, 251, 377, 283, 425, 319, 479, 719, 1079, 1619, 2429, 911, 1367, 2051, 3077, 577, 433, 325, 61, 23, 35, 53, 5, 1 with 41 terms (without 27), hence fr^[37] = 23 < 27 and a(13) = 37. - Wolfdieter Lang, Feb 20 2019
MATHEMATICA
nextOddK[n_]:=Module[{m=3n+1}, While[EvenQ[m], m=m/2]; m]; dt[n_]:=Module[{m=n, cnt=0}, If[n>1, While[m=nextOddK[m]; cnt++; m>n]]; cnt]; Table[dt[n], {n, 1, 301, 2}]
CROSSREFS
KEYWORD
nonn
AUTHOR
T. D. Noe, Sep 08 2006
STATUS
approved