This site is supported by donations to The OEIS Foundation. Please make a donation to keep the OEIS running. We are now in our 55th year. In the past year we added 12000 new sequences and reached 8000 citations (which often say "discovered thanks to the OEIS"). We need to raise money to hire someone to manage submissions, which would reduce the load on our editors and speed up editing. Other ways to donate

 Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 A122458 "Dropping time" of the reduced Collatz iteration starting with 2n+1. 13
 0, 2, 1, 4, 1, 3, 1, 4, 1, 2, 1, 3, 1, 37, 1, 35, 1, 2, 1, 5, 1, 3, 1, 34, 1, 2, 1, 3, 1, 4, 1, 34, 1, 2, 1, 32, 1, 3, 1, 5, 1, 2, 1, 3, 1, 28, 1, 5, 1, 2, 1, 26, 1, 3, 1, 19, 1, 2, 1, 3, 1, 5, 1, 9, 1, 2, 1, 4, 1, 3, 1, 4, 1, 2, 1, 3, 1, 25, 1, 13, 1, 2, 1, 18, 1, 3, 1, 5, 1, 2, 1, 3, 1, 4, 1, 8, 1, 2, 1, 5 (list; graph; refs; listen; history; text; internal format)
 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... 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 with 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 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^ = 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 Cf. A000265, A060445, A075677 (one step of the reduced Collatz iteration), A075680. Sequence in context: A242885 A181157 A095248 * A329644 A256578 A127461 Adjacent sequences:  A122455 A122456 A122457 * A122459 A122460 A122461 KEYWORD nonn AUTHOR T. D. Noe, Sep 08 2006 STATUS approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

Last modified December 7 20:25 EST 2019. Contains 329848 sequences. (Running on oeis4.)