A122437 Allowable values of the "dropping time" of the Collatz (3x+1) iteration.

1, 3, 6, 8, 11, 13, 16, 19, 21, 24, 26, 29, 32, 34, 37, 39, 42, 44, 47, 50, 52, 55, 57, 60, 63, 65, 68, 70, 73, 75, 78, 81, 83, 86, 88, 91, 94, 96, 99, 101, 104, 106, 109, 112, 114, 117, 119, 122, 125, 127, 130, 132, 135, 138, 140, 143, 145, 148, 150, 153, 156, 158, 161

Offset: 1

Author

T. D. Noe, Sep 06 2006

Keywords

• nice
• nonn

Comments

Only these numbers appear in A060445, which tabulates the "dropping time" of odd numbers. Note that all even numbers have a "dropping time" of 1.

a(n) is also the number of binary digits of 6^(n-1); for example, a(4)=8 since 6^(4-1)=216 in binary is 11011000, an 8-digit number. - Julio Cesar de la Yncera, Mar 28 2009

A positive integer (x) is an allowable value if and only if (x-1)/(1+log(2)/log(3)) - floor(x/(1+log(2)/log(3))) is not negative. - K. Spage, Oct 22 2009

Here the word "allowable" means that it is necessary for a sequence of iterates starting from odd value m to arrive at a value x = f^{floor(1+n+n*log(3)/log(2))}(m) < m, where n gives the number of odds in such a sequence including m, to have undergone precisely floor(1+n+n*log(3)/log(2)) iterations of f, where f(2*m)=m, f(2*m+1)=6*m+4. However, the formula for a(n+1) does not fully account for the order of odds and evens in such a sequence because it does not account for the effects of the "+1". Thus it is unknown whether it maximizes the value x for all values m. For example, fix m = 1 and the "+1" is enough to give the trivial cycle. So it is possible that for some m we have f^{floor(1+n+n*log(3)/log(2))}(m) >= m. - Jeffrey R. Goodwin, Aug 24 2011

The indices of the powers of 3 in A006899. - Ruud H.G. van Tol, Nov 02 2022

Links

• T. D. Noe, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A022921 (number of 2^m between 3^n and 3^(n+1)), A122442 (least k having dropping time a(n)).

Cf. A006899.

Programs

Mathematica

Floor[1+Range[0,100]*(1+Log[2,3])] (* T. D. Noe, Sep 08 2006 *)
Map[Length[RealDigits[ #, 2][[1]]] &, Table[10^i, {i, 0, 50}]] (* Julio Cesar de la Yncera, Mar 28 2009 *)

PARI

a(n)=logint(3^(n-1),2)+n \\ Ruud H.G. van Tol, Nov 04 2022

Formula

a(1) = 1, a(n+1) = a(n) + A022921(n-1) + 1.

a(n+1) = floor(1 + n + n*log(3)/log(2)). - T. D. Noe, Sep 08 2006

a(n) = floor((1 + log(2)/log(3))*A020914(n-1)). - K. Spage, Oct 22 2009

a(n) = A020914(n-1) + n - 1. - K. Spage, Oct 23 2009 [corrected by Ruud H.G. van Tol, Nov 03 2022]

a(n) = a(n-1)+2 if 3^(n-1) < 2^(a(n-1)+2-(n-1)); a(n) = a(n-1)+3 otherwise. - V. Barbera, Aug 12 2025

Extensions

Comment corrected and edited by Jon E. Schoenfield, Feb 27 2014