login
This site is supported by donations to The OEIS Foundation.
Logo

Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A122437 Allowable values of the "dropping time" of the Collatz (3x+1) iteration. 8
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 (list; graph; refs; listen; history; internal format)
OFFSET

1,2

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. Example a(4)=8 since 6^3=216 and 216 in binary is 11011000 and the length of that binary number is 8. [From Julio de la Yncera (ynceraj(AT)gmail.com), 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. [From K. Spage (kevspage2001(AT)yahoo.co.uk), 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 counts 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]

LINKS

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

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 (noe(AT)sspectra.com), Sep 08 2006

a(n) = floor((1+log(2)/log(3))*A020914(n-1)) [From K Spage (kevspage2001(AT)yahoo.co.uk), Oct 22 2009]

MATHEMATICA

Floor[1+Range[0, 100]*(1+Log[2, 3])] - T. D. Noe (noe(AT)sspectra.com), Sep 08 2006

Map[Length[RealDigits[ #, 2][[1]]] &, Table[10^i, {i, 0, 50}]] [From Julio de la Yncera (ynceraj(AT)gmail.com), Mar 28 2009]

CROSSREFS

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

Cf. a(n) = A020914(n+1)+n-1 [From K. Spage (kevspage2001(AT)yahoo.co.uk), Oct 23 2009]

Sequence in context: A190085 A047219 A139477 * A090848 A184657 A157017

Adjacent sequences:  A122434 A122435 A122436 * A122438 A122439 A122440

KEYWORD

nice,nonn

AUTHOR

T. D. Noe (noe(AT)sspectra.com), Sep 06 2006

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Transforms | Puzzles | Hot | Classics
Recent Additions | More pages | Superseeker | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement .

Last modified February 17 07:41 EST 2012. Contains 205998 sequences.