OFFSET
1,2
COMMENTS
Original name: In the '3x + 1' problem we would expect a positive integer n to be approximately equal to 2^j / 3^k where j is the number of halving steps and k is the number of '3x + 1' steps required to reach 1. The numbers in this sequence are those for which 2^j / (n*3^k) sets a new record.
Eric Roosendaal calls 2^j / (n*3^k) the residue of n and conjectures that 993 yields the highest residue. - T. D. Noe, Apr 08 2007
It has been verified that the next value, if it exists, is larger than 2^32 ~ 4.3e9. We do not need an "escape clause" (as for A006577, A006666, A006667, ...) in this sequence since the unlikely case of a possibly undefined ratio is irrelevant for the list of records. - M. F. Hasler, May 07 2018
LINKS
Eric Roosendaal, On the 3x+1 Problem
PROG
(PARI) (c(n, c=[0, 0])=while(n>1, bittest(n, 0)&&c[1]++&&(n=n*3+1)&&next; n\=2; c[2]++); c); m=1; for(n=1, oo, m<<(t=c(n))[2]>n*3^t[1]||next; m=n*3^t[1]/2^t[2]; print1(n", ")) \\ M. F. Hasler, May 07 2018
CROSSREFS
KEYWORD
more,nonn
AUTHOR
Paul Boddington, Apr 04 2007
EXTENSIONS
New name from M. F. Hasler, May 07 2018
STATUS
approved