A127789 Record indices of 2^h(n)/(3^t(n)*n), where h and t are the number of halving resp. tripling steps in the '3x + 1' problem.

1, 3, 7, 9, 505, 559, 745, 993

Offset: 1

Author

Paul Boddington, Apr 04 2007

Keywords

• more
• nonn

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

Crossrefs

Cf. A006370 (Collatz map), A014682 (condensed version using (3n+1)/2).

Cf. A006666 (halving steps), A006667 (tripling steps), A006577 (total).

Programs

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

Extensions

New name from M. F. Hasler, May 07 2018