A304174 Decimal expansion of 2^61/(3^32*993), the conjectured maximal residue in the Collatz 3x+1 problem.

1, 2, 5, 3, 1, 4, 2, 1, 4, 4, 3, 9, 5, 0, 6, 8, 0, 5, 0, 1, 6, 5, 4, 9, 5, 2, 9, 7, 8, 3, 9, 0, 4, 6, 1, 4, 2, 4, 8, 6, 1, 5, 3, 6, 5, 9, 7, 3, 9, 6, 5, 1, 3, 6, 9, 2, 7, 6, 3, 0, 4, 6, 5, 5, 5, 7, 3, 6, 7, 5, 8, 6, 4, 8, 9, 7, 4, 7, 8, 3, 0, 0, 8, 7, 8, 4, 0, 1, 1, 2, 8, 5, 4, 9, 9, 7, 5, 3, 1, 3, 5, 3, 6, 7, 7, 4, 7, 9, 1, 5, 6, 1, 6, 0, 2, 5, 5, 1, 8, 5, 9, 2, 4, 3, 8, 4, 5, 7, 7, 8

Offset: 1

Author

M. F. Hasler (following an idea of Michel Lagneau), May 07 2018

Keywords

• nonn
• cons

Comments

The residue of n in the 3x+1 problem is defined as the ratio 2^h(n)/(3^t(n)*n), where h = A006666 is the number of halving steps, and t = A006667 is the number of tripling steps. It is conjectured that n = 993 yields the highest possible residue. See e.g. the Roosendaal page, and A127789 for indices of record residues.

Examples

			res(993) = 1.253142144395068050165495297839461424861536597396513692763...
		

Links

• Paolo Xausa, Table of n, a(n) for n = 1..10000
• Eric Roosendaal, On the 3x + 1 problem, last modified on April 6, 2018.
• Index entries for sequences related to 3x+1 (or Collatz) problem

Crossrefs

Cf. A006370 (Collatz map), A014682 (condensed version), A127789 (indices of record residues).

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

Programs

Mathematica

First[RealDigits[2^61/(3^32*993), 10, 100]] (* Paolo Xausa, Mar 10 2024 *)

PARI

2^61/(3^32*993.) \\ Or, to find this value experimentally:
(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];printf("res(%d) = %f\n",n,1./m )) \\ M. F. Hasler, May 07 2018

Formula

res(993) = 2^61/(3^32*993).