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

1,2

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.

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.

Formula

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

Example

res(993) = 1.253142144395068050165495297839461424861536597396513692763...

Mathematica

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

Prog

(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

Crossrefs

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

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

Sequence in context: A062706 A059217 A021802 * A011192 A011356 A021399

Adjacent sequences: A304171 A304172 A304173 * A304175 A304176 A304177

Keyword

nonn,cons

Author

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

Status

approved