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
KEYWORD
nonn,cons
AUTHOR
M. F. Hasler (following an idea of Michel Lagneau), May 07 2018
STATUS
approved