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%I #16 Mar 10 2024 19:27:57
%S 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,
%T 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,
%U 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
%N Decimal expansion of 2^61/(3^32*993), the conjectured maximal residue in the Collatz 3x+1 problem.
%C 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.
%H Paolo Xausa, <a href="/A304174/b304174.txt">Table of n, a(n) for n = 1..10000</a>
%H Eric Roosendaal, <a href="http://www.ericr.nl/wondrous/index.html">On the 3x + 1 problem</a>, last modified on April 6, 2018.
%F res(993) = 2^61/(3^32*993).
%e res(993) = 1.253142144395068050165495297839461424861536597396513692763...
%t First[RealDigits[2^61/(3^32*993), 10, 100]] (* _Paolo Xausa_, Mar 10 2024 *)
%o (PARI) 2^61/(3^32*993.) \\ Or, to find this value experimentally:
%o (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
%Y Cf. A006370 (Collatz map), A014682 (condensed version), A127789 (indices of record residues).
%Y Cf. A006666 (halving steps), A006667 (tripling steps), A006577 (total).
%K nonn,cons
%O 1,2
%A _M. F. Hasler_ (following an idea of _Michel Lagneau_), May 07 2018
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