%I #5 Feb 11 2014 19:05:33
%S 13,6,235,342,41,5735,90746642
%N a(n) = least positive integer solution k to h(k) = h(k-1)+h(k-2)+...+h(k-n), where h(n) is the length of n, f(n), f(f(n)), ...., 1 in the Collatz (or 3x + 1) problem. (The earliest "1" is meant.)
%C 1. Recall that f(n) = n/2 if n is even; = 3n + 1 if n is odd. 2. Problem: Is a(n) defined for all n, that is, does a positive integer solution k to h(n) = h(k-1)+h(k-2)+...+h(k-n) always exist?
%e k = 235 is the least k satisfying h(k) = h(k-1)+h(k-2)+h(k-3), so a(3) = 235.
%K more,nonn
%O 1,1
%A _Joseph L. Pe_, Dec 31 2002
%E a(7) from _Donovan Johnson_, Nov 14 2010
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