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%I #12 Jul 24 2021 13:30:28
%S 0,0,3,1,0,4,0,2,7,0,5,5,5,0,8,3,0,8,3,0,6,6,1,6,0,6,3,1,9,9,4,4,11,0,
%T 9,9,0,4,12,1,0,7,7,7,5,2,12,7,9,0,7,7,15,4,0,2,28,10,10,10,0,5,15,5,
%U 36,12,3,0,18,10,18,10,7,0,5,5,13,13,13,2,18
%N a(n) is the number of steps the Collatz trajectory of -n takes to reach a cycle, or -1 if no cycle is ever reached.
%C The analog of A139399 for negative starting values.
%C Is a(n) > -1 for all n?
%F a(n) = A224166(n) - A224254(n) = 1 + A224183(n) - A224254(n). - _Pontus von Brömssen_, Jul 24 2021
%e For n = 5: The trajectory of -5 starts -5, -14, -7, -20, -10, -5, -14, -7, -20, -10, -5, ..., with -5 already being part of the cycle {-5, -14, -7, -20, -10}, so a(5) = 0.
%e For n = 6: The trajectory of -6 starts -6, -3, -8, -4, -2, -1, -2, -1, -2, -1, -2, ..., reaching a term of the cycle {-2, -1} after 4 steps, so a(6) = 4.
%o (PARI) a006370(n) = if(n%2==0, n/2, 3*n+1)
%o trajectory(n, terms) = my(v=[n]); while(#v < terms, v=concat(v, a006370(v[#v]))); v
%o a(n) = my(t, v=[]); for(k=1, oo, t=trajectory(-n, k); for(x=1, #t, if(x < #t && t[x]==t[#t], return(x-1))))
%Y Cf. A139399, A224166, A224183, A224254.
%K sign
%O 1,3
%A _Felix Fröhlich_, Jul 14 2021
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