A346369 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.
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, 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, 36, 12, 3, 0, 18, 10, 18, 10, 7, 0, 5, 5, 13, 13, 13, 2, 18
Offset: 1
Keywords
Examples
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.
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.
Programs
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PARI
a006370(n) = if(n%2==0, n/2, 3*n+1) trajectory(n, terms) = my(v=[n]); while(#v < terms, v=concat(v, a006370(v[#v]))); v 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))))
Comments