A290101 a(n) = dropping time for the modified Collatz problem, where x -> 3x+1 if x is odd, and x -> either x/2 or 3x+1 if x is even (minimal number of any such steps to reach a lower number than the starting value n); a(1) = 0 by convention.
0, 1, 6, 1, 3, 1, 11, 1, 3, 1, 8, 1, 3, 1, 11, 1, 3, 1, 6, 1, 3, 1, 8, 1, 3, 1, 16, 1, 3, 1, 19, 1, 3, 1, 6, 1, 3, 1, 11, 1, 3, 1, 8, 1, 3, 1, 16, 1, 3, 1, 6, 1, 3, 1, 8, 1, 3, 1, 11, 1, 3, 1, 19, 1, 3, 1, 6, 1, 3, 1, 13, 1, 3, 1, 8, 1, 3, 1, 13, 1, 3, 1, 6, 1, 3, 1, 8, 1, 3, 1, 11, 1, 3, 1, 13, 1, 3, 1, 6, 1, 3, 1, 16, 1, 3, 1, 8, 1, 3
Offset: 1
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Examples
Starting from n = 27, the following is a shortest path leading to a value smaller than 27: 27 -> 82 -> 41 -> 124 -> 373 -> 1120 -> 560 -> 280 -> 140 -> 70 -> 35 -> 106 -> 53 -> 160 -> 80 -> 40 -> 20. It has 16 steps, thus a(27) = 16. Note the 3x+1 step from 124 to 373 which is not allowed in the ordinary Collatz problem.
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Programs
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PARI
A290101(n) = { if(1==n,return(0)); my(S, k); S=[n]; k=0; while( S[1]>=n, k++; S=vecsort( concat(apply(x->3*x+1, S), apply(x->x\2, select(x->x%2==0, S) )), , 8); ); k } \\ After Max Alekseyev's code for A127885
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Python
from sympy import flatten def ok(n, L): return any(i < n for i in L) def a(n): if n==1: return 0 L=[n] i = 0 while not ok(n, L): L=set(flatten([[3*k + 1, k//2] if k%2==0 else 3*k + 1 for k in L])) i+=1 return i print([a(n) for n in range(1, 121)]) # Indranil Ghosh, Aug 31 2017
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