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 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. 3
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,3 COMMENTS In contrast to the "3x+1" problem (see A006577, A102419), here you are free to choose either step if x is even. The sequence counts the minimum number of optimally chosen steps which leads to a value smaller than the value we started from. LINKS Antti Karttunen, Table of n, a(n) for n = 1..10000 FORMULA a(n) <= A102419(n). a(n) <= A127885(n) [apart from any hypothetical -1's in A127885]. EXAMPLE 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. PROG (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 (Python) from sympy import flatten def ok(n, L):     for i in L: return 1 if i

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Last modified October 18 14:44 EDT 2019. Contains 328161 sequences. (Running on oeis4.)