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A359207 Number of steps to reach 0 starting with n in the map x->A359194(x) (binary complement of 3n), or -1 if 0 is never reached. 10
0, 1, 2, 11, 12, 1, 10, 3, 4, 13, 2, 19, 80, 9, 2, 15, 16, 81, 14, 11, 12, 1, 6, 83, 8, 73, 22, 79, 7572, 5, 18, 75, 76, 7573, 74, 7, 12, 17, 10, 3, 4, 13, 2, 7571, 4, 85, 78, 15, 96, 21, 5498, 91, 72, 13, 6, 7, 56, 13, 82, 3, 20, 5, 98, 15, 16, 21, 14, 7 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,3
COMMENTS
It is unknown whether each positive starting integer eventually reaches 0.
From Jon E. Schoenfield, Dec 21 2022: (Start)
a(n) == n (mod 4).
a(n) = 1 iff 3*n + 1 = 4^k for some integer k. (End)
All but 10 values under 10^7 have been run to 0. Each of the remaining 10 requires over 2*10^12 steps. They're all in one group that reaches the same high value (nearly 8 million bits wide) after about 2*10^12 steps. The smallest value in this group is 3417582. - Tim Peters, Jun 14 2023
LINKS
EXAMPLE
a(7) = 3 because it takes 3 steps to reach 0: (7, 10, 1, 0).
MATHEMATICA
f[n_] := FromDigits[BitXor[1, IntegerDigits[3*n, 2]], 2]; Array[-1 + Length@ NestWhileList[f, #, # != 0 &] &, 68, 0] (* Michael De Vlieger, Dec 21 2022, faster function by Hans Havermann *)
PROG
(Python)
def f(n): return 1 if n == 0 else (m:=3*n)^((1 << m.bit_length())-1)
def a(n):
i, fi = 0, n
while fi != 0: i, fi = i+1, f(fi)
return i
print([a(n) for n in range(68)]) # Michael S. Branicky, Dec 20 2022
(PARI) f(n) = if(n, bitneg(n, exponent(n)+1), 1); \\ A035327
a(n) = my(nb=0, m=n); while (m, m=f(3*m); nb++); nb; \\ Michel Marcus, Dec 21 2022
CROSSREFS
Sequence in context: A090009 A153706 A201187 * A068225 A367810 A043080
KEYWORD
nonn,base
AUTHOR
Joshua Searle, Dec 20 2022
STATUS
approved

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Last modified May 28 15:12 EDT 2024. Contains 372916 sequences. (Running on oeis4.)