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A359207
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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.
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10
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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
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OFFSET
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0,3
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COMMENTS
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It is unknown whether each positive starting integer eventually reaches 0.
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
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LINKS
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EXAMPLE
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a(7) = 3 because it takes 3 steps to reach 0: (7, 10, 1, 0).
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MATHEMATICA
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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 *)
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PROG
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(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
(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
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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