A359194 -id:A359194 - cp's OEIS frontend

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.

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

Offset: 0

Joshua Searle, Dec 20 2022

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

			a(7) = 3 because it takes 3 steps to reach 0: (7, 10, 1, 0).

Rémy Sigrist, Table of n, a(n) for n = 0..10000
Tim Peters, Seeds under 2*10^6 that take more than 10^8 steps to reach 0
Tim Peters, Seeds under 10^7 that take more than 10^9 steps to reach 0
Joshua Searle, Collatz-inspired sequences.

Cf. A035327, A359194, A359208, A359209, A359255.

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 *)

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

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

A359208 Maximum value reached when starting from n during iteration of the map x->A359194(x) (binary complement of 3n), or -1 if infinite.

Original entry on oeis.org

0, 1, 2, 300, 300, 5, 300, 10, 10, 300, 10, 300, 328536, 300, 21, 300, 300, 328536, 300, 300, 300, 21, 72, 328536, 300, 328536, 661, 328536, 123130640068522377168864228132316865867184046004226894, 40, 300, 328536, 328536

Offset: 0

Joshua Searle, Dec 20 2022

It is unknown whether any terms are -1. The next term a(33) is equal to a(28), a 54-digit number. a(425720) is >= 2.09 * 10^114778, unresolved after 10^10 iterations.

a(425720) = 7.14... * 10^179246. - Joshua Searle, Jan 10 2023

			a(3) = 300 because the largest term in the iterated sequence: (3, 6, 13, 24, 55, 90, 241, 300, 123, 142, 85, 0) is 300.

Rémy Sigrist, Table of n, a(n) for n = 0..10000
Joshua Searle, Collatz-inspired sequences

Cf. A359194, A359207, A359209, A359215, A359218, A359219, A359220, A359221, A359222, A359255.

f[n_] := BitXor[3 n, 2^IntegerPart[Log2[3 n] + 1] - 1]; Table[Max@ NestWhileList[f, n, # != 0 &], {n, 0, 32}] (* Michael De Vlieger, Dec 21 2022 *)

f(n) = if(n, bitneg(n, exponent(n)+1), 1); \\ A035327
a(n) = my(x=n, m=n); while (m, m=f(3*m); if (m>x, x=m)); x; \\ Michel Marcus, Dec 21 2022

def f(n): return 1 if n == 0 else (m:=3*n)^((1 << m.bit_length())-1)
def a(n):
    i, fi, m = 0, n, n
    while fi != 0: i, fi, m = i+1, f(fi), max(m, fi)
    return m
print([a(n) for n in range(33)]) # Michael S. Branicky, Dec 20 2022

A359209 Numbers that under iteration of the map x->A359194(x) (binary complement of 3n) until 0 is reached never exceed the initial term.

Original entry on oeis.org

0, 1, 2, 5, 10, 21, 39, 40, 42, 71, 72, 78, 85, 142, 150, 157, 163, 167, 168, 170, 285, 291, 300, 303, 311, 313, 315, 316, 317, 319, 320, 321, 322, 327, 328, 329, 331, 333, 334, 335, 336, 338, 339, 340, 341, 569, 571, 572, 573, 575, 576, 577, 578, 579

Offset: 1

Joshua Searle, Dec 20 2022

If it can be shown that all iterations of A359194 eventually reach one of the terms of this sequence, it would prove that all such trajectories are finite.
From M. F. Hasler, Dec 26 2022: (Start)
(1) If A359194(x) = a(k) for some a(k) < x, then x is also in the sequence.
(2) The orbit of any x under iterations of A359194 is finite if and only if it reaches a term of this sequence.
(3) No term has binary digits starting with '11...', i.e., all terms > 1 have binary digits starting '10...'.
(4) If the 3rd binary digit of a(n) is a '1', it cannot be followed by another bit 1, so all terms > 5 have a binary representation of the form '100...' or '1010...'
(5) A substring of '11', i.e., two consecutive bits 1, in a term of this sequence, is necessarily preceded by an earlier substring '00', i.e., two consecutive bits 0 to the left of the '11'. [This implies (3) and (4).] (End)

			40 is a term since its trajectory is 40 -> 7 -> 10 -> 1 -> 0, which never exceeds 40.

Cf. A035327, A359194, A359207, A359208, A359255.

bc:= n -> 2^(1+ilog2(n))-1-n: bc(0):= 1:
filter:= proc(n) local x;
  x:= n;
  while x <> 0 do
    x:= bc3(x);
    if x > n then return false fi;
  od;
  true
end proc:
select(filter, [$0..1000]); # Robert Israel, Dec 22 2022

f[n_] := BitXor[3 n, 2^IntegerPart[Log2[3 n] + 1] - 1]; Select[Range[0, 200], Function[n, AllTrue[NestWhileList[f, n, # != 0 &], # <= n &]]] (* Michael De Vlieger, Dec 21 2022 *)

f(n) = if(n, bitneg(n, exponent(n)+1), 1); \\ A035327
isok(m) = my(km=m); while (m, m=f(3*m); if (m>km, return(0))); return(1); \\ Michel Marcus, Dec 21 2022

def f(n): return 1 if n == 0 else (m:=3*n)^((1 << m.bit_length())-1)
def ok(n):
    i, fi, m = 0, n, n
    while fi != 0 and m <= n: i, fi, m = i+1, f(fi), max(m, fi)
    return m <= n
print([k for k in range(580) if ok(k)]) # Michael S. Branicky, Dec 20 2022

A359215 Number of terms in S(n) that did not appear in previous trajectories, where S(n) is the trajectory of the mappings of x->A359194(x) starting with n and stopping when 0 is reached, -1 if 0 is never reached.

Original entry on oeis.org

0, 1, 1, 11, 1, 1, 0, 2, 1, 1, 0, 6, 78, 0, 2, 0, 0, 1, 0, 1, 1, 0, 0, 3, 0, 0, 11, 0, 7571, 2, 0, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 3, 0, 3, 77, 0, 5419, 1, 0, 1, 4, 0, 1, 0, 0, 2, 2, 0, 2, 0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 1

Offset: 0

Michael De Vlieger, Dec 21 2022

"Branch length" of n->A359194(n).
a(0) = 0 since n = 0.
Let m be the first term in S(n) that has appeared in S(k), k < n. A359218(n) = m.
Analogous to A222118 which instead regards the Collatz function A006318.

			a(0) = 0 since n = 0.
a(1) = 1 since S(1) = {1, 0}, but m = 0 appeared in S(0).
a(2) = 1 since S(2) = {2, 1, 0}, but m = 1 appeared in S(1).
a(3) = 11 since S(3) = {3, 6, 13, 24, 55, 90, 241, 300, 123, 142, 85, 0}, but m = 0 appeared in S(0).
a(4) = 1 since S(4) = {4, 3, ...} but 3 appears in S(3), etc.

Michael De Vlieger, Table of n, a(n) for n = 0..16384
Michael De Vlieger, Log log scatterplot of a(n) and b(n), n = 1..2^14, b(n) = A359207(n) in dark blue, a(n) in red, highlighting where a(n) = b(n) in green.

Cf. A222118, A359194, A359207, A359218.

c[] = -1; c[0] = 0; f[n] := FromDigits[BitXor[1, IntegerDigits[3*n, 2]], 2]; Table[(Map[If[c[#1] == -1, Set[c[#1], #2]] & @@ # &, Partition[#, 2, 1]]; -1 + Length[#]) &@ NestWhileList[f, n, c[#] == -1 &], {n, 0, 120}]

A359218 Let S(n) be the sequence obtained through the mapping of x->A359194(x) starting with n and stopping when 0 is reached, -1 if 0 is never reached. a(n) = m if appears in S(k), k < n, otherwise -1.

Original entry on oeis.org

0, 0, 1, 0, 3, 0, 6, 1, 7, 4, 10, 9, 10, 13, 0, 15, 16, 12, 18, 6, 3, 21, 22, 12, 24, 25, 3, 27, 21, 7, 30, 31, 31, 28, 34, 22, 19, 37, 13, 39, 40, 4, 1, 43, 123, 58, 46, 4, 187, 49, 27, 102, 52, 96, 42, 55, 87, 57, 58, 21, 30, 61, 48, 63, 64, 60, 66, 54, 51, 69

Offset: 0

Michael De Vlieger, Dec 21 2022

By convention, a(0) = 0 since n = 0.

Regarding A359215(n), this is the value m that had appeared in S(k), k < n.

			a(1) = 0 since S(1) = {1, 0}, but m = 0 appeared in S(0).
a(2) = 1 since S(2) = {2, 1, ...}, but m = 1 appeared in S(1).
a(3) = 0 since S(3) = {3, 6, 13, 24, 55, 90, 241, 300, 123, 142, 85, 0}, but m = 0 appeared in S(0).
a(4) = 3 since S(4) = {4, 3, ...} but 3 appears in S(3), etc.
a(5) = 0 since S(5) = {5, 0}, but 0 appears in S(0).
a(6) = 6 since 6 appears in F(3).
a(7) = 1 since S(7) = {7, 10, 1, ...} but 1 appears in S(1).
a(8) = 7 since S(8) = {8, 7, ...} but 7 appears in S(7)
a(9) = 4 since S(9) = {9, 4, ...} but 4 appears in S(4).
a(10) = 10 since 10 appears in S(7).
a(11) = 9 since S(11) = {11, 30, 37, 16, 15, 18, 9, ...} but 9 appears in S(9).
a(12) = 10 since S(12) = {12, 27, ..., 39, 10, ...} but 10 appears in S(7), etc.

Michael De Vlieger, Table of n, a(n) for n = 0..16384
Michael De Vlieger, Scatterplot of a(n) for n = 0..2^14 and for a(n) <= 2^14, showing a curious pattern that scales with n = 2^k.

Cf. A359194, A359207, A359215.

c[] = -1; c[0] = 0; f[n] := BitXor[3 n, 2^IntegerPart[Log2[3 n] + 1] - 1]; Table[(Map[If[c[#1] == -1, Set[c[#1], #2]] & @@ # &, Partition[#, 2, 1]]; Last@ #) &@ NestWhileList[f, n, c[#] == -1 &], {n, 0, 120}]

A359255 Number of steps to reach a maximum starting with n in the map x->A359194(x) (binary complement of 3n), or -1 if n goes to infinity.

Original entry on oeis.org

0, 0, 0, 7, 8, 0, 6, 1, 2, 9, 0, 15, 28, 5, 1, 11, 12, 29, 10, 7, 8, 0, 2, 31, 4, 21, 5, 27, 2962, 1, 14, 23, 24, 2963, 22, 3, 8, 13, 6, 0, 0, 9, 0, 2961, 2, 33, 26, 11, 74, 4, 1591, 69, 20, 5, 3, 3, 34, 9, 30, 1, 16, 1, 76, 11

Offset: 0

Hans Havermann, Dec 22 2022

The corresponding maxima reached are in A359208. The indices at which zeros appear here are in A359209.

Hans Havermann, Table of n, a(n) for n = 0..10000

Cf. A359194, A359207, A359208, A359209.

f[n_] := FromDigits[BitXor[1, IntegerDigits[3*n, 2]], 2]; Table[c=m=0; n=r=i; While[n!=0, c++; n=f[n]; If[n>r, m=c; r=n]]; m, {i, 0, 63}]

def f(n): return 1 if n == 0 else (m:=3*n)^((1 << m.bit_length())-1)
def a(n):
    i, fi, m, mi = 0, n, n, 0
    while fi != 0:
        i, fi = i+1, f(fi)
        if fi > m: m, mi = fi, i
    return mi
print([a(n) for n in range(64)]) # Michael S. Branicky, Dec 22 2022

A359219 Starting numbers that require more iterations of the map x->A359194(x) (binary complement of 3n) to reach 0 than any smaller number.

Original entry on oeis.org

0, 1, 2, 3, 4, 9, 11, 12, 17, 23, 28, 33, 74, 86, 180, 227, 350, 821, 3822, 4187, 5561, 6380, 6398, 22174, 22246, 26494, 34859, 49827, 70772, 103721, 104282, 204953, 213884, 225095, 407354, 425720

Offset: 1

Joshua Searle, Dec 21 2022

425720 after 10^10 iterations has not yet reached 0 and in general it is unknown whether every starting number does reach 0.

A359207(425720) = 87037147316. - Martin Ehrenstein, Jan 02 2023

			3 is a term because it requires 11 iterations to reach 0, which is more than any starting number less than 3.
0: (0) -- 0 terms
1: (1, 0) -- 1 term
2: (2, 1, 0) -- 2 terms
3: (3, 6, 13, 24, 55, 90, 241, 300, 123, 142, 85, 0) -- 11 terms.

Joshua Searle, Collatz-inspired sequences

Cf. A035327, A359194, A359207, A359208, A359209, A359215, A359218, A359220, A359221, A359222.

from itertools import count, islice
def f(n): return 1 if n == 0 else (m:=3*n)^((1 << m.bit_length())-1)
def iters(n):
    i, fi = 0, n
    while fi != 0: i, fi = i+1, f(fi)
    return i
def agen(): # generator of terms
    record = -1
    for m in count(0):
        v = iters(m)
        if v > record: yield m; record = v
print(list(islice(agen(), 18))) # Michael S. Branicky, Dec 21 2022

a(27)-a(36) from Tom Duff (SeqFan mailing list, Dec 19 2022)

A359220 Number of steps to reach 0 from A359219(n) where A359219 are the starting numbers that require more iterations in the map x->A359194(x) than any smaller number.

Original entry on oeis.org

0, 1, 2, 11, 12, 13, 19, 80, 81, 83, 7572, 7573, 7574, 7578, 7580, 664475, 664882, 3180929, 3180930, 3180931, 3181981, 3181988, 3182002, 3182226, 120796790, 556068798, 556068799, 556068871, 556068872, 572086553, 572086610, 1246707529, 1246707552, 1246707555, 1246707602

Offset: 1

Joshua Searle, Dec 21 2022

It is unknown whether all starting numbers reach 0; the next term, a(36) depends on whether 425720 ever reaches 0 (see A359207). It remains nonzero after 10^10 iterations.
A359207(425720) = 87037147316. Calculated by Tom Duff (12/16/22) - Joshua Searle, Jan 10 2023
a(4) - a(7) only differ by a small fraction of their starting terms. The same is true for the terms in the intervals a(8) - a(10), a(11) - a(15), a(16) - a(17) and a(18) - a(24). It may also be true for a(26) - a(29), a(30) - a(31) and a(32) - a(35).

			a(4) is the step count from the starting number A359219(4) = 3: (3, 6, 13, 24, 55, 90, 241, 300, 123, 142, 85, 0) -- 11 steps, hence a(4) = 11.

Joshua Searle, Collatz-inspired sequences

Step counts of A359219.

Cf. A035327, A359194, A359207, A359208, A359209, A359215, A359218, A359219, A359221.

from itertools import count, islice
def f(n): return 1 if n == 0 else (m:=3*n)^((1 << m.bit_length())-1)
def iters(n):
    i, fi = 0, n
    while fi != 0: i, fi = i+1, f(fi)
    return i
def agen(): # generator of terms
    record = -1
    for m in count(0):
        v = iters(m)
        if v > record: yield v; record = v
print(list(islice(agen(), 18))) # Michael S. Branicky, Dec 21 2022

a(27) and beyond from Tom Duff (SeqFan mailing list, Dec 19 2022)

A359221 Starting numbers which reach a new record high value when iterating the map x->A359194(x) (binary complement of 3n).

Original entry on oeis.org

0, 1, 2, 3, 12, 28, 227, 821, 22246, 26494, 204953, 425720

Offset: 1

Joshua Searle, Dec 22 2022

It is unknown whether all starting numbers reach 0.

103721 is not a term of this sequence despite having a trajectory of record length, because its maximum of 2.42...*10^14081 is lower than the previous record holder.

			Let S(x) = iteration sequence of A359194 starting with x; then
S(0) = (0), maximum = 0;
S(1) = (1, 0), maximum = 1;
S(2) = (2, 1, 0), maximum = 2;
S(3) = (3, 6, 13, 24, 55, 90, 241, 300, 123, 142, 85, 0), maximum = 300;
Since S(3) contains a higher maximum than any lower positive starting integer, 3 is a term of this sequence.

Joshua Searle, Collatz-inspired sequences

Cf. A035327, A359194, A359207, A359208, A359209, A359215, A359218, A359219, A359220, A359222, A359255.

from itertools import count, islice
def f(n): return 1 if n == 0 else (m:=3*n)^((1 << m.bit_length())-1)
def itersmax(n):
    i, fi, m = 0, n, n
    while fi != 0: i, fi, m = i+1, f(fi), max(m, fi)
    return i, m
def agen(): # generator of terms
    record = -1
    for m in count(0):
        v, mx = itersmax(m)
        if mx > record:
            yield m # use mx to obtain values
            record = mx
print(list(islice(agen(), 8))) # Michael S. Branicky, Dec 22 2022

A359222 Number of steps to reach 0 from A359221(n) (Starting numbers that reach a new record high value during iteration by the map x->A359194(x)).

Original entry on oeis.org

0, 1, 2, 11, 80, 7572, 664475, 3180929, 120796790, 556068798, 1246707529, 87037147316

Offset: 1

Joshua Searle, Dec 29 2022

a(12) found by Tom Duff (26 Dec 2022).

It is unknown whether all starting numbers reach 0.

			a(4) is the step count from the starting number A359221(4) = 3: (3, 6, 13, 24, 55, 90, 241, 300, 123, 142, 85, 0) -- 11 steps, hence a(4) = 11.

Joshua Searle, Collatz-inspired sequences

Step counts of A359221.

Cf. A035327, A359194, A359207, A359208, A359209, A359215, A359218, A359219, A359220, A359255.

from itertools import count, islice
def f(n): return 1 if n == 0 else (m:=3*n)^((1 << m.bit_length())-1)
def itersmax(n):
    i, fi, m = 0, n, n
    while fi != 0: i, fi, m = i+1, f(fi), max(m, fi)
    return i, m
def agen(): # generator of terms
    record = -1
    for m in count(0):
        v, mx = itersmax(m)
        if mx > record:
            yield v # use m to obtain starting numbers
            record = mx
print(list(islice(agen(), 8))) # Michael S. Branicky, Dec 29 2022

cp's OEIS Frontend

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.

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A359208 Maximum value reached when starting from n during iteration of the map x->A359194(x) (binary complement of 3n), or -1 if infinite.

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A359209 Numbers that under iteration of the map x->A359194(x) (binary complement of 3n) until 0 is reached never exceed the initial term.

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A359215 Number of terms in S(n) that did not appear in previous trajectories, where S(n) is the trajectory of the mappings of x->A359194(x) starting with n and stopping when 0 is reached, -1 if 0 is never reached.

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A359218 Let S(n) be the sequence obtained through the mapping of x->A359194(x) starting with n and stopping when 0 is reached, -1 if 0 is never reached. a(n) = m if appears in S(k), k < n, otherwise -1.

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A359255 Number of steps to reach a maximum starting with n in the map x->A359194(x) (binary complement of 3n), or -1 if n goes to infinity.

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A359219 Starting numbers that require more iterations of the map x->A359194(x) (binary complement of 3n) to reach 0 than any smaller number.

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A359220 Number of steps to reach 0 from A359219(n) where A359219 are the starting numbers that require more iterations in the map x->A359194(x) than any smaller number.

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