A289674 -id:A289674 - cp's OEIS frontend

A284116 a(n) = largest number of distinct words arising in Post's tag system {00, 1101} applied to a binary word w, over all starting words w of length n, or a(n) = -1 if there is a word w with an unbounded trajectory.

4, 7, 6, 7, 22, 23, 24, 25, 30, 31, 34, 421, 422, 423, 422, 423, 424, 2169, 2170, 2171, 2170, 2171, 2172, 2165, 2166, 2167, 24566, 24567, 24568, 24567, 24568, 24569, 24568, 24569, 24570, 253513, 253514, 342079, 342080, 342083, 342084, 342103, 20858070

Offset: 1

Jeffrey Shallit, Mar 20 2017

nonn

Post's tag system {00, 1101} maps a word w over {0,1} to w', where if w begins with 0, w' is obtained by appending 00 to w and deleting the first three letters, or if w begins with 1, w' is obtained by appending 1101 to w and deleting the first three letters.
The empty word is included in the count.
It is an important open question to decide if there is any word whose orbit grows without limit. - N. J. A. Sloane, Jul 30 2017, based on an email from Allan C. Wechsler
Comment from Don Reble, Aug 01 2017: For n <= 57, all words reach the empty word or a cycle. - N. J. A. Sloane, Aug 01 2017
From David A. Corneth, Aug 02 2017: (Start)
A word w can be described by the pair (c, d) where c is the length of w and d is the number represented by the binary word w. Then 0 <= d < 2^c.
Appending a word ww of m letters to w is the same as setting d to 2^m * w + ww. Preserving only the rightmost q digits of w is the same as setting w to w mod 2^q.
Lastly, we're only really interested in the 1st, 4th, 7th, ... leftmost digits. The others could without loss of generality be set to 0. This can be done with bitand(x, y), with y in A033138.
Therefore this problem can be formulated as follows: Let w = (c, d).
Then if d < 2^(c - 1), w' = (c - 1, bitand(4*d, floor(2^(c + 1) / 7)))
else (if (d >= 2^(c - 1)), w' = (c + 1, bitand(16*d + 13, floor(2^(c + 3) / 7))).
To find a(n), it would be enough to check values d in A152111 with n binary digits and c = n.
(End)
a(110) >= 43913328040672, from w = (100)^k, k=110. - N. J. A. Sloane, Oct 23 2017, based on Lars Blomberg's work on A291792.

			Suppose n=1. Then w = 0 ->000 -> w' = empty word, and w = 1 -> 11101 -> w' = 01 -> 0100 -> w'' = 0 -> 000 -> w''' = empty word. So a(1) = 4 by choosing w = 1.
For n = 5 the orbit of the word 10010 begins 10010, 101101, 1011101, ..., 0000110111011101, and the next word in the orbit has already appeared. The orbit consists of 22 distinct words.
From _David A. Corneth_, Aug 02 2017: (Start)
The 5-letter word w = 10100 can be described as (a, b) = (5, 20). This is equivalent to (5, bitand(20, floor(2^7 / 7))) = (5, bitand(20, 18)) = (5, 16).
As 16 >= 2^(5-1), w' = (5 + 1, bitand(16*16 + 13, floor(2^(5 + 3) / 7))) = (6, bitand(279, 36)) = (6, 4). w'' = w = (5, 16) so 10100 ~ 10000 ends in a period. (End)
Words w that achieve a(1) through a(7) are 1, 10, 100, 0001, 10010, 100000, 0001000. - _N. J. A. Sloane_, Aug 17 2017

John Stillwell, Elements of Mathematics: From Euclid to Goedel, Princeton, 2016. See page 100, Post's tag system.

Peter R. J. Asveld, On a Post's System of Tag. Bulletin of the EATCS 36 (1988), 96-102.
Liesbeth De Mol, Tracing unsolvability. A historical, mathematical and philosophical analysis with a special focus on tag systems, Ph.D. Thesis, Universiteit Gent, 2007.
Liesbeth De Mol, Tag systems and Collatz-like functions Theoret. Comput. Sci. 390 (2008), no. 1, 92-101.
Liesbeth De Mol, On the complex behavior of simple tag systems—an experimental approach, Theoret. Comput. Sci. 412 (2011), no. 1-2, 97-112.
Emil L. Post, Formal Reductions of the General Combinatorial Decision Problem, Amer. J. Math. 65, 197-215, 1943. See page 204.
Don Reble, Python program for A289670.
Shigeru Watanabe, Periodicity of Post's normal process of tag, in Jerome Fox, ed., Proceedings of Symposium on Mathematical Theory of Automata, New York, April 1962, Polytechnic Press, Polytechnic Institute of Brooklyn, 1963, pp. 83-99. [Annotated scanned copy]

Cf. A033138, A152111, A284119, A284121, A289670, A289671, A289672, A289673, A289674, A289675, A289676, A289677.
Cf. also A290436-A290441, A290741, A290742, A291792, A291793, A291794, A291795, A291796.
For the 3-shift tag systems {00,1101}, {00, 1011}, {00, 1110}, {00, 0111} see A284116, A291067, A291068, A291069 respectively (as well as the cross-referenced entries mentioned there).

Table[nmax = 0;
 For[i = 0, i < 2^n, i++, lst = {};
  w = IntegerString[i, 2, n];
  While[! MemberQ[lst, w],
   AppendTo[lst, w];
   If[w == "", Break[]];
   If[StringTake[w, 1] == "0", w = StringDrop[w <> "00", 3],
    w = StringDrop[w <> "1101", 3]]];
nmax = Max[nmax, Length[lst]]]; nmax, {n, 1, 12}] (* Robert Price, Sep 26 2019 *)
(* Or, using the (c,d) procedure: *)
 Table[nmax = 0;
 For[i = 0, i < 2^n, i++,
  c = n; d = i; lst = {};
  While[! MemberQ[lst, {c, d}],
   AppendTo[lst, {c, d}];
   If[c == 0,  Break[]];
   If[ d < 2^(c - 1),
    d = BitAnd[4*d, 2^(c - 1) - 1]; c--,
    d = BitAnd[16*d + 13, 2^(c + 1) - 1]; c++]];
nmax = Max[nmax, Length[lst]]]; nmax, {n, 1, 12}] (* Robert Price, Sep 26 2019 *)

a(19)-a(43) from Lars Blomberg, Apr 09 2017

Edited by N. J. A. Sloane, Jul 29 2017 and Oct 23 2017 (adding escape clause in case an infinite trajectory exists)

A289670 Consider the Post tag system defined in A284116; a(n) = number of binary words of length n which terminate at the empty word.

Original entry on oeis.org

2, 2, 4, 8, 16, 16, 64, 128, 192, 320, 512, 768, 2560, 6656, 12288, 21504, 36864, 81920, 176128, 327680, 638976, 1392640, 2326528, 4194304, 9568256, 17301504, 30408704, 65536000, 121110528, 220200960, 484442112, 962592768, 1837105152, 4026531840, 8304721920, 16206790656, 34712059904, 70934069248, 140190416896

Offset: 1

N. J. A. Sloane, Jul 29 2017

nonn

The orbit of a word may terminate at the empty word (this sequence and A289675), or enter a cycle (A289671, A289672, A289674), or grow without limit (it is not known if this ever happens).

			For length n=2, there are two words which terminate at the empty word, 00 and 01. For example, 00 -> 0 -> empty word. See A289675 for further examples.

Don Reble, Table of n, a(n) for n = 1..57
Don Reble, Python program for A289670.

Cf. A284116, A284119, A284121, A289671, A289672, A289673, A289674, A289675.

with(StringTools):
# Post's tag system applied once to w
# The empty string is represented by -1.
f1:=proc(w) local L,t0,t1,ws,w2;
t0:="00"; t1:="1101"; ws:=convert(w,string);
if ws[1]="0" then w2:=Join([ws,t0],""); else w2:=Join([ws,t1],"");  fi;
L:=length(w2); if L <= 3 then return(-1); fi;
w2[4..L]; end;
# Post's tag system repeatedly applied to w (valid for |w| <= 11).
# Returns number of steps to reach empty string, or 999 if w cycles
P:=proc(w) local ws,i,M; global f1;
ws:=convert(w,string); M:=1;
for i from 1 to 38 do
M:=M+1; ws:=f1(ws); if ws = -1 then return(M); fi;
od; 999; end;
# Count strings of length n which terminate and which cycle
a0:=[]; a1:=[];
for n from 1 to 11 do
lprint("starting length ",n);
ter:=0; noter:=0;
for n1 from 0 to 2^n-1 do
t1:=convert(2^n+n1,base,2); t2:=[seq(t1[i],i=1..n)];
map(x->convert(x,string),t2); t3:=Join(%,""); t4:=P(%);
if t4=999 then noter:=noter+1; else ter:=ter+1; fi;
od;
a0:=[op(a0),ter]; a1:=[op(a1),noter];
od:
a0; a1;

Table[ne = 0;
 For[i = 0, i < 2^n, i++, lst = {};
  w = IntegerString[i, 2, n];
  While[! MemberQ[lst, w],
   AppendTo[lst, w];
   If[w == "", ne++; Break[]];
   If[StringTake[w, 1] == "0", w = StringDrop[w <> "00", 3],
    w = StringDrop[w <> "1101", 3]]]];
ne, {n, 1, 12}] (* Robert Price, Sep 26 2019 *)

a(12)-a(57) from Don Reble, Jul 30 2017 and Aug 01 2017; a(12)-a(39) confirmed by Sean A. Irvine, Jul 30 2017.

A289671 Consider the Post tag system defined in A284116; a(n) = number of binary words of length n which terminate in a cycle.

Original entry on oeis.org

0, 2, 4, 8, 16, 48, 64, 128, 320, 704, 1536, 3328, 5632, 9728, 20480, 44032, 94208, 180224, 348160, 720896, 1458176, 2801664, 6062080, 12582912, 23986176, 49807360, 103809024, 202899456, 415760384, 853540864, 1663041536, 3332374528, 6752829440, 13153337344, 26055016448

Offset: 1

N. J. A. Sloane, Jul 29 2017

nonn

For n such that no binary word of length n has an infinite orbit under the Post tag system (cf. A284116), which includes all n <= 57, a(n) + A289670(n) = 2^n.

			For length n=2, there are two words which cycle, 10 and 11: 10 -> 101 -> 1101 -> 11101 -> 011101 -> 10100 -> 001101 -> 10100, which has entered a cycle.

Don Reble, Table of n, a(n) for n = 1..57

Cf. A284116, A284119, A284121, A289670-A289674.

A289675 lists the initial words that terminate at the empty string.

Maple
```
See A289670.
```

Table[ne = 0;
For[i = 0, i < 2^n, i++, lst = {};
  w = IntegerString[i, 2, n];
  While[! MemberQ[lst, w],
   AppendTo[lst, w];
   If[w == "", ne++; Break[]];
   If[StringTake[w, 1] == "0", w = StringDrop[w <> "00", 3],
    w = StringDrop[w <> "1101", 3]]]];
2^n - ne, {n, 1, 12}] (* Robert Price, Sep 26 2019 *)

A289673 Take n-th string over {1,2} in lexicographic order and apply the Post tag system described in A284116 (but adapted to the alphabet {1,2}) just once.

Original entry on oeis.org

-1, 12, 1, 1, 212, 212, 11, 11, 11, 11, 2212, 2212, 2212, 2212, 111, 211, 111, 211, 111, 211, 111, 211, 12212, 22212, 12212, 22212, 12212, 22212, 12212, 22212, 1111, 1211, 2111, 2211, 1111, 1211, 2111, 2211, 1111, 1211, 2111, 2211, 1111, 1211, 2111, 2211, 112212

Offset: 1

N. J. A. Sloane, Jul 29 2017

sign

Post's tag system maps a word w over {1,2} to w', where if w begins with 1, w' is obtained by appending 11 to w and deleting the first three letters, or if w begins with 2, w' is obtained by appending 2212 to w and deleting the first three letters.
The empty word is denoted by -1.
We work over {1,2} rather than the official alphabet {0,1} because of the prohibition in the OEIS of terms (other than 0 itself) which begin with 0.

			The initial words are:
1,2,11,12,21,22,111,112,121,122,211,212,221,222,1111,...
Applying the tag system over {1,2} these become:
-1, 12, 1, 1, 212, 212, 11, 11, 11, 11, 2212, 2212, 2212, 2212, 111, ...
If we were working over {0,1} the initial strings would be:
0,1,00,01,10,11,000,001,010,011,100,101,110,111,0000,...
and applying the tag system over {0,1} described in A284116 these would become:
-1, 01, 0, 0, 101, 101, 00, 00, 00, 00, 1101, 1101, 1101, 1101, 000, ...

Chai Wah Wu, Table of n, a(n) for n = 1..10000

Cf. A284116, A284119, A284121, A289670, A289671, A289672, A289674, A289675.

A289675 Consider the Post tag system described in A284116 (but adapted to the alphabet {1,2}) ; sequence lists the words that terminate in the empty word.

Original entry on oeis.org

1, 2, 11, 12, 111, 112, 121, 122, 1111, 1121, 1211, 1221, 2111, 2121, 2211, 2221, 11111, 11112, 11121, 11122, 11211, 11212, 11221, 11222, 12111, 12112, 12121, 12122, 12211, 12212, 12221, 12222, 111111, 111112, 111121, 111122, 112111, 112112, 112121, 112122, 121111, 121112, 121121, 121122, 122111, 122112, 122121, 122122

Offset: 1

N. J. A. Sloane, Jul 30 2017

nonn

Post's tag system maps a word w over {1,2} to w', where if w begins with 1, w' is obtained by appending 11 to w and deleting the first three letters, or if w begins with 2, w' is obtained by appending 2212 to w and deleting the first three letters.
Under this Post tag system, some words when iterated end at the empty word, others go into cycles, and others may have an orbit which grows without limit. See A289670 and A289671 for the counts of the first two types. This sequence gives a list of the words that end at the empty word.
We work over {1,2} rather than the official alphabet {0,1} because of the prohibition in the OEIS of terms (other than 0 itself) which begin with 0.
Stillwell (2016, page 100) remarks that Post was unable to find an algorithm to determine which words belong to this sequence, "and in fact this particular `halting problem' remains unsolved to this day".

			Working over the more usual alphabet {0,1}, the following are the orbits of the first few words that terminate at the empty word:
[0, -1]
[1, 01, 0, -1]
[00, 0, -1]
[01, 0, -1]
[000, 00, 0, -1]
[001, 00, 0, -1]
[010, 00, 0, -1]
[011, 00, 0, -1]
[0000, 000, 00, 0, -1]
[0010, 000, 00, 0, -1]
[0100, 000, 00, 0, -1]
[0110, 000, 00, 0, -1]
[1000, 01101, 0100, 000, 00, 0, -1]
[1010, 01101, 0100, 000, 00, 0, -1]
[1100, 01101, 0100, 000, 00, 0, -1]
[1110, 01101, 0100, 000, 00, 0, -1]
[00000, 0000, 000, 00, 0, -1]
...
Writing the initial words in this list over {1,2} rather than {0,1} gives the sequence.

John Stillwell, Elements of Mathematics: From Euclid to Goedel, Princeton, 2016. See page 100, Post's tag system.

Rémy Sigrist, Table of n, a(n) for n = 1..25000
Emil L. Post, Formal Reductions of the General Combinatorial Decision Problem, Amer. J. Math. 65, 197-215, 1943. See page 204.

Cf. A284116, A284119, A284121, A289670, A289671, A289672, A289673, A289674.

A289675 = {};
Do[For[i = 0, i < 2^n, i++, lst = {};
   w = IntegerString[i, 2, n];
   While[! MemberQ[lst, w],
    AppendTo[lst, w];
    If[w == "", AppendTo[A289675, IntegerString[i, 2, n]]; Break[]];
    If[StringTake[w, 1] == "0", w = StringDrop[w <> "00", 3],
     w = StringDrop[w <> "1101", 3]]]], {n, 6}];
Map[StringReplace[#, {"1" -> "2", "0" -> "1"}] &, A289675]
(* Robert Price, Sep 26 2019 *)

A289677 a(n) = A289671(n)/2^f(n), where f(n) = 2*floor((n-1)/3) + ((n+2) mod 3) = A004523(n).

Original entry on oeis.org

0, 1, 1, 2, 2, 3, 4, 4, 5, 11, 12, 13, 22, 19, 20, 43, 46, 44, 85, 88, 89, 171, 185, 192, 366, 380, 396, 774, 793, 814, 1586, 1589, 1610, 3136, 3106, 3130, 6123, 6078, 6103, 12088, 12147, 12229, 24283, 24534, 24736, 49040, 49482, 49879, 99031, 99792, 100747, 200444, 201892, 203765, 405931, 408478, 411403

Offset: 1

N. J. A. Sloane, Aug 01 2017

nonn

This is the number of distinct binary words w of length n that eventually cycle under the Post tag system (see A284116, A289670) reduced to take into account the observation made by Don Reble that (if the bits of w are labeled from the left starting at bit 0) bits 1,2,4,5,7,8,... (not a multiple of 3) are "junk DNA" and have no effect on the outcome.

Cf. A284116, A284119, A284121, A289670, A289671, A289672, A289673, A289674, A289675, A289676.

Cf. also A290436-A290441.

from _future_ import division
def A289677(n):
    c, k, r, n2, cs, ts = 0, 1+(n-1)//3, 2**((n-1) % 3), 2**(n-1), set(), set()
    for i in range(2**k):
        j, l = int(bin(i)[2:],8)*r, n2
        traj = set([(l,j)])
        while True:
            if j >= l:
                j = j*16+13
                l *= 2
            else:
                j *= 4
                l //= 2
            if l == 0:
                ts |= traj
                break
            j %= 2*l
            if (l,j) in traj:
                c += 1
                cs |= traj
                break
            if (l,j) in cs:
                c += 1
                break
            if (l,j) in ts:
                break
            traj.add((l,j))
    return c # Chai Wah Wu, Aug 03 2017

Corrected by Don Reble, Aug 01 2017 (there were errors in A289671).

A289672 Consider the Post tag system defined in A284116; a(n) = maximum, taken over all binary words w of length n which terminate in a cycle, of the number of words in the orbit of w.

Original entry on oeis.org

4, 3, 4, 7, 8, 7, 14, 15, 14, 15, 16, 15, 24, 25, 28, 29, 30, 35, 38, 39, 38, 39, 38

Offset: 1

N. J. A. Sloane, Jul 29 2017

The terminating empty word is included in the count.

			For length n=2, there are two words which cycle, 10 and 11:
10 -> 101 -> 1101 -> 11101 -> 011101 -> 10100 -> 001101 -> 10100, which has entered a cycle.

Cf. A284116, A284119, A284121, A289670, A289671, A289673, A289674, A289675.

# Uses procedures f1 and P from A289670.
# Count strings of length n which terminate and which cycle
# Print max length to reach empty word (mx)
mx:=[];
for n from 1 to 11 do
lprint("starting length ",n);
m:=0;
for n1 from 0 to 2^n-1 do
t1:=convert(2^n+n1,base,2); t2:=[seq(t1[i],i=1..n)];
map(x->convert(x,string),t2); t3:=Join(%,""); t4:=P(%);
if t4 <> 999 then if t4>m then m:=t4; fi; fi;
od;
mx:=[op(mx),m];
od:
mx;

a(12)-a(23) from Indranil Ghosh, Jul 30 2017

A289676 a(n) = A289670(n)/2^f(n), where f(n) = 2*floor((n-1)/3) + ((n+2) mod 3).

Original entry on oeis.org

2, 1, 1, 2, 2, 1, 4, 4, 3, 5, 4, 3, 10, 13, 12, 21, 18, 20, 43, 40, 39, 85, 71, 64, 146, 132, 116, 250, 231, 210, 462, 459, 438, 960, 990, 966, 2069, 2114, 2089, 4296, 4237, 4155, 8485, 8234, 8032, 16496, 16054, 15657, 32041, 31280, 30325, 61700, 60252, 58379, 118357, 115810, 112885

Offset: 1

N. J. A. Sloane, Aug 01 2017

nonn

This is the number of distinct binary words w of length n that terminate under the Post tag system (see A284116, A289670) reduced to take into account the observation made by Don Reble that (if the bits of w are labeled from the left starting at bit 0) bits 1,2,4,5,7,8,... (not a multiple of 3) are "junk DNA" and have no effect on the outcome.

Cf. A284116, A284119, A284121, A289670, A289671, A289672, A289673, A289674, A289675, A289677.

Cf. also A290436-A290441.

from _future_ import division
def A289676(n):
    c, k, r, n2, cs, ts = 0, 1+(n-1)//3, 2**((n-1) % 3), 2**(n-1), set(), set()
    for i in range(2**k):
        j, l = int(bin(i)[2:],8)*r, n2
        traj = set([(l,j)])
        while True:
            if j >= l:
                j = j*16+13
                l *= 2
            else:
                j *= 4
                l //= 2
            if l == 0:
                c += 1
                ts |= traj
                break
            j %= 2*l
            if (l,j) in traj:
                cs |= traj
                break
            if (l,j) in cs:
                break
            if (l,j) in ts:
                c += 1
                break
            traj.add((l,j))
    return c # Chai Wah Wu, Aug 03 2017

Corrected by Don Reble, Aug 01 2017 (there were errors in A289670)

A346040 a(n) is 1w' converted to decimal, where the binary word w' is the result of applying Post's tag system {00,1101} to the binary word w, where 1w is n converted to binary (the leftmost 1 acts as a delimiter).

Original entry on oeis.org

1, 1, 5, 2, 2, 13, 13, 4, 4, 4, 4, 29, 29, 29, 29, 8, 12, 8, 12, 8, 12, 8, 12, 45, 61, 45, 61, 45, 61, 45, 61, 16, 20, 24, 28, 16, 20, 24, 28, 16, 20, 24, 28, 16, 20, 24, 28, 77, 93, 109, 125, 77, 93, 109, 125, 77, 93, 109, 125, 77, 93, 109, 125, 32, 36, 40

Offset: 1

Carlos Gómez-Ambrosi, Jul 02 2021

Post's tag system maps a word w over {0,1} to w', where if w begins with 0, w' is obtained by appending 00 to w and deleting the first three letters, or if w begins with 1, w' is obtained by appending 1101 to w and deleting the first three letters.
The empty word is included in the count.
It is an important open question to decide whether there is any word whose orbit grows without limit.
Note that there is a one-to-one correspondence between positive integers and binary words (including the empty word), given by n (decimal) = 1w (binary) -> w.
With alphabet {0,1} replaced by {1,2}, the above correspondence is given by A007931, and a step of the tag system by A289673.
The present sequence allows for looking into Post's tag system "numerically", instead of "combinatorially".

			n = 22 (decimal) = 10110 (binary) = 1w ->
                w = 0110 ->
                    011000 ->
                  w' = 000 ->
                1w' = 1000 (binary) = 8 (decimal) = a(22)
n = 25 (decimal) = 11001 (binary) = 1w ->
                w = 1001 ->
                    10011101 ->
                  w' = 11101 ->
                1w' = 111101 (binary) = 61 (decimal) = a(25)

Carlos Gómez-Ambrosi, Table of n, a(n) for n = 1..10000
Liesbeth De Mol, Tracing unsolvability: a mathematical, historical and philosophical analysis with a special focus on tag systems, Ph.D. Thesis, University of Ghent, 2007. See also.
Emil L. Post, Formal reductions of the general combinatorial decision problem, Amer. J. Math. 65 (1943), 197-215. See also. Post's tag system {00,1101} appears on page 204.

Cf. A289673 (alphabet 1,2).

Cf. A284116, A284119, A284121, A289670, A289671, A289672, A289674, A289675, A291792, A291793, A291794, A291795, A291796, A291798, A291799, A291800, A291801, A291802, A337537.

function m = A346040(n)
if n == 1
    m = 1;
else
    s = dec2bin(n);
    if strcmp(s(2),'0')
        t = [s '00'];
    else
        t = [s '1101'];
    end
    t(2) = [];
    t(2) = [];
    t(2) = [];
    m = bin2dec(t);
end
end

a(n) = if(n==1,1, my(k=logint(n,2)); if(bittest(n,k-1), n=n<<4+13;k++, n<<=2;k--); bitand(n,bitneg(0,k)) + 1<Kevin Ryde, Jul 02 2021

def a(n):
    if n == 1:
        return 1
    else:
        s = n.digits(2)
        s.reverse()
        if s[1] == 0:
            t = s + [0,0]
        else:
            t = s + [1,1,0,1]
        del(t[1])
        del(t[1])
        del(t[1])
        return sum(t[k]*2^(len(t)-1-k) for k in srange(0,len(t)))

a(n) = delete(append(n)), where:
append(1) = 1;
append(n) = 2^(2 + 2 * floor((n - 2^k)/2^(k-1))) * n + 13 * floor((n - 2^k)/2^(k-1)) if n > 1, where k = floor(log_2(n));
delete(n) = n + 2^t * (1 - floor(n/2^t)), where t = max(floor(log_2(n))-3,0).
In the expression for append(n), floor((n - 2^k)/2^(k-1)) is the second-highest bit in the binary expansion of n, which is A079944, with offset 2.

cp's OEIS Frontend

A284116 a(n) = largest number of distinct words arising in Post's tag system {00, 1101} applied to a binary word w, over all starting words w of length n, or a(n) = -1 if there is a word w with an unbounded trajectory.

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A289670 Consider the Post tag system defined in A284116; a(n) = number of binary words of length n which terminate at the empty word.

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A289671 Consider the Post tag system defined in A284116; a(n) = number of binary words of length n which terminate in a cycle.

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A289673 Take n-th string over {1,2} in lexicographic order and apply the Post tag system described in A284116 (but adapted to the alphabet {1,2}) just once.

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A289675 Consider the Post tag system described in A284116 (but adapted to the alphabet {1,2}) ; sequence lists the words that terminate in the empty word.

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A289677 a(n) = A289671(n)/2^f(n), where f(n) = 2*floor((n-1)/3) + ((n+2) mod 3) = A004523(n).

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A289672 Consider the Post tag system defined in A284116; a(n) = maximum, taken over all binary words w of length n which terminate in a cycle, of the number of words in the orbit of w.

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A289676 a(n) = A289670(n)/2^f(n), where f(n) = 2*floor((n-1)/3) + ((n+2) mod 3).

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