cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Previous Showing 21-29 of 29 results.

A336327 Period of orbit of Post's tag system ({0,1},{(0,0),(1,01101)},3,100^n).

Original entry on oeis.org

0, 4, 0, 4450, 0, 4450, 0, 0, 0, 4450, 0, 0, 6, 0, 0, 4450, 0, 0, 0, 0, 910, 4450, 0, 4450, 910, 4450, 0, 4450, 0, 4450, 910, 0, 0, 4450, 0, 4450, 910, 4, 0, 4, 0, 0, 910, 0, 6, 0, 910, 0, 0, 4450, 0, 4450, 910, 4450, 0, 292, 0, 4450, 0, 0, 910, 4450, 6, 4450
Offset: 1

Views

Author

A.H.M. Smeets, Jul 17 2020

Keywords

Comments

In general a tag as defined by Emil Leon Post, is given by a 4-tuple (Sigma,AF,n,w0), where Sigma is some (nonempty) set of symbols called the alphabet, AF is the associated function (sometimes also called set of production rules) AF: Sigma -> Sigma*, n is the deletion number and w0 the initial string.
From the starting sequence we obtain a new string in each step by adjoining the string associated to the prefix symbol of the string, where after the prefix n symbols are removed from the string.
The decision problem is: will the tag end up in an empty string, a(n) = 0 or not, a(n) <> 0?
a(n) is an even number. Proof: for each cycle the number of associations (productions) 0 -> 0 must equal the number of associations (productions) 1 -> 01101 applied within a cycle.
For n=85 the tag is hard to solve by a brute force method, similar to the tag for n=110 with associated function {(0,00),(1,1101)} as reported in A284119.
Also period of orbit of Post's tag system ({0,1},{(0,0),(0,11010)},3,100^(n-1)).

Crossrefs

Programs

  • Python
    def step(w):
        i = 0
        while w[0] != alfabet[i]:
            i = i+1
        w = w+suffix[i]
        return w[n:len(w)]
    alfabet, suffix, n, ws, w0, m = "01", ["0", "01101"], 3, "100", "", 0
    while m >= 0:
        w0, m = w0+ws, m+1
        w, ww, i, a = w0, w0, 0, 0
        while w != "" and a == 0:
            w, i = step(w), i+1
            if i%100000 == 0:
                ww = w
            else:
                if w == ww or w == "":
                    if w != "":
                        a = i%100000
                    print(m, a)

Formula

Observed: if n is even then a(n) in {0, 4, 292, 4450}, if n is odd then a(n) in {0, 6, 910}.

A291798 Take n-th word over {1,2} listed in A291797 and apply the Post tag system described in A284116 (but adapted to the alphabet {1,2}); a(n) = preperiod (or threshold) of orbit of that word.

Original entry on oeis.org

3, 4, 6, 2, 17, 15, 9, 11, 13, 14, 16, 12, 24, 10, 12, 19, 14, 23, 21, 4, 13, 9, 25, 11, 29, 27, 3, 7, 420, 25, 15, 24, 28, 6, 17, 19, 28, 4, 26, 21, 23, 419, 6, 417, 46, 24, 8, 5, 421, 26, 28, 415, 12, 413, 10, 44, 27, 10, 40, 22, 84, 411, 18, 20, 29, 9, 11
Offset: 1

Views

Author

N. J. A. Sloane, Sep 04 2017

Keywords

Comments

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.
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.
This is an analog of A284119 for the words in A291797.

Crossrefs

Extensions

a(31)-a(67) from Lars Blomberg, Sep 08 2017

A302202 a(n) = number of distinct words arising in Post's tag system {00, 1101} applied to the word (100)^n , or a(n) = -1 if this word has an unbounded trajectory.

Original entry on oeis.org

6, 21, 16, 31, 412, 57, 2156, 859, 382, 2811, 372, 2609, 704, 37913, 618, 155, 1008, 2407, 1210, 629, 5286, 1779, 1468, 4346275, 4130, 3247, 7024, 3891, 14638, 7025, 4570, 4329, 147694, 1863, 11126, 81147, 20210, 3853, 116020, 7641, 6494
Offset: 1

Views

Author

Lars Blomberg and N. J. A. Sloane, Apr 19 2018

Keywords

Comments

a(n) = A284119(n) + A284121(n) (or -1 if the trajectory is unbounded).
a(n) is a lower bound on A284116(3n) (assuming that there is no starting string with an unbounded trajectory).
On October 30 2020, a student in my CS 360 class, Zhiping Cai, found that for n = 70051, 96938660265781 (96.9 trillion) steps are needed. See https://github.com/zcai1/posts-problem . I am not sure this has been independently verified. - Jeffrey Shallit, Sep 13 2021

Crossrefs

A337537 Period of orbit of Post's tag system ({0,1},{(0,0101100),(1,11000111100000)},10,(1+0^9)^n).

Original entry on oeis.org

7, 7, 7, 7, 7, 308, 7, 308, 308, 112, 308, 308, 140, 308, 140, 3429251, 140, 308, 140, 802613, 3429251, 140, 140, 3429251, 802613, 3429251, 3429251, 3429251, 3429251, 3429251, 140, 140, 802613, 3429251, 802613, 802613, 140, 802613, 140, 802613, 802613, 3429251
Offset: 1

Views

Author

A.H.M. Smeets, Aug 31 2020

Keywords

Comments

In general a tag as defined by Emil Leon Post, is given by a 4-tuple (Sigma,AF,n,w0), where Sigma is some (nonempty) set of symbols called the alphabet, AF is the associated function (sometimes also called set of production rules) AF: Sigma -> Sigma*, n is the deletion number and w0 the initial string.
From the starting sequence we obtain a new string in each step by adjoining the string associated to the prefix symbol of the string, where after the prefix n symbols are removed from the string.
The decision problem is: will the tag end up in an empty string, a(n) = 0 or not, a(n) <> 0?
This tag system was proposed by Liesbeth De Mol (p. 329).
a(n) == 0 (mod 7). Proof: for each cycle four times the number of associations (productions) 0 -> 0101100 must equal three times the number of associations (productions) 1 -> 11000111100000 applied within a cycle.

Crossrefs

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

Views

Author

Carlos Gómez-Ambrosi, Jul 02 2021

Keywords

Comments

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".

Examples

			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)
		

Crossrefs

Programs

  • MATLAB
    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
    
  • PARI
    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
  • Sage
    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)))
    

Formula

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.

A290437 a(n) = A289676(3*n+2).

Original entry on oeis.org

1, 2, 4, 4, 13, 18, 40, 71, 132, 231, 459, 990, 2114, 4237, 8234, 16054, 31280, 60252, 115810
Offset: 0

Views

Author

N. J. A. Sloane, Aug 02 2017

Keywords

Comments

No formulas or recurrences are known for the important sequences A289670 and A289671. The essence of these two sequences is captured in the six entries A290436-A290441. Any numerical properties of these would be most welcome.

Crossrefs

A290438 a(n) = A289676(3*n).

Original entry on oeis.org

1, 1, 3, 3, 12, 20, 39, 64, 116, 210, 438, 966, 2089, 4155, 8032, 15657, 30325, 58379, 112885
Offset: 1

Views

Author

N. J. A. Sloane, Aug 02 2017

Keywords

Comments

No formulas or recurrences are known for the important sequences A289670 and A289671. The essence of these two sequences is captured in the six entries A290436-A290441. Any numerical properties of these would be most welcome.

Crossrefs

A290439 a(n) = A289677(3*n+1).

Original entry on oeis.org

0, 2, 4, 11, 22, 43, 85, 171, 366, 774, 1586, 3136, 6123, 12088, 24283, 49040, 99031, 200444, 405931
Offset: 0

Views

Author

N. J. A. Sloane, Aug 02 2017

Keywords

Comments

No formulas or recurrences are known for the important sequences A289670 and A289671. The essence of these two sequences is captured in the six entries A290436-A290441. Any numerical properties of these would be most welcome.

Crossrefs

A290440 a(n) = A289677(3*n+2).

Original entry on oeis.org

1, 2, 4, 12, 19, 46, 88, 185, 380, 793, 1589, 3106, 6078, 12147, 24534, 49482, 99792, 201892, 408478
Offset: 0

Views

Author

N. J. A. Sloane, Aug 02 2017

Keywords

Comments

No formulas or recurrences are known for the important sequences A289670 and A289671. The essence of these two sequences is captured in the six entries A290436-A290441. Any numerical properties of these would be most welcome.

Crossrefs

Previous Showing 21-29 of 29 results.