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).

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

Author

Carlos Gómez-Ambrosi, Jul 02 2021

Keywords

• nonn
• base
• easy

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)
		

Links

• 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.

Crossrefs

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.

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.