A340407 a(n) gives the number of side-branches that will be passed, if A342369 is used to trace the Collatz tree backward starting at 6*n-2 with n > 0.

2, 5, 1, 4, 5, 1, 4, 2, 1, 2, 3, 1, 5, 5, 1, 6, 2, 1, 2, 5, 1, 3, 3, 1, 3, 2, 1, 2, 4, 1, 6, 10, 1, 5, 2, 1, 2, 3, 1, 5, 8, 1, 4, 2, 1, 2, 4, 1, 3, 3, 1, 3, 2, 1, 2, 18, 1, 5, 4, 1, 6, 2, 1, 2, 3, 1, 4, 4, 1, 5, 2, 1, 2, 7, 1, 3, 3, 1, 3, 2, 1, 2, 7, 1, 4, 9, 1, 4, 2, 1

Offset: 1

Author

Thomas Scheuerle, Mar 24 2021

Keywords

• nonn

Comments

Recursion into A342369 means tracing the Collatz tree backward, starting at k = A342369(6*n-2), then k = A342369(k) until k is divisible by 3. At each A342369(k) = 3*m - 1, a new side-branch is connected which would start at 6*m-2. If A342369(k) reached a value divisible by three no further side-branches will be found.

This sequence is a rearrangement of A087088 such that all values at positions divisible by 3 are unchanged.

Examples

			n = 2:
6*n-2 = 10.
A342369(10) = 20. -> 7*3 - 1 -> A side-branch is connected.
A342369(20) = 13.
A342369(13) = 26. -> 9*3 - 1 -> A side-branch is connected.
A342369(26) = 17. -> 6*3 - 1 -> A side-branch is connected.
A342369(17) = 11. -> 4*3 - 1 -> A side-branch is connected.
A342369(11) = 7.
A342369(7) = 14. -> 5*3 - 1 -> A side-branch is connected.
A342369(14) = 9. -> divisible by 3 we stop here.
-> We found 5 connected side-branches, a(2) = 5.
		

Links

• Thomas Scheuerle, Table of n, a(n) for n = 1..5000
• Index entries for sequences related to 3x+1 (or Collatz) problem

Crossrefs

Cf. A014682, A342369, A342261, A087088.

Programs

MATLAB

function a = A340407( max_n )
    for n = 1:max_n
        c = 0;
        s = 6*n -2;
        while mod(s,3) ~= 0
            s = A342369( s );
            if mod(s,3) == 2
                c = c+1;
            end
        end
        a(n) = c;
    end
end
function b = A342369( n )
    if mod(n,3) == 2
        b = (2*n - 1)/3;
    else
        b = 2*n;
    end
end

Formula

a(n) > 0.

a(3*n) = 1.

a(9*n - b) = 2, b = {1, 8} row 2 of A342261. ( a(A056020(n)) = 2 ).

a(27*n - b) = 3, b = {2, 4, 5, 16} row 3 of A342261.

a(81*n - b) = 4, b = {13, 14, 22, 34, 38, 52, 74, 77} row 4 of A342261.

a(3^k*n - b) = k, b = row k of A342261.

( Sum_{k=1..j} a(k) )/j lim_{j->infinity} = 3 = Sum_{k=1..infinity} k*2^(k-1)/3^k.