A340407 -id:A340407 - cp's OEIS frontend

A342261 Irregular triangular array T(n,k) = m read by rows. Row n lists all solutions m < 3^n, where A340407(3^n*j - m) = n is true for all j > 0, sorted in ascending order.

0, 1, 8, 2, 4, 5, 16, 13, 14, 22, 34, 38, 52, 74, 77, 20, 25, 40, 50, 88, 130, 146, 173, 185, 203, 209, 223, 229, 230, 238, 241, 130, 146, 173, 185, 203, 209, 223, 229, 230, 238, 241, 41, 61, 76, 104, 106, 121, 128, 157, 254, 266, 292, 311, 403, 412, 430, 445, 454, 493

Offset: 1

Thomas Scheuerle, Mar 26 2021

Each row n has 2^(n-1) values.
In all rows other than the first row of T(n,k), there are exactly 2^(n-2) numbers of the form 3*p + 1 and the same number of numbers of the form 3*q - 1.
Each integer has a unique representation of the form 3^n*j - T(n,k).

			Triangle T(n,k) begins:
  0;
  1,   8;
  2,   4,  5, 16;
  13, 14, 22, 34, 38, 52, 74, 77;

Cf. A340407.

function t = A342261 (max_row)
    maxtest = 10;
    d = A340407(maxtest*3^max_row);
    for row = 1:max_row
        m = 0;
        for k = 1:2^(row-1)
            test = d((1:maxtest)*(3^row)-m);
            while ~all(test == test(1))||(test(1) ~= row)
                m = m+1;
                test = d((1:maxtest)*(3^row)-m);
            end
            t(row,k) = m;
            t = t+1;
        end
    end
end
function d = A340407 (max_p)
    for p = 1:max_p
        s = 6*p -2;
        c = 0;
        while mod(s,3) ~= 0
            s = A342369( s );
           if mod(s,3) == 2
                c = c+1;
            end
        end
        d(p) = c;
    end
end
function b = A342369( n )
    if mod(n,3) == 2
        b = (2*n - 1)/3;
    else
        b = 2*n;
    end
end

A189637 Positions of 1 in A116178; complement of A189636.

Original entry on oeis.org

3, 6, 8, 9, 12, 15, 17, 18, 21, 23, 24, 26, 27, 30, 33, 35, 36, 39, 42, 44, 45, 48, 50, 51, 53, 54, 57, 60, 62, 63, 66, 68, 69, 71, 72, 75, 77, 78, 80, 81, 84, 87, 89, 90, 93, 96, 98, 99, 102, 104, 105, 107, 108, 111, 114, 116, 117, 120, 123, 125, 126, 129, 131, 132, 134, 135, 138, 141, 143, 144, 147, 149, 150, 152, 153, 156, 158, 159, 161, 162, 165, 168, 170, 171, 174

Offset: 1

Clark Kimberling, Apr 24 2011

nonn

See A116178.

Michael De Vlieger, Table of n, a(n) for n = 1..10000
F. M. Dekking, Permutations of N generated by left-right filling algorithms, arXiv:2001.08915 [math.CO], 2020.

Cf. A116178, A189636, A189638, A254046, A340407.

t = Nest[Flatten[# /. {0->{0,0,1}, 1->{0,1,1}}] &, {0}, 5] (* A116178 *)
f[n_] := t[[n]]
Flatten[Position[t, 0]] (* A189636 *)
Flatten[Position[t, 1]] (* A189637 *)
s[n_] := Sum[f[i], {i, 1, n}]; s[0] = 0;
Table[s[n], {n, 1, 120}] (* A189638 *)

A340407(a(n)) = A254046(a(n)). Discovered with help from sequencedb.net. - Thomas Scheuerle, Jun 20 2022

A342842 All positive integer solutions m of equation A342369^k(6p - 2) = m3 + 2, sorted by p and k in ascending order, p has higher priority than k. p and k are positive integers. "^k" means recursion here.

Original entry on oeis.org

2, 1, 6, 8, 5, 3, 4, 10, 14, 9, 12, 16, 18, 24, 32, 21, 28, 22, 26, 17, 11, 7, 30, 40, 34, 38, 25, 42, 56, 37, 46, 50, 33, 44, 29, 19, 54, 72, 96, 128, 85, 58, 62, 41, 27, 36, 48, 64, 66, 88, 70, 74, 49, 78, 104, 69, 92, 61, 82, 86, 57, 76, 90, 120, 160, 94, 98, 65, 43

Offset: 1

Thomas Scheuerle, Mar 24 2021

nonn

It is conjectured that this sequence is a permutation of the positive integers. If it does not contain all positive integers, then there exists a number of the form q = p*6 - 2, where no solution for j*3 - 1 = A006370^k(q) can be found for any j and any k. Such an example is not yet known.
If the sequence were to contain a positive integer more than once, this would mean that A340407 contains a term of uncountable size, which is not the case.
Let us assume here that this sequence is a permutation, then let a'(m) be the inverse permutation, such that a'(a(n)) = n.
Let p = A006370^k(6*(a(n) + 1) - 2) and choose k such that p is of the form m*6 + 4, then a'((p + 2)/6 - 1) < n.
Infinitely many formulas can be developed from this template: a(Sum_{k=1..3^d*n - b} A340407(k) + c) = e*n - f. c is here in the range 0 to d-1 if d-1 > 0. b can be any element of row d in A342261. For all combinations of d, b and c we may find a suitable e and f.

Index entries for sequences related to 3x+1 (or Collatz) problem

Cf. A342369, A340407, A006370, A342261.

function a = A342842( max_p )
    c = 1;
    for p = 1:max_p
        s = 6*p -2;
        while mod(s,3) ~= 0
            s = A342369( s );
            if mod(s,3) == 2
                a(c) = (s-2)/3;
                c = c+1;
            end
        end
    end
end
function b = A342369( n )
    if mod(n,3) == 2
        b = (2*n - 1)/3;
    else
        b = 2*n;
    end
end

a(1 + Sum_{k=1..n-1} A340407(k)) = 4*n-2.
a(Sum_{k=1..9*n-8} A340407(k)) = 24*n-23.
a(Sum_{k=1..9*n-1} A340407(k)) = 48*n-8.
a(n) = 8*(10^m - 1)/3 + 1 if n = Sum_{k=1..10^m} A340407(k).
a(n) = 4*10^m - 2 if n = -1 + Sum_{k=1..10^m} A340407(k).
a(n) = 4*10^m - 6 if n = -2 + Sum_{k=1..10^m} A340407(k).
a(n) = 5*10^m + (10^(n - 1) - 1)/3 - 13
if n = -3 + Sum_{k=1..10^m} A340407(k).
a(n) = 4*10^m - 10 if n = -4 + Sum_{k=1..10^m} A340407(k).

cp's OEIS Frontend

A342261 Irregular triangular array T(n,k) = m read by rows. Row n lists all solutions m < 3^n, where A340407(3^n*j - m) = n is true for all j > 0, sorted in ascending order.

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A189637 Positions of 1 in A116178; complement of A189636.

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A342842 All positive integer solutions m of equation A342369^k(6p - 2) = m3 + 2, sorted by p and k in ascending order, p has higher priority than k. p and k are positive integers. "^k" means recursion here.

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cp's OEIS Frontend

A342261 Irregular triangular array T(n,k) = m read by rows. Row n lists all solutions m < 3^n, where A340407(3^n*j - m) = n is true for all j > 0, sorted in ascending order.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

MATLAB

A189637 Positions of 1 in A116178; complement of A189636.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

A342842 All positive integer solutions m of equation A342369^k(6*p - 2) = m*3 + 2, sorted by p and k in ascending order, p has higher priority than k. p and k are positive integers. "^k" means recursion here.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

MATLAB

Formula

A342842 All positive integer solutions m of equation A342369^k(6p - 2) = m3 + 2, sorted by p and k in ascending order, p has higher priority than k. p and k are positive integers. "^k" means recursion here.