cp's OEIS Frontend

a(n) is the number of odd numbers in row A070165(2*n+1) and also in row A070168(2*n+1), for n >= 0 (if it turns out that a row 2*n+1 has infinite row length one puts a(n) = 0).

Coefficients T(m,j) in the array A259663 are least residues in congruence classes T(m,j) mod 2^(m+j). T"(m,j) denotes all members of that class.

Reduced Collatz sequences (i.e., reduced sequences) are standard Collatz sequences excluding even terms. Row n in triangle A256598 shows the reduced sequence starting with 2n+1.

Let every positive odd number z = T"(m,j)_q, where q is the quotient of z in T"(m,j). For example, T(2,3) = 7 in A259663, so T"(2,3) contains all numbers == 7 (mod 32). So z = T"(2,3)_0 = 7, z = T"(2,3)_1 = 39, z = T"(2,3)_2 = 71, etc.

T(n,k) is the q-value of A256598(n,k) (see Example below). Thus, each row n is defined here as a "quotient trajectory" for the reduced sequence with starting term 2n+1.

n-th row has A008908(n) entries (unless some n never reaches 1, in which case the triangle ends with an infinite row). [Escape clause added by N. J. A. Sloane, Jun 06 2017]

A216059(n) is the smallest number not occurring in n-th row; see also A216022.

Comment on the mp3 file from Gordon Charlton (the recording artist Beat Frequency). The piece uses the first 3242 terms (i.e. the first 100 hailstone sequences), with pitch modulus 36, duration modulus 2. Its musicality stems from the many repetitions and symmetries within the sequence, and in particular the infrequency of multiples of 3. This means that when the pitch modulus is a multiple of 12 the notes are predominantly in the symmetric octatonic scale, known to modern classical composers as the second of Messiaen's modes of limited transposition, and to jazz musicians as half-whole diminished. - N. J. A. Sloane, Jan 30 2019

The even-indexed terms a(2i+2) = 6i+5 = A016969(i), i >= 0 [Comment corrected by Bob Selcoe, Apr 06 2015]. The odd-indexed terms are the same as A067745. Note that this sequence is A016789 with all factors of 2 removed from each term. Also note that a(4i-1) = a(i). No multiple of 3 is in this sequence. See A075680 for the number of iterations of R required to yield 1.

From Bob Selcoe, Apr 06 2015: (Start)

All numbers in this sequence appear infinitely often.

From Eq. 1 and Eq. 2 in Formulas: Eq. 1 is used with 1/3 of the numbers in this sequence, Eq. 2 is used with 2/3 of the numbers.

(End)

Empirical: For arbitrary m, Sum_{n=2..A007583(m)} (a(n) - a(n-1)) = 0. - Fred Daniel Kline, Nov 23 2015

From Wolfdieter Lang, Dec 07 2021: (Start)

Only positive numbers congruent to 1 or 5 modulo 6 appear.

i) For the sequence entry with value A016921(m), for m >= 0, that is, a value from {1, 7, 13, ...}, the indices n are given by the row of array A178415(2*m+1, k), for k >= 1.

ii) For the sequence entry with value A007528(m), for m >= 1, that is, a value from {5, 11, 17, ...}, the indices n are given by the row of array A178415(2*m, k), for k >= 1.

See also the array A347834 with permuted row numbers and columns k >= 0. (End)

For the definition of this array A see the formula section.

The rows of A appear in a draft by Immmo O. Kerner in eqs. (1) and (2) as so-called horizontal sequences (horizontale Folgen). Thanks to Dr. A. Eckert for sending me this paper.

This array with entry A(k, n) becomes equal to the array T with T(n, k) given in A178415 by using a permutation of the rows, and changing the offset: A(k, n) = T(pe(k), n+1), with pe(3*(L+1)) = 4*(L+1), pe(1+3*L) = 1 + 2*L, pe(2+3*L) = 2*(1 + 2*L), for L >= 0. This permutation appears in A265667.

A proper sub-array is A238475(n, k) = A(1 + 3*(k-1), n-1), for k >= 1 and n >= 1.

In the directed Collatz tree with nodes labeled with only positive odd numbers (see A256598 for the paths), here called CTodd, the level L = 0 (on the top) has the node with label 1 as root. Because 1 -> 1 there is an arrow (a 1-cycle or loop) at the root. The level L = 1 consists of the nodes with labels A(1, n), for n >= 1, and each node is connected to 1 by a downwards directed arrow. The next levels for L >= 2 are obtained using the successor rule (used also by Kerner): S(u) = (4*u - 1)/3 if u == 1 (mod 3), (2*u - 1)/3 if u == 5 (mod 3), and there is no successor S(u) = empty if u = 3 (mod 6), that is, this node is a leaf.

However, each node with label u on level L >= 1, except a leaf, has as successors at level L + 1 not only the node with S(u) but all the nodes with labels A(S(u), n), for n >= 0.

In this way each node (also the root) of this CTodd has in-degree 1 and infinite out-degree (for L >= 2 there are infinitely many infinite outgoing arrows). All nodes with label A(k, n) with n >= 1, have the same precursor as the node A(k,0) in this tree for each k >= 1.

Except for the loop (1-cycle) for the root 1 there are no cycles in this directed tree CTodd.

That each number N = 5 + 8*K, for K >= 0 appears in array A for some column n >= 1 uniquely can be proved, using the fact of strictly increasing rows and columns, by showing that the columns n = 1, 2, ..., c contain all positive integers congruent to 5 modulo 8 except those of the positive congruence class A(1, c+1) modulo 2^(2*c+3) by induction on c. [added Dec 05 2021]

Row index k for numbers congruent to 5 modulo 8: Each number N = 5 + 8*K, for K >= 0, from A004770 is a member of row k of the array A starting with element A(k, 0) = (2*A065883(2 + 3*N) - 1)/3. For this surjective map see A347840. [simplified Dec 05 2021]

The Collatz conjecture can be reduced to the conjecture that in this rooted and directed tree CTodd each positive odd number appears as a label once, that is, all entries of the array A appear.

In the following, let F^(k)(x) denote k-fold iteration of F and defined by the recurrence F^(k)(x) = F(F^(k-1)(x)), k > 0, with initial condition F^(0)(x) = x, and let S^(k)(n) denote k-fold iteration of S and defined by the recurrence S^(k)(n) = S(S^(k-1)(n)), k > 0, with initial condition S^(0)(n) = n, where F and S are as defined above.

Theorem 1: For each x, there exists a j>0 such that F^(j)(x) == 1 (mod 4).

Theorem 2: S(n) = m if and only if S(4*n-2) = m.

Conjecture 1: For each n, there exists a k such that S^(k)(n) = 1.

Theorem 3: Conjecture 1 is equivalent to the 3x+1 conjecture.

Theorem 4: The sequence {log(S(n))/log(n)}_{n>1} is bounded with least upper bound equal to log(3)/log(2).

[I have proved Theorems 1--4 (along with several lemmas) and am trying to finish typesetting the draft containing the proofs but had been too ill to finish that work until now. The draft also contains the derivation of the function S from properties of the known function F (A075677). When that paper is completed (hopefully within two weeks) I will then upload it to the links section and delete this comment.]

Collatz conjecture is equal to the claim that in each column 1 will eventually appear. See also arrays A372287 and A372288.

Definitions: Let v(y) denote the 2-adic valuation of y. Let N_1 denote the set of odd natural numbers. Let F : N_1 -> N_1 be the map defined by F(x) = (3*x + 1)/2^v(3*x + 1) (cf. A075677). Let F^(k)(x) denote k-fold iteration of F and defined by the recurrence F^(k)(x) = F(F^(k-1)(x)), k>0, with initial condition F^(0)(x) = x.

This triangle can be constructed by restricting the initial values to the numbers 4*n - 3, iterating F until 1 is reached (assuming the 3x+1 conjecture) and removing all iterates not congruent to 1 modulo 4. Equivalently, for each n, this is accomplished by iterating (until 1 is reached, assuming the 3x+1 conjecture) the function S defined in A257480 to get the triangle A253676, and finally taking T(n,k) = 4*A253676(n,k) - 3.

Conjecture: For each natural number n, there exists a k >= 0, such that F^k(4*n - 3) = 1.

Theorem 1: Conjecture 1 is equivalent to the 3x+1 (or Collatz) conjecture.

Proof: See A257480.