cp's OEIS Frontend

Theorem: For all indices n and k such that n + k > 2, log(A(n,k))/log(A257499(n,k)) < log_2(3).

Conjecture: Arranging the sequence in ascending order gives A189707 (positions of 0 in A189706).

Theorem: The sequence contains (i) a subset of equivalence class 0 modulo 4 comprising all numbers congruent to 0 modulo 32 and no others; (ii) no numbers congruent to 1 modulo 4; (iii) a subset of numbers congruent to 2 modulo 4; (iv) all numbers of congruence class 3 modulo 4.

Conjecture: A254312 is a permutation of this sequence.

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

Conjecture (now Lemma 1): The sequence is a permutation of the odd natural numbers.

Proof from Max Alekseyev, Apr 29 2015:

Reformulating the conjecture, we need to prove that for any integer m >= 0, the equation (1 + 2^n*(6*k - 3 + 2*(-1)^n))/3 = 2*m + 1 has a unique solution in integers n,k >= 1. Simplifying a bit, we have

(1) 2^n*(6*k - 3 + 2*(-1)^n) = 6*m + 2.

Since the factor (6*k - 3 + 2*(-1)^n) is odd, n is uniquely defined by n = A007814(6*m+2). Since 6*m+2 is even, we have n>=1. Dividing (1) by 2^n and rearranging, we further get

(2) 6*k = (6*m + 2)/2^n + 3 - 2*(-1)^n.

To prove the uniqueness of k, it remains to prove that the r.h.s of (2) is divisible by 6. To that end, the value of n implies that (6*m + 2)/2^n is odd; hence the r.h.s. of (2) is even and thus divisible by 2. Now, taking the r.h.s. modulo 3, we get

(6*m + 2)/2^n + 3 - 2*(-1)^n == 2/(-1)^n + 0 - 2*(-1)^n == 0 (mod 3);

so the r.h.s. of (2) is also divisible by 3. Therefore k is uniquely defined by

k = ((6*m + 2)/2^n + 3 - 2*(-1)^n)/6.

Finally, it is easy to see that (6*m + 2)/2^n >= 1, so k >= 1.

QED

Let v(y) denote the 2-adic valuation of y (see A007814). For x an odd natural number, define the function F(x) = (3*x+1)/2^v(3*x+1) (see A075677). Let F^(j)(x) denote k-fold iteration of F and defined by the recurrence F^(j)(x) = F(F^(j-1)(x)), j>0, with initial condition F^(0)(x) = x.

Lemma 2: The following statements are equivalent. (i) Row n of A is the set of all odd m such that F^(n)(4*m-3) == 1 (mod 4); (ii) Row n of A is the set of all odd m such that v(1+F(4m-3)) = n.

a(n) is the first successor in the 3x+1 trajectory of 4*n-3 that is congruent to 1 mod 4. - Ruud H.G. van Tol, Jul 16 2023

Conjecture: The array A contains without duplication all natural numbers m such that m < S(m), where the function S is as defined in A257480; i.e., the sequence is a permutation of A254311.

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.

The sequence is a permutation of the natural numbers.

Theorem: Let v(y) denote the 2-adic valuation of y. For x an odd natural number, let F(x) = (3*x+1)/2^v(3*x+1) (see A075677). Row n of A is the set of all natural numbers m such that v(1+F(4*(2*m-1)-3)) = n.

This is a fractal sequence: removing all entries a(n) with indices n == 0,1 or 3 (mod 4) and reindexing yields the original sequence (see Thm 1 (iii)). This sequence also contains A001511 (the ruler sequence) as a subsequence (see Thm 1 (i)).

THEOREM 1. For all natural numbers n, the following hold: (i) a(4*n) = A001511(n); (ii) a(2*n-1) = a(12*n-8) = A001511(3*n-2); (iii) a(4*n-2) = a(n).

Proof. Let n be a natural number. For part (i), we have F(4*4*n-3) = (3*(16*n-3)+1)/2^v(3*(16*n-3)+1) = (48*n-8)/2^v(48*n-8) = 6*n-1, hence a(4*n) = v(1 + (6*n-1)) = v(6*n) = v(3*2*n) = v(2*n) = A001511(n); for part (ii), v(1+F(4*(2*n-1)-3)) = v(1+(24*n-20)/2^v(24*n-20)) = v(1+(6*n-5)) = v(6*n-4) and, similarly, v(1 + F(4*(12*n-8)-3)) = v(1+(144*n-104)/2^v(144*n-104)) = v(6*n-4), so a(2*n-1) = a(12*n-8), as claimed, and finally note that v(6*n-4) = v(2*(3*n-2)) = A001511(3*n-2); for part (iii), the claim follows from the fact that F(4*(4*n-2)-3) = (48*n-32)/2^v(48*n-32) = (3*n-2)/2^v(3*n-2) = F(4*n-3). QED