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If the Fibonacci pairs are kept in the natural order (F(n),F(n+1)), it appears that the first term of the pair occurs in A002487 at the index given by A061547(n).

Equals row sums of triangle A177954. - Gary W. Adamson, May 15 2010

Starting at n=3, begin subtracting from (2^(n-1)-1)/2^(n-1): 3/4 - 1/2 = 1/4 with 1+4=5=a(3); 7/8 - 1/4 = 5/8 with 5+8=13=a(4); 15/16 - 5/8 = 5/16 with 5+16=21= a(5); 31/32 - 5/16 = 21/32 with 21+32=53=a(6); 63/64 - 21/32 = 21/64 with 21+64=85=a(7) and so on. For n odd in the first fraction (2^(n-1)-1)/2^(n-1), the result approaches 1/3, and for n even in the first fraction, the result approaches 2/3. - J. M. Bergot, May 08 2015

Also, the decimal representation of the x-axis, from the left edge to the origin, of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 678", based on the 5-celled von Neumann neighborhood, initialized with a single black (ON) cell at stage zero. See A283641. - Robert Price, Mar 12 2017

Construction: take the binary expansion of 3n+1 (A016777(n)), and substitute "01" for all trailing 0-bits that follow after its odd part (= A067745(1+n)), of which there are A371093(n) in total. See the examples.

Sequence is a permutation of the odd natural numbers.

Let N_1 denote the set of odd natural numbers, and let |y|_2 denote 2-adic valuation of y. Define the map F : N_1 -> N_1 by F(x) = (3*x + 1)/2^|3*x+1|_2 (cf. A075677). Then row n of A is the set of all x in N_1 for which |3*x + 1|_2 = A371093(x) = n. Hence F(A(n,k)) = 6*k - 3 - 2*(-1)^n.

The drib tree is an infinite binary tree labeled with rational numbers. It is generated by the following iterative process: start with the rational 1; for the left subtree increment and then take the reciprocal of the current rational; for the right subtree interchange the order of the two steps: take the reciprocal and then increment. Like the Stern-Brocot and the Bird tree, the drib tree enumerates the positive rationals: A162911(n)/A162912(n).

From Yosu Yurramendi, Jul 11 2014: (Start)

If the terms (n>0) are written as an array (left-aligned fashion) with rows of length 2^m, m = 0,1,2,3,...

1,

2, 1,

3, 1, 3,2,

5, 2, 4,3,4,1, 5,3,

8, 3, 7,5,5,1, 7,4,7,3,5,4, 7,2, 8,5,

13,5,11,8,9,2,12,7,9,4,6,5,10,3,11,7,11,4,10,7,6,1,9,5,12,5,9,7,11,3,13,8,

then the sum of the m-th row is 3^m (m = 0,1,2,), and each column k is a Fibonacci sequence (a(2^(m+2)+k) = a(2^(m+1)+k) + a(2^m+k), m = 0,1,2,..., k = 0,1,2,...,2^m-1).

If the rows are written in a right-aligned fashion:

1,

2,1,

3,1, 3,2,

5,2,4,3, 4,1, 5,3,

8,3, 7,5,5,1,7,4, 7,3,5,4, 7,2, 8,5,

13,5,11,8,9,2,12,7,9,4,6,5,10,3,11,7,11,4,10,7,6,1,9,5,12,5,9,7,11,3,13,8,

then each column k also is a Fibonacci sequence.

If the sequence is considered by blocks of length 2^m, m = 0,1,2,..., the blocks of this sequence are the reverses of blocks of A162911 ( a(2^m+k) = A162911(2^(m+1)-1-k), m = 0,1,2,..., k = 0,1,2,...,2^m-1). (End)

From Bob Selcoe, Jun 03 2015: (Start)

For k >= 1: all numbers congruent to A002450(k) mod 2^(2k+1) and A072197(k) mod 4^(k+1) not congruent to 0 mod 3. Equivalently, for k >= 3: all numbers congruent to A096773(k) mod 2^k not congruent to 0 mod 3.

Conjecture: at least one number in this sequence must appear in all Collatz sequences.

(End)

The sequence is composed of all numbers in congruence classes T(n,1) mod 2^(n+k) in A259663 (i.e., T"(1) in array T259663(n,k)) not congruent to 0 mod 3. Therefore the conjecture above is true (see A259663 for additional explanation). - Bob Selcoe, Jul 15 2017

Closure of {1} under the map (x,y)->2x+3y [Klarner-Rado, see Lagarias (2016), p. 755]. - N. J. A. Sloane, Oct 06 2016

The above conjecture is true: this is because even numbers and odd numbers divisible by 3 will lead to the set of odd numbers not divisible by 3. Odd numbers of the form 4k - 1 can also be ignored, as this consists of odd numbers that grow between themselves and the next odd term through Collatz iteration. No infinite sequence of growth between consecutive odd terms is possible, so all numbers of the form 4k - 1 will lead to an odd number that shrinks between itself and the next odd number. All numbers 4k - 1 will lead to a number in 4k - 3, the odd numbers that shrink between themselves and the following odd term. What we are left after that elimination is this sequence. - Aidan Simmons, Feb 25 2019

a(n) is the number of nonzero terms of A096773 that are found in the interval [n, A371094(n)], inclusive.

Zeros occur in the same locations where 1's occur in array A372287.

The ratio a(n+1)/a(n) approaches 10 for even n and 2/5 for odd n as n->infinity.

Let N_1 be the set of odd natural numbers and v(y) the 2-adic valuation of y. Define the map F : N_1 -> N_1 by F(x) = (3*x+1)/2^v(3*x+1) (see A075677). Let F^(k)(x) denote k-fold iteration of F, with recurrence F^(k)(x) = F(F^(k-1)(x)), k > 0, and initial condition F^(0)(x) = x. Then, for n>0, a(n) is the least m such that F^(n)(4*m-3) == 1 (mod 4). Cf. A257499.

Let k == 1 mod 4, and k(r) be the r-th iteration at which k appears in a Collatz sequence. When n >= 2 and k(r) == [2^(n+1) - a(n)] mod 2^(n+1), then n is the number of halving steps following k(r+1). For instance, since a(5) = 11, there are 5 halving steps following k(r+1) when k(r) == 53 mod 64, because 2^(5+1) = 64 and 64-11 = 53; e.g., k(r) = 117: 117 -> 352 -> 176 -> 88 -> 44 -> 22 -> 11. - Bob Selcoe, Feb 09 2017

If interpreted as a flat sequence a(j), we obtain a(j+1)-2a(j)= -3, 4, -1, -8, 8, -13, 16, -16, 16, -5, -32, 32, -32, 32, -53, 64, ... which is essentially the negative values of A096773 padded by groups of one, then two, then three etc. signed elements of A098354.