cp's OEIS Frontend

Let u(k), v(k), w(k) be the 3 sequences defined by u(1)=1, v(1)=0, w(1)=0 and u(k+1)=u(k)+v(k)-w(k), v(k+1)=u(k)-v(k)+w(k), w(k+1)=-u(k)+v(k)+w(k); let M(k)=Max(u(k),v(k),w(k)); then a(n)=M(2n)=M(2n-1). - Benoit Cloitre, Mar 25 2002

Also the number of words of length 2n generated by the two letters s and t that reduce to the identity 1 by using the relations ssssss=1, tt=1 and stst=1. The generators s and t along with the three relations generate the dihedral group D6=C2xD3. - Jamaine Paddyfoot (jay_paddyfoot(AT)hotmail.com) and John W. Layman, Jul 08 2002

Binomial transform of A025192. - Paul Barry, Apr 11 2003

Number of walks of length 2n+1 between two adjacent vertices in the cycle graph C_6. Example: a(1)=3 because in the cycle ABCDEF we have three walks of length 3 between A and B: ABAB, ABCB and AFAB. - Emeric Deutsch, Apr 01 2004

Numbers of the form 1 + Sum_{i=1..m} 2^(2*i-1). - Artur Jasinski, Feb 09 2007

Prime numbers of the form 1+Sum[2^(2n-1)] are in A000979. Numbers x such that 1+Sum[2^(2n-1)] is prime for n=1,2,...,x is A127936. - Artur Jasinski, Feb 09 2007

Related to A024493(6n+1), A131708(6n+3), A024495(6n+5). - Paul Curtz, Mar 27 2008

Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=-6, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n-1)=(-1)^(n-1)*charpoly(A,2). - Milan Janjic, Feb 21 2010

Number of toothpicks in the toothpick structure of A139250 after 2^n stages. - Omar E. Pol, Feb 28 2011

Numbers whose binary representation is "10" repeated (n-1) times with "11" appended on the end, n >= 1. For example 171 = 10101011 (2). - Omar E. Pol, Nov 22 2012

a(n) is the smallest number for which A072219(a(n)) = 2*n+1. - Ramasamy Chandramouli, Dec 22 2012

An Engel expansion of 2 to the base b := 4/3 as defined in A181565, with the associated series expansion 2 = b + b^2/3 + b^3/(3*11) + b^4/(3*11*43) + .... Cf. A007051. - Peter Bala, Oct 29 2013

The positive integer solution (x,y) of 3*x - 2^n*y = 1, n>=0, with smallest x is (a(n/2), 2) if n is even and (a((n-1)/2), 1) if n is odd. - Wolfdieter Lang, Feb 15 2014

The smallest positive number that requires at least n additions and subtractions of powers of 2 to be formed. See Puzzling StackExchange link. - Alexander Cooke Jul 16 2023

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)

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

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.

The limit of the local maxima, lim_{m->inf} A081252(m)/m^2 = 1/10. For local minima cf. A081253.

Row sums of the triangle A181971. - Reinhard Zumkeller, Jul 09 2012

Every odd number occurs uniquely in this array. See A178414.

Remainders for classes m of integers n (mod 2^(m+1)). After applying one Collatz (3x+1)-transformation to the so-classified integers the result can be written in two classes (mod 6) only.

This classifying scheme covers all positive integers.

With one 3x+1-transformation T(x;p) := x' = (3x+1)/2^p, all numbers x, described in the form, with the free parameter i >= 0, x = i*2^N + a(N) result in x', describable by the two classes with the same parameter i:

x' = i*6 + 1 (for odd N>2), or x' = i*6 + 5 (for even N). Thus

x = 4*i + 3 -> x' = 6*i + 5, x = 8*i + 1 -> x' = 6*i + 1,

x = 16*i + 13 -> x' = 6*i + 5, x = 32*i + 5 -> x' = 6*i + 1,

x = 64*i + 53 -> x' = 6*i + 5, x = 128*i + 21 -> x' = 6*i + 1,

....

all with "i" as a free parameter >= 0 covering all positive integers.

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 reciprocalize the current rational; for the right subtree interchange the order of the two steps: the rational is first reciprocalized and then incremented. Like the Stern-Brocot and the Bird tree, the drib tree enumerates all 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,

1, 2,

2, 3,1, 3,

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

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

...

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

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

1

1, 2

2, 3,1, 3

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

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

...

then each column k also is a Fibonacci-type 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 A162912 (a(2^m+k) = A162912(2^(m+1)-1-k), m = 0,1,2,..., k = 0..2^m-1).

(End)

From Yosu Yurramendi, Jan 12 2017: (Start)

a(2^(m+2m' ) + A020988(m')) = A000045(m+1), m>=0, m'>=0

a(2^(m+2m'+1) + A020989(m')) = A000045(m+3), m>=0, m'>=0

a(2^(m+2m' ) - 1 - A002450(m')) = A000045(m+1), m>=0, m'>=0

a(2^(m+2m'+1) - 1 - A072197(m'-1)) = A000045(m+3), m>=0, m'>0

a(2^(m+1) -1) = A000045(m+2), m>=0. (End)

These are the integers N which on application of the Collatz function yield the number 11. The Collatz function: if N is an odd number then (3N+1)/2^r yields a positive odd integer for some value of r (which in this case is 11).

These numbers reach 11 in Collatz function iteration after 2(n+1) steps and so end in 1 after exactly 2n+18 steps. - Lambert Klasen (lambert.klasen(AT)gmx.de), Nov 08 2004

Numbers whose binary representation is 111 together with n - 1 times 01. For example, a(4) = 469 = 111010101 (2). - Omar E. Pol, Nov 24 2012

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