cp's OEIS Frontend

Although this is a list, it has offset 0 for both historical and mathematical reasons.

Numbers whose set of base-4 digits is a subset of {0,1}. - Ray Chandler, Aug 03 2004, corrected by M. F. Hasler, Oct 16 2018

Numbers k such that the sum of the base-2 digits of k = sum of the base-4 digits of k. - Clark Kimberling

Numbers having the same representation in both binary and negabinary (A039724). - Eric W. Weisstein

This sequence has many other interesting and useful properties. Every term k corresponds to a unique pair i,j with k = a(i) + 2*a(j) (i=A059905(n), j=A059906(n)) -- see A126684. Every list of numbers L = [L1,L2,L3,...] can be encoded uniquely by "recursive binary interleaving", where f(L) = a(L1) + 2*a(f([L2,L3,...])) with f([])=0. - Marc LeBrun, Feb 07 2001

This may be described concisely using the "rebase" notation b[n]q, which means "replace b with q in the expansion of n", thus "rebasing" n from base b into base q. The present sequence is 2[n]4. Many interesting operations (e.g., 10[n](1/10) = digit reverse, shifted) are nicely expressible this way. Note that q[n]b is (roughly) inverse to b[n]q. It's also natural to generalize the idea of "basis" so as to cover the likes of F[n]2, the so-called "fibbinary" numbers (A003714) and provide standard ready-made images of entities obeying other arithmetics, say like GF2[n]2 (e.g., primes = A014580, squares = the present sequence, etc.). - Marc LeBrun, Mar 24 2005

a(n) is also equal to the product n X n formed using carryless binary multiplication (A059729, A063010). - Henry Bottomley, Jul 03 2001

Numbers k such that A004117(k) is odd. - Pontus von Brömssen, Nov 25 2008

Fixed point of the morphism: 0 -> 01; 1 -> 45; 2 -> 89; ...; n -> (4n)(4n+1), starting from a(0)=0. - Philippe Deléham, Oct 22 2011

If n is even and present, so is n+1. - Robert G. Wilson v, Oct 24 2014

Also: interleave binary digits of n with 0's. (Equivalent to the "rebase" interpretation above.) - M. F. Hasler, Oct 16 2018

Named after the Austrian-Canadian mathematician Leo Moser (1921-1970) and the Dutch mathematician Nicolaas Govert de Bruijn (1918-2012). - Amiram Eldar, Jun 19 2021

Conjecture: The sums of distinct powers of k > 2 can be constructed as the following (k-1)-ary rooted tree. For each n the tree grows and a(n) is then the total number of nodes. For n = 1, the root of the tree is added. For n > 1, if n is odd one leaf of depth n-2 grows one child. If n is even all leaves of depth >= (n - 1 - A000225(A001511(n/2))) grow the maximum number of children. An illustration is provided in the links. - John Tyler Rascoe, Oct 09 2022

The representation of the compositions (for fixed n) is as lists of parts, the order between individual compositions (for the same n) is (list-)reversed lexicographic; see the example by Omar E. Pol. - Joerg Arndt, Sep 03 2013

This is the standard ordering for compositions in this database; it is similar to the Mathematica ordering for partitions (A080577). Other composition orderings include A124734 (similar to the Abramowitz & Stegun ordering for partitions, A036036), A108244 (similar to the Maple partition ordering, A080576), etc (see crossrefs).

Factorize each term in A057335; sequence records the values of the resulting exponents. It also runs through all possible permutations of multiset digits.

This can be regarded as a table in two ways: with each composition as a row, or with the compositions of each integer as a row. The first way has A000120 as row lengths and A070939 as row sums; the second has A001792 as row lengths and A001788 as row sums. - Franklin T. Adams-Watters, Nov 06 2006

This sequence includes every finite sequence of positive integers. - Franklin T. Adams-Watters, Nov 06 2006

Compositions (or ordered partitions) are also generated in sequence A101211. - Alford Arnold, Dec 12 2006

The equivalent sequence for partitions is A228531. - Omar E. Pol, Sep 03 2013

The sole partition of zero has no components, not a single component of length one. Hence the first nonempty row is row 1. - Franklin T. Adams-Watters, Apr 02 2014 [Edited by Andrey Zabolotskiy, May 19 2018]

See sequence A261300 for another version where the terms of each composition are concatenated to form one single integer: (0, 1, 2, 11, 3, 21, 12, 111,...). This also shows how the terms can be obtained from the binary numbers A007088, cf. Arnold's first Example. - M. F. Hasler, Aug 29 2015

The k-th composition in the list is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This is described as the standard ordering used in the OEIS, although the sister sequence A228351 is also sometimes considered to be canonical. Both sequences define a bijective correspondence between nonnegative integers and integer compositions. - Gus Wiseman, May 19 2020

First differences of A030303 = positions of bits 1 in the concatenation A030190 (= A030302) of numbers written in binary (A007088). - Indices of record values (= first occurrence of n) are given by A005183: a(A005183(n)) = n, cf. FORMULA for more. - M. F. Hasler, Oct 12 2020

The geometric mean approaches the Somos constant (A112302). - Jwalin Bhatt, Feb 10 2025

Nonnegative integers with no decimal digit > 2. Thus nonnegative integers in base 10 whose quadrupling by normal addition or multiplication requires no carry operation. - Rick L. Shepherd, Jun 25 2009

A permutation of the squarefree numbers A005117. The missing positive numbers are in A013929. - Alois P. Heinz, Sep 06 2014

From Antti Karttunen, Apr 18 & 19 2017: (Start)

Because a(n) toggles the parity of n there are neither fixed points nor any cycles of odd length.

Conjecture: there are no finite cycles of any length. My grounds for this conjecture: any finite cycle in this sequence, if such cycles exist at all, must have at least one member that occurs somewhere in A285319, the terms that seem already to be quite rare. Moreover, any such a number n should satisfy in addition to A019565(n) < n also that A048675^{k}(n) is squarefree, not just for k=0, 1 but for all k >= 0. As there is on average a probability of only 6/(Pi^2) = 0.6079... that any further term encountered on the trajectory of A048675 is squarefree, the total chance that all of them would be squarefree (which is required from the elements of A019565-cycles) is soon minuscule, especially as A048675 is not very tightly bounded (many trajectories seem to skyrocket, at least initially). I am also assuming that usually there is no significant correlation between the binary expansions of n and A048675(n) (apart from their least significant bits), or, for that matter, between their prime factorizations.

See also the slightly stronger conjecture in A285320, which implies that there would neither be any two-way infinite cycles.

If either of the conjectures is false (there are cycles), then certainly neither sequence A285332 nor its inverse A285331 can be a permutation of natural numbers. (End)

The conjecture made in A087207 (see also A288569) implies the two conjectures mentioned above. A further constraint for cycles is that in any A019565-trajectory which starts from a squarefree number (A005117), every other term is of the form 4k+2, while every other term is of the form 6k+3. - Antti Karttunen, Jun 18 2017

The sequence satisfies the exponential function identity, a(x + y) = a(x) * a(y), whenever x and y do not have a 1-bit in the same position, i.e., when A004198(x,y) = 0. See also A283475. - Antti Karttunen, Oct 31 2019

The above identity becomes unconditional if binary exclusive OR, A003987(.,.), is substituted for addition, and A059897(.,.), a multiplicative equivalent of A003987, is substituted for multiplication. This gives us a(A003987(x,y)) = A059897(a(x), a(y)). - Peter Munn, Nov 18 2019

Also the Heinz number of the binary indices of n, where the Heinz number of a sequence (y_1,...,y_k) is prime(y_1)*...*prime(y_k), and a number's binary indices (A048793) are the positions of 1's in its reversed binary expansion. - Gus Wiseman, Dec 28 2022

Nonnegative integers with no decimal digit > 3. Thus nonnegative integers in base 10 whose tripling (trebling) by normal addition or multiplication requires no carry operation. - Rick L. Shepherd, Jun 25 2009

Interpreted in base 10: a(x)+a(y) = a(z) => x+y = z. The converse is not true in general. - Karol Bacik, Sep 27 2012

From Rick L. Shepherd, Jun 25 2009: (Start)

Nonnegative integers with no decimal digit > 4.

Thus nonnegative integers in base 10 whose doubling by normal addition or multiplication requires no carry operation. (End)

It appears that this sequence corresponds to the numbers n for which twice the sum of digits of n is the sum of digits of 2*n. - Rémy Sigrist, Nov 22 2009

Also numbers without 9 as a digit.

Complement of A011539: A102683(a(n)) = 0; A068505(a(n)) != a(n)). - Reinhard Zumkeller, Dec 29 2011

Might be called the "Lichtenberg sequence" after Georg Christoph Lichtenberg, who discussed it in 1769 in connection with the Chinese Rings puzzle (baguenaudier). - Andreas M. Hinz, Feb 15 2017

Number of steps to change from a binary string of n 0's to n 1's using a Gray code. - Jon Stadler (jstadler(AT)coastal.edu)

Popular puzzles such as Spin-Out and The Brain Puzzler are based on the Gray binary system and require a(n) steps to complete for some number n.

Conjecture: {a(n)} also gives all j for which A048702(j) = A000217(j); i.e., if we take the a(n)-th triangular number (a(n)^2 + a(n))/2 and multiply it by 3, we get a(n)-th even-length binary palindrome A048701(a(n)) concatenated from a(n) and its reverse. E.g., a(4) = 10, which is 1010 in binary; the tenth triangular number is 55, and 55*3 = 165 = 10100101 in binary. - Antti Karttunen, circa 1999. (This has been now proved by Paul K. Stockmeyer in his arXiv:1608.08245 paper.) - Antti Karttunen, Aug 31 2016

Number of ways to tie a tie of n or fewer half turns, excluding mirror images. Also number of walks of length n or less on a triangular lattice with the following restrictions; given l, r and c as the lattice axes. 1. All steps are taken in the positive axis direction. 2. No two consecutive steps are taken on the same axis. 3. All walks begin with l. 4. All walks end with rlc or lrc. - Bill Blewett, Dec 21 2000

a(n) is the minimal number of vertices to be selected in a vertex-cover of the balanced tree B_n. - Sen-peng Eu, Jun 15 2002

A087117(a(n)) = A038374(a(n)) = 1 for n > 1; see also A090050. - Reinhard Zumkeller, Nov 20 2003

Intersection of A003754 and A003714; complement of A107907. - Reinhard Zumkeller, May 28 2005

Equivalently, numbers m whose binary representation is effectively, for some number k, both the lazy Fibonacci and the Zeckendorf representation of k (in which case k = A022290(m)). - Peter Munn, Oct 06 2022

a(n+1) gives row sums of Riordan array (1/(1-x), x(1+2x)). - Paul Barry, Jul 18 2005

Total number of initial 01's in all binary words of length n+1. Example: a(3) = 5 because the binary words of length 4 that start with 01 are (01)00, (01)(01), (01)10 and (01)11 and the total number of initial 01's is 5 (shown between parentheses). a(n) = Sum_{k >= 0} k*A119440(n+1, k). - Emeric Deutsch, May 19 2006

In Norway we call the 10-ring puzzle "strikketoy" or "knitwear" (see link). It takes 682 moves to free the two parts. - Hans Isdahl, Jan 07 2008

Equals A002450 and A020988 interlaced. - Zak Seidov, Feb 10 2008

For n > 1, let B_n = the complete binary tree with vertex set V where |V| = 2^n - 1. If VC is a minimum-size vertex cover of B_n, Sen-Peng Eu points out that a(n) = |VC|. It also follows that if IS = V\VC, a(n+1) = |IS|. - K.V.Iyer, Apr 13 2009

Starting with 1 and convolved with [1, 2, 2, 2, ...] = A000295. - Gary W. Adamson, Jun 02 2009

a(n) written in base 2 is sequence A056830(n). - Jaroslav Krizek, Aug 05 2009

This is the sequence A(0, 1; 1, 2; 1) of the family of sequences [a,b:c,d:k] considered by G. Detlefs, and treated as A(a,b;c,d;k) in the W. Lang link given below. - Wolfdieter Lang, Oct 18 2010

From Vladimir Shevelev, Jan 30 2012, Feb 13 2012: (Start)

1) Denote by {n, k} the number of permutations of 1, ..., n with the up-down index k (for definition, see comment in A203827). Then max_k{n, k} = {n, a(n)} = A000111(n).

2) a(n) is the minimal number > a(n-1) with the Hamming distance d_H(a(n-1), a(n)) = n. Thus this sequence is the Hamming analog of triangular numbers 0, 1, 3, 6, 10, ... (End)

From Hieronymus Fischer, Nov 22 2012: (Start)

Represented in binary form each term after the second one contains every previous term as a substring.

The terms a(2) = 2 and a(3) = 5 are the only primes. Proof: For even n we get a(n) = 2*(2^(2*n) - 1)/3, which shows that a(n) is even, too, and obviously a(n) > 2 for all even n > 2. For odd n we have a(n) = (2^(n+1) - 1)/3 = (2^((n+1)/2) - 1) * (2^((n+1)/2) + 1)/3. Evidently, at least one of these factors is divisible by 3, both are greater than 6, provided n > 3. Hence it follows that a(n) is composite for all odd n > 3.

Represented as a binary number, a(n+1) has exactly n prime substrings. Proof: Evidently, a(1) = 1_2 has zero and a(2) = 10_2 has 1 prime substring. Let n > 1. Written in binary, a(n+1) is 101010101...01 (if n + 1 is odd) and is 101010101...10 (if n + 1 is even) with n + 1 digits. Only the 2- and 3-digits substrings 10_2 (=2) and 101_2 (=5) are prime substrings. All the other substrings are nonprime since every substring is a previous term and all terms unequal to 2 and 5 are nonprime. For even n + 1, the number of prime substrings equal to 2 = 10_2 is (n+1)/2, and the number of prime substrings equal to 5 = 101_2 is (n-1)/2, makes a sum of n. For odd n + 1 we get n/2 for both, the number of 2's and 5's prime substrings, in any case, the sum is n. (End)

Number of different 3-colorings for the vertices of all triangulated planar polygons on a base with n+2 vertices if the colors of the two base vertices are fixed. - Patrick Labarque, Feb 09 2013

A090079(a(n)) = a(n) and A090079(m) <> a(n) for m < a(n). - Reinhard Zumkeller, Feb 16 2013

a(n) is the number of length n binary words containing at least one 1 and ending in an even number (possibly zero) of 0's. a(3) = 5 because we have: 001, 011, 100, 101, 111. - Geoffrey Critzer, Dec 15 2013

a(n) is the number of permutations of length n+1 having exactly one descent such that the first element of the permutation is an even number. - Ran Pan, Apr 18 2015

a(n) is the sequence of the last row of the Hadamard matrix H(2^n) obtained via Sylvester's construction: H(2) = [1,1;1,-1], H(2^n) = H(2^(n-1))*H(2), where * is the Kronecker product. - William P. Orrick, Jun 28 2015

Conjectured record values of A264784: a(n) = A264784(A155051(n-1)). - Reinhard Zumkeller, Dec 04 2015. (This is proved by Paul K. Stockmeyer in his arXiv:1608.08245 paper.) - Antti Karttunen, Aug 31 2016

Decimal representation of the x-axis, from the origin to the right edge, of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 131", based on the 5-celled von Neumann neighborhood. See A279053 for references and links. - Robert Price, Dec 05 2016

For n > 4, a(n-2) is the second-largest number in row n of A127824. - Dmitry Kamenetsky, Feb 11 2017

Conjecture: a(n+1) is the number of compositions of n with two kinds of parts, n and n', where the order of the 1 and 1' does not matter. For n=2, a(3) = 5 compositions, enumerated as follows: 2; 2'; 1,1; 1',1 = 1',1; 1',1'. - Gregory L. Simay, Sep 02 2017

Conjecture proved by recognizing the appropriate g.f. is x/(1 - x)(1 - x)(1 - 2*x^2 - 2x^3 - ...) = x/(1 - 2*x - x^2 + 2x^3). - Gregory L. Simay, Sep 10 2017

a(n) = 2^(n-1) + 2^(n-3) + 2^(n-5) + ... a(2*k -1) = A002450(k) is the sum of the powers of 4. a(2*k) = 2*A002450(k). - Gregory L. Simay, Sep 27 2017

a(2*n) = n times the string [10] in binary representation, a(2*n+1) = n times the string [10] followed with [1] in binary representation. Example: a(7) = 85 = (1010101) in binary, a(8) = 170 = (10101010) in binary. - Ctibor O. Zizka, Nov 06 2018

Except for 0, these are the positive integers whose binary expansion has cuts-resistance 1. For the operation of shortening all runs by 1, cuts-resistance is the number of applications required to reach an empty word. Cuts-resistance 2 is A329862. - Gus Wiseman, Nov 27 2019

From Markus Sigg, Sep 14 2020: (Start)

Let s(k) be the length of the Collatz orbit of k, e.g. s(1) = 1, s(2) = 2, s(3) = 5. Then s(a(n)) = n+3 for n >= 3. Proof by induction: s(a(3)) = s(5) = 6 = 3+3. For odd n >= 5 we have s(a(n)) = s(4*a(n-2)+1) = s(12*a(n-2)+4)+1 = s(3*a(n-2)+1)+3 = s(a(n-2))+2 = (n-2)+3+2 = n+3, and for even n >= 4 this gives s(a(n)) = s(2*a(n-1)) = s(a(n-1))+1 = (n-1)+3+1 = n+3.

Conjecture: For n >= 3, a(n) is the second largest natural number whose Collatz orbit has length n+3. (End)

From Gary W. Adamson, May 14 2021: (Start)

With offset 1 the sequence equals the numbers of 1's from n = 1 to 3, 3 to 7, 7 to 15, ...; of A035263; as shown below:

..1 3 7 15...

..1 0 1 1 1 0 1 0 1 0 1 1 1 0 1...

..1.....2...........5......................10...; a(n) = Sum_(k=1..2n-1)A035263(k)

.....1...........2.......................5...; as to zeros.

..1's in the Tower of Hanoi game represent CW moves On disks in the pattern:

..0, 1, 2, 0, 1, 2, ... whereas even numbered disks move in the pattern:

..0, 2, 1, 0, 2, 1, ... (End)

Except for 0, numbers that are repunits in Gray-code representation (A014550). - Amiram Eldar, May 21 2021

From Gus Wiseman, Apr 20 2023: (Start)

Also the number of nonempty subsets of {1..n} with integer median, where the median of a multiset is the middle part in the odd-length case, and the average of the two middle parts in the even-length case. For example, the a(1) = 1 through a(4) = 10 subsets are:

{1} {1} {1} {1}

{2} {2} {2}

{3} {3}

{1,3} {4}

{1,2,3} {1,3}

{2,4}

{1,2,3}

{1,2,4}

{1,3,4}

{2,3,4}

The complement is counted by A005578.

For mean instead of median we have A051293, counting empty sets A327475.

For normal multisets we have A056450, strongly normal A361202.

For partitions we have A325347, strict A359907, complement A307683.

(End)

In binary form, because the sequence 10101_2... keeps repeating, one can represent a(n) = floor(0.(10)2 * 2^n). Because 0.(10)_2 = 0.(10)_2 * 4 - 2, then 0.(10)_2 = 2/3. Substituting this value in the first expression, we obtain a(n) = floor(2^(n+1)/3). This also agrees with the formula of Paul Barry, Oct 08 2005. - _Giovanni Ciriani, Mar 31 2025

3 does not divide binomial(2s, s) if and only if s is a member of this sequence, where binomial(2s, s) = A000984(s) are the central binomial coefficients.

This is the lexicographically earliest increasing sequence of nonnegative numbers that contains no arithmetic progression of length 3. - Robert Craigen (craigenr(AT)cc.umanitoba.ca), Jan 29 2001

In the notation of A185256 this is the Stanley Sequence S(0,1). - N. J. A. Sloane, Mar 19 2010

Complement of A074940. - Reinhard Zumkeller, Mar 23 2003

Sums of distinct powers of 3. - Ralf Stephan, Apr 27 2003

Numbers n such that central trinomial coefficient A002426(n) == 1 (mod 3). - Emeric Deutsch and Bruce E. Sagan, Dec 04 2003

A039966(a(n)+1) = 1; A104406(n) = number of terms <= n.

Subsequence of A125292; A125291(a(n)) = 1 for n>1. - Reinhard Zumkeller, Nov 26 2006

Also final value of n - 1 written in base 2 and then read in base 3 and with finally the result translated in base 10. - Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), Jun 23 2007

a(n) modulo 2 is the Thue-Morse sequence A010060. - Dennis Tseng, Jul 16 2009

Also numbers such that the balanced ternary representation is the same as the base 3 representation. - Alonso del Arte, Feb 25 2011

Fixed point of the morphism: 0 -> 01; 1 -> 34; 2 -> 67; ...; n -> (3n)(3n+1), starting from a(1) = 0. - Philippe Deléham, Oct 22 2011

It appears that this sequence lists the values of n which satisfy the condition sum(binomial(n, k)^(2*j), k = 0..n) mod 3 <> 0, for any j, with offset 0. See Maple code. - Gary Detlefs, Nov 28 2011

Also, it follows from the above comment by Philippe Lallouet that the sequence must be generated by the rules: a(1) = 0, and if m is in the sequence then so are 3*m and 3*m + 1. - L. Edson Jeffery, Nov 20 2015

Add 1 to each term and we get A003278. - N. J. A. Sloane, Dec 01 2019

The name "Fibbinary" is due to Marc LeBrun.

"... integers whose binary representation contains no consecutive ones and noticed that the number of such numbers with n bits was fibonacci(n)". [posting to sci.math by Bob Jenkins (bob_jenkins(AT)burtleburtle.net), Jul 17 2002]

From Benoit Cloitre, Mar 08 2003: (Start)

A number m is in the sequence if and only if C(3m, m) (or equally, C(3m, 2m)) is odd.

a(n) == A003849(n) (mod 2). (End)

Numbers m such that m XOR 2*m = 3*m. - Reinhard Zumkeller, May 03 2005. [This implies that A003188(2*a(n)) = 3*a(n) holds for all n.]

Numbers whose base-2 representation contains no two adjacent ones. For example, m = 17 = 10001_2 belongs to the sequence, but m = 19 = 10011_2 does not. - Ctibor O. Zizka, May 13 2008

m is in the sequence if and only if the central Stirling number of the second kind S(2*m, m) = A007820(m) is odd. - O-Yeat Chan (math(AT)oyeat.com), Sep 03 2009

A000120(3*a(n)) = 2*A000120(a(n)); A002450 is a subsequence.

Every nonnegative integer can be expressed as the sum of two terms of this sequence. - Franklin T. Adams-Watters, Jun 11 2011

Subsequence of A213526. - Arkadiusz Wesolowski, Jun 20 2012

This is also the union of A215024 and A215025 - see the Comment in A014417. - N. J. A. Sloane, Aug 10 2012

The binary representation of each term m contains no two adjacent 1's, so we have (m XOR 2m XOR 3m) = 0, and thus a two-player Nim game with three heaps of (m, 2m, 3m) stones is a losing configuration for the first player. - V. Raman, Sep 17 2012

Positions of zeros in A014081. - John Keith, Mar 07 2022

These numbers are similar to Fibternary numbers A003726, Tribbinary numbers A060140 and Tribternary numbers. This sequence is a subsequence of Fibternary numbers A003726. The number of Fibbinary numbers less than any power of two is a Fibonacci number. We can generate this sequence recursively: start with 0 and 1; then, if x is in the sequence add 2x and 4x+1 to the sequence. The Fibbinary numbers have the property that the n-th Fibbinary number is even if the n-th term of the Fibonacci word is a. Respectively, the n-th Fibbinary number is odd (of the form 4x+1) if the n-th term of the Fibonacci word is b. Every number has a Fibbinary multiple. - Tanya Khovanova and PRIMES STEP Senior, Aug 30 2022

This is the ordered set S of numbers defined recursively by: 0 is in S; if x is in S, then 2*x and 4*x + 1 are in S. See Kimberling (2006) Example 3, in references below. - Harry Richman, Jan 31 2024