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When n > 0 is written as k*2^j with k odd then k = A000265(n) and j = A007814(n), so: when n is written as k*2^j - 1 with k odd then k = A000265(n+1) and j = A007814(n+1), when n > 1 is written as k*2^j + 1 with k odd then k = A000265(n-1) and j = A007814(n-1).

Also denominator of 2^n/n (numerator is A075101(n)). - Reinhard Zumkeller, Sep 01 2002

Slope of line connecting (o, a(o)) where o = (2^k)(n-1) + 1 is 2^k and (by design) starts at (1, 1). - Josh Locker (joshlocker(AT)macfora.com), Apr 17 2004

Numerator of n/2^(n-1). - Alexander Adamchuk, Feb 11 2005

From Marco Matosic, Jun 29 2005: (Start)

"The sequence can be arranged in a table:

1

1 3 1

1 5 3 7 1

1 9 5 11 3 13 7 15 1

1 17 9 19 5 21 11 23 3 25 13 27 7 29 15 31 1

Every new row is the previous row interspaced with the continuation of the odd numbers.

Except for the ones; the terms (t) in each column are t+t+/-s = t_+1. Starting from the center column of threes and working to the left the values of s are given by A000265 and working to the right by A000265." (End)

This is a fractal sequence. The odd-numbered elements give the odd natural numbers. If these elements are removed, the original sequence is recovered. - Kerry Mitchell, Dec 07 2005

2k + 1 is the k-th and largest of the subsequence of k terms separating two successive equal entries in a(n). - Lekraj Beedassy, Dec 30 2005

It's not difficult to show that the sum of the first 2^n terms is (4^n + 2)/3. - Nick Hobson, Jan 14 2005

In the table, for each row, (sum of terms between 3 and 1) - (sum of terms between 1 and 3) = A020988. - Eric Desbiaux, May 27 2009

This sequence appears in the analysis of A160469 and A156769, which resemble the numerator and denominator of the Taylor series for tan(x). - Johannes W. Meijer, May 24 2009

Indices n such that a(n) divides 2^n - 1 are listed in A068563. - Max Alekseyev, Aug 25 2013

From Alexander R. Povolotsky, Dec 17 2014: (Start)

With regard to the tabular presentation described in the comment by Marco Matosic: in his drawing, starting with the 3rd row, the first term in the row, which is equal to 1 (or, alternatively the last term in the row, which is also equal to 1), is not in the actual sequence and is added to the drawing as a fictitious term (for the sake of symmetry); an actual A000265(n) could be considered to be a(j,k) (where j >= 1 is the row number and k>=1 is the column subscript), such that a(j,1) = 1:

1

1 3

1 5 3 7

1 9 5 11 3 13 7 15

1 17 9 19 5 21 11 23 3 25 13 27 7 29 15 31

and so on ... .

The relationship between k and j for each row is 1 <= k <= 2^(j-1). In this corrected tabular representation, Marco's notion that "every new row is the previous row interspaced with the continuation of the odd numbers" remains true. (End)

Partitions natural numbers to the same equivalence classes as A064989. That is, for all i, j: a(i) = a(j) <=> A064989(i) = A064989(j). There are dozens of other such sequences (like A003602) for which this also holds: In general, all sequences for which a(2n) = a(n) and the odd bisection is injective. - Antti Karttunen, Apr 15 2017

From Paul Curtz, Feb 19 2019: (Start)

This sequence is the truncated triangle:

1, 1;

3, 1, 5;

3, 7, 1, 9;

5, 11, 3, 13, 7;

15, 1, 17, 9, 19, 5;

21, 11, 23, 3, 25, 13, 27;

7, 29, 15, 31, 1, 33, 17, 35;

...

The first column is A069834. The second column is A213671. The main diagonal is A236999. The first upper diagonal is A125650 without 0.

c(n) = ((n*(n+1)/2))/A069834 = 1, 1, 2, 2, 1, 1, 4, 4, 1, 1, 2, 2, 1, 1, 8, 8, 1, 1, ... for n > 0. n*(n+1)/2 is the rank of A069834. (End)

As well as being multiplicative, a(n) is a strong divisibility sequence, that is, gcd(a(n),a(m)) = a(gcd(n,m)) for n, m >= 1. In particular, a(n) is a divisibility sequence: if n divides m then a(n) divides a(m). - Peter Bala, Feb 27 2019

a(n) is also the map n -> A026741(n) applied at least A007814(n) times. - Federico Provvedi, Dec 14 2021

Starting with a(1) = 0 mirror all initial 2^k segments and increase by one.

a(n) gives the net rotation (measured in right angles) after taking n steps along a dragon curve. - Christopher Hendrie (hendrie(AT)acm.org), Sep 11 2002

This sequence generates A082410: (0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, ...) and A014577; identical to the latter except starting 1, 1, 0, ...; by writing a "1" if a(n+1) > a(n); if not, write "0". E.g., A014577(2) = 0, since a(3) < a(2), or 1 < 2. - Gary W. Adamson, Sep 20 2003

Starting with 1 = partial sums of A034947: (1, 1, -1, 1, 1, -1, -1, 1, 1, 1, ...). - Gary W. Adamson, Jul 23 2008

The composer Per Nørgård's name is also written in the OEIS as Per Noergaard.

Can be used as a binomial transform operator: Let a(n) = the n-th term in any S(n); then extract 2^k strings, adding the terms. This results in the binomial transform of S(n). Say S(n) = 1, 3, 5, ...; then we obtain the strings: (1), (3, 1), (3, 5, 3, 1), (3, 5, 7, 5, 3, 5, 3, 1), ...; = the binomial transform of (1, 3, 5, ...) = (1, 4, 12, 32, 80, ...). Example: the 8-bit string has a sum of 32 with a distribution of (1, 3, 3, 1) or one 1, three 3's, three 5's, and one 7; as expected. - Gary W. Adamson, Jun 21 2012

Considers all positive odd numbers as nodes of a graph. Two nodes are connected if and only if the sum of the two corresponding odd numbers is a power of 2. Then a(n) is the distance between 2n + 1 and 1. - Jianing Song, Apr 20 2019

Fractal sequence obtained from powers of 2.

k occurs at (2*k-1)*A000079(m), m >= 0. - Robert G. Wilson v, May 23 2006

Sequence is T^(oo)(1) where T is acting on a word w = w(1)w(2)..w(m) as follows: T(w) = "1"w(1)"2"w(2)"3"(...)"m"w(m)"m+1". For instance T(ab) = 1a2b3. Thus T(1) = 112, T(T(1)) = 1121324, T(T(T(1))) = 112132415362748. - Benoit Cloitre, Mar 02 2009

Note that iterating the post-numbering operator U(w) = w(1) 1 w(2) 2 w(3) 3... produces the same limit sequence except with an additional "1" prepended, i.e., 1,1,1,2,1,3,2,4,... - Glen Whitney, Aug 30 2023

In the binary expansion of n, first swallow all zeros from the right, then add 1, and swallow the now-appearing 0 bit as well. - Ralf Stephan, Aug 22 2013

Although A264646 and this sequence initially agree in their digit-streams, they differ after 48 digits. - N. J. A. Sloane, Nov 20 2015

"[This is a] fractal because we get the same sequence after we delete from it the first appearance of all positive integers" - see Cobeli and Zaharescu link. - Robert G. Wilson v, Jun 03 2018

From Peter Munn, Jun 16 2022: (Start)

The sequence is the list of positive integers interleaved with the sequence itself. Provided the offset is suitable (which is the case here) a term of such a self-interleaved sequence is determined by the odd part of its index. Putting some of the formulas given here into words, a(n) is the position of the odd part of n in the list of odd numbers.

Applying the interleaving transform again, we get A110963.

(End)

Omitting all 1's leaves A131987 + 1. - David James Sycamore, Jul 26 2022

a(n) is also the smallest positive number not among the terms between a(a(n-1)) and a(n-1) inclusive (with a(0)=1 prepended). - Neal Gersh Tolunsky, Mar 07 2023

The name "Van Eck's sequence" is due to N. J. A. Sloane, not the author!

Note that it is obvious from the definition that a(n) < n for all n. - N. J. A. Sloane, Jun 20 2019. Even stronger, a(n)+a(n+1) < n for all n, since the a(n+1) implies that a(n-a(n+1)) = a(n). - Jan Ritsema van Eck, Jun 30 2019

Starting with a number different from 0, for instance with 1, gives a different but similar sequence. See A171911-A171918 for examples.

Examination of the first 10^6 terms suggests that lim sup a(n)/n = 1. Cf. A171866/A171867. - David Applegate and N. J. A. Sloane, Oct 18 2010

From Jan Ritsema van Eck, Oct 25 2010: (Start)

Theorem: There are infinitely many zeros.

Proof: Suppose not. Then from a certain point on, no new terms appear, so the sequence is bounded and nonzero. Let M be the maximal term. Any block of length M determines the rest of the sequence.

But there are only M^M different blocks of length M containing the numbers 1 through M.

So a block must repeat, and so the sequence eventually becomes periodic. The periodic part does not contain any zeros.

Suppose the period has length p, and starts at term r, with a(r)=x, ..., a(r+p-1)=z, a(r+p)=x, ... There is another z after q <= p steps, which is immediately followed by q.

But this q implies that a(r-1)=z. Therefore the periodic part really began at step r-1.

Repeating this shows that the periodic part starts at a(1). But a(1)=0, and the periodic part cannot contain a 0. Contradiction. (End)

An alternative wording of the definition: For n>=1, if there exists an m < n such that a(m) = a(n), take the largest such m, otherwise take m = n; set a(n+1) = n-m. Start with a(1)=0. - Arie Bos, Dec 10 2010

Conjectures: (i) lim sup a(n)/n = 1; (ii) gaps between 0's are about log_10 n; (iii) every number eventually appears. - N. J. A. Sloane, in lecture "The OEIS: The Major Problems", Conference for 50th Anniversary of the OEIS, Rutgers University, Oct 10 2014. (Added Jun 16 2019.)

Conjecture: every pair of nonnegative integers (x,y) other than (1,1) and (x,x+1) for x>0 appear as consecutive entries (that is, a(i) = x, a(i+1) = y, for some i). - László Kozma, Aug 09 2016. Correction from Tomas Rokicki, Jun 19 2019: The pair (x,x+1) only occurs at (0,1), as it would imply distinct values x positions previously.

As mentioned in an earlier comment, for any k >= 0 there is a "van Eck" sequence E(k) starting with k and extended using the same rule (cf. A171911-A171918). The initial E(k)(1) = k is followed by at least k initial terms from this sequence A181391 = E(0): E(k)(n+1) = E(0)(n) for all n <= k and beyond, as long as E(0)(n) != k. - M. F. Hasler, Jun 11 2019

Comment from Jordan Linus, circa Jun 16 2019, from an online comment added to the Haran-Sloane video: (Start)

Theorem: limsup a(n)/sqrt(n) >= 1.

Proof: Whenever a(n)=0, either there have been sqrt(n) zeros in the sequence, thus sqrt(n) new distinct numbers (and at least one bigger than sqrt(n)), or there have been fewer than sqrt(n) zeros in the sequence, and thus there is a gap of at least sqrt(n) between two zeros (and the term after the second zero is at least sqrt(n)). So either way there is a term >= sqrt(n). (End)

The long-term behavior of the sequence E(k) appears to be the same for all k, although the individual numbers differ. Empirically modeled up to 2^25 terms of E(k) for k between 0 and 9. - Po-chia Chen, Jun 18 2019

After searching the first 318 billion entries, the smallest number not appearing is 645315850; the smallest pair not of the form (1,1), (0,a), or (x,x+1) not appearing is (268,5). - Tomas Rokicki, Jun 19 2019

Theorem: i-a(i) is unique for all i, a(i)>0. Put differently: i-a(i)<>j-a(j) for all i,j, a(i)>0 and j>i. Proof: if a(i-1)=a(j-1)=x, a(j) can be at most j-i because there is by definition not another x between a(j-1) and a(j-a(j)-1). If a(i-1)<>a(j-1), it follows that a(i-a(i)-1)<>a(j-a(j)-1). Either way i-a(i)<>j-a(j). A special case of this is that pairs (x,x+1) cannot occur for x>0 (as remarked above); similarly, triples (x,y,x+2) cannot occur and so on. Also, since a(i-a(i+1))=a(i) by definition, i-a(i+1)-a(i) is unique for all i, a(i+1) and a(i)>0. A simple example is that triples (x,y,x+1) cannot occur for y>0. Many other "impossible patterns" can be derived. - Jan Ritsema van Eck, Jul 22 2019

After 10^12 terms, the smallest number not appearing is 1732029957; the smallest pair not of the form (1,1), (0,a), or (x,x+1) not appearing is (528,5); there have been 90689534032 zeros. - Benjamin Chaffin, Sep 11 2019

Similar to E(k) above, the sequence E(k,l,...,m) could be defined as the sequence starting with k,l,...,m and continuing using Van Eck's rule. For example, E(1,1) is 1,1,1,... and has period 1. The smallest possible period greater than 1 is 42, attained by E(37, 42, 7, 42, 2, 5, 22, 42, 4, 11, 42, 3, 21, 42, 3, 3, 1, 25, 38, 42, 6, 25, 4, 14, 42, 5, 20, 42, 3, 13, 42, 3, 3, 1, 17, 36, 42, 6, 17, 4, 17, 2). More info: https://redd.it/dbdhpj - Michiel De Muynck, Sep 30 2019

If the rule a(n+1)=0 (when a(n) has not been seen before) is replaced by a(n+1)=a(a(n)) and a(0) is set to 0, then the resulting sequence is A025480. - David James Sycamore, Nov 01 2019

From Yosu Yurramendi, Jun 25 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,

1,2,3,3,

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

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

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

then the sum of the m-th row is 3^m (m = 0,1,2,), each column k is constant, and the constants are from A007306, denominators of Farey (or Stern-Brocot) tree fractions (see formula).

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

1,

1,2,

1, 2,3,3,

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

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

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

then each column is an arithmetic sequence. The differences of the arithmetic sequences also give the sequence A007306 (see formula). The first terms of columns are from A007305 itself (a(A004761(n+1)) = a(n), n>0), and the second ones from A049448 (a(A004761(n+1)+2^A070941(n)) = A049448(n), n>0). (End)

If the sequence is considered in blocks of length 2^m, m = 0,1,2,..., the blocks are the reverse of the blocks of A047679: (a(2^m+1+k) = A047679(2^(m+1)-2-k), m = 0,1,2,..., k = 0,1,2,...,2^m-1). - Yosu Yurramendi, Jun 30 2014

Numerators are A007305.

Write n in binary; list run lengths; add 1 to last run length; make into continued fraction. Sequence gives denominator of fraction obtained.

From Reinhard Zumkeller, Dec 22 2008: (Start)

For n > 1: a(n) = if A025480(n-1) != 0 and A025480(n) != 0 then = a(A025480(n-1)) + a(A025480(n)) else if A025480(n)=0 then a(A025480(n-1))+0 else 1+a(A025480(n-1));

a(n) = A007305(A054429(n)+2) and a(A054429(n)) = A007305(n+2);

A153036(n+1) = floor(A007305(n+2)/a(n)). (End)

From Yosu Yurramendi, Jun 25 2014 and Jun 30 2014: (Start)

If the terms are written as an array a(m, k) = a(2^(m-1)-1+k) with m >= 1 and k = 0, 1, ..., 2^(m-1)-1:

1,

2,1,

3,3, 2, 1,

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

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

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

then the sum of the m-th row is 3^(m-1), and each column is an arithmetic sequence. The differences of these arithmetic sequences give the sequence A007306(k+1). The first terms of columns are 1 for k = 0 and a(k-1) for k >= 1.

In a row reversed version A(m, k) = a(m, m-(k+1)):

1

1,2

1,2,3,3,

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

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

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

each column k >= 0 is constant, namely A007306(k+1).

This row reversed version coincides with the array for A007305 (see the Jun 25 2014 comment there). (End)

Looking at the plot, the sequence clearly shows a fractal structure. (The repeating pattern oddly resembles the [first completed] facade of the Sagrada Familia!) - Daniel Forgues, Nov 15 2019

The a(n) appeared in the analysis of A220002, a sequence related to the Catalan numbers.

The first Maple program makes use of a program by Peter Luschny for the calculation of the a(n) values. The second Maple program shows that this sequence has a beautiful internal structure, see the first formula, while the third Maple program makes optimal use of this internal structure for the fast calculation of a(n) values for large n.

The cross references lead to sequences that have the same internal structure as this sequence.

In a Josephus problem such as A006257, a(n) is the order in which the person originally first in line is eliminated.

The number of remaining survivors after the person originally first in line has been eliminated, i.e., n-a(n), gives the fractal sequence A025480.

For the linear version, see A225489.

First column of array is powers of 2, first row is odd numbers, other cells are products of these two, so every positive integer appears exactly once. [Comment edited to match the definition. - L. Edson Jeffery, Jun 05 2015]

An analogous N X N <-> N bijection based, not on the binary, but on the Fibonacci number system, is given by the Wythoff array A035513.

As an array, this sequence (hence also A135764) is the dispersion of the even positive integers. For the definition of dispersion, see the link "Interspersions and Dispersions." The fractal sequence of this dispersion is A003602. - Clark Kimberling, Dec 03 2010

Place the numbers 1..N (N>=2) on a circle and cyclically mark the 2nd unmarked number until all N numbers are marked. The order in which the N numbers are marked defines a permutation; N is a J_2-prime if this permutation consists of a single cycle of length N.

The resulting permutation can be written as p(m,N) = (2N+1-||2N+1-m||)/2 (1 <= m <= N), where ||x|| is the odd number such that x/||x|| is a power of 2. E.g., ||16||=1 and ||120||=15.

No formula is known for a(n): the J_2-primes have been found by exhaustive search (however, see the CROSS-REFERENCES). But we have: (1) N is J_2-prime iff p=2N+1 is a prime number and +2 generates Z_p^* (the multiplicative group of Z_p). (2) N is J_2-prime iff p=2N+1 is a prime number and exactly one of the following holds: (a) N == 1 (mod 4) and +2 generates Z_p^* but -2 does not, (b) N == 2 (mod 4) and both +2 and -2 generate Z_p^*.