cp's OEIS Frontend

Least positive k such that m^k + 1 divides m^n + 1 (with fixed base m). - Vladimir Baltic, Mar 25 2002

To construct the sequence: start with 1, concatenate 1, 1 and double last term gives 1, 2. Concatenate those 2 terms, 1, 2, 1, 2 and double last term 1, 2, 1, 2 -> 1, 2, 1, 4. Concatenate those 4 terms: 1, 2, 1, 4, 1, 2, 1, 4 and double last term -> 1, 2, 1, 4, 1, 2, 1, 8, etc. - Benoit Cloitre, Dec 17 2002

a(n) = gcd(seq(binomial(2*n, 2*m+1)/2, m = 0 .. n - 1)) (odd numbered entries of even numbered rows of Pascal's triangle A007318 divided by 2), where gcd() denotes the greatest common divisor of a set of numbers. Due to the symmetry of the rows it suffices to consider m = 0 .. floor((n-1)/2). - Wolfdieter Lang, Jan 23 2004

Equals the continued fraction expansion of a constant x (cf. A100338) such that the continued fraction expansion of 2*x interleaves this sequence with 2's: contfrac(2*x) = [2; 1, 2, 2, 2, 1, 2, 4, 2, 1, 2, 2, 2, 1, 2, 8, 2, ...].

Simon Plouffe observes that this sequence and A003484 (Radon function) are very similar, the difference being all zeros except for every 16th term (see A101119 for nonzero differences). Dec 02 2004

This sequence arises when calculating the next odd number in a Collatz sequence: Next(x) = (3*x + 1) / A006519, or simply (3*x + 1) / BitAnd(3*x + 1, -3*x - 1). - Jim Caprioli, Feb 04 2005

a(n) = n if and only if n = 2^k. This sequence can be obtained by taking a(2^n) = 2^n in place of a(2^n) = n and using the same sequence building approach as in A001511. - Amarnath Murthy, Jul 08 2005

Also smallest m such that m + n - 1 = m XOR (n - 1); A086799(n) = a(n) + n - 1. - Reinhard Zumkeller, Feb 02 2007

Number of 1's between successive 0's in A159689. - Philippe Deléham, Apr 22 2009

Least number k such that all coefficients of k*E(n, x), the n-th Euler polynomial, are integers (cf. A144845). - Peter Luschny, Nov 13 2009

In the binary expansion of n, delete everything left of the rightmost 1 bit. - Ralf Stephan, Aug 22 2013

The equivalent sequence for partitions is A194446. - Omar E. Pol, Aug 22 2013

Also the 2-adic value of 1/n, n >= 1. See the Mahler reference, definition on p. 7. This is a non-archimedean valuation. See Mahler, p. 10. Sometimes called 2-adic absolute value of 1/n. - Wolfdieter Lang, Jun 28 2014

First 2^(k-1) - 1 terms are also the heights of the successive rectangles and squares of width 2 that are adjacent to any of the four sides of the toothpick structure of A139250 after 2^k stages, with k >= 2. For example: if k = 5 the heights after 32 stages are [1, 2, 1, 4, 1, 2, 1, 8, 1, 2, 1, 4, 1, 2, 1] respectively, the same as the first 15 terms of this sequence. - Omar E. Pol, Dec 29 2020

Moebius transform of A093811(n). a(n) = A093811(n) * A008683(n), where operation * denotes Dirichlet convolution, namely b(n) * c(n) = Sum_{d|n} b(d) * c(n/d). Simultaneously holds Dirichlet multiplication: a(n) * A000012(n) = A093811(n). - Jaroslav Krizek, Mar 22 2009

Apart from the 0's, all terms are in A002473. Further, for all m in A002473 there is some n such that a(n) = m, see A096867. - Charles R Greathouse IV, Sep 29 2013

a(n) = 0 asymptotically almost surely, namely for all n except for the set of numbers without digit '0'; this set is of density zero, since it is less and less probable to have no '0' as the number of digits of n grows. (See also A054054.) - M. F. Hasler, Oct 11 2015

a(n) is the number of allowable transition rules for passing from one change to the next (on n-1 bells) in the English art of bell-ringing. This is also the number of involutions in the symmetric group S_{n-1} which can be represented as a product of transpositions of consecutive numbers from {1, 2, ..., n-1}. Thus for n = 6 we have a(6) from (12), (12)(34), (12)(45), (23), (23)(45), (34), (45), for instance. See my 1983 Math. Proc. Camb. Phil. Soc. paper. - Arthur T. White, letter to N. J. A. Sloane, Dec 18 1986

Number of permutations p of {1, 2, ..., n-1} such that max|p(i) - i| = 1. Example: a(4) = 2 since only the permutations 132 and 213 of {1, 2, 3} satisfy the given condition. - Emeric Deutsch, Jun 04 2003 [For a(5) = 4 we have 2143, 1324, 2134 and 1243. - Jon Perry, Sep 14 2013]

Number of 001-avoiding binary words of length n-3. a(n) is the number of partitions of {1, ..., n-1} into two blocks in which only 1- or 2-strings of consecutive integers can appear in a block and there is at least one 2-string. E.g., a(6) = 7 because the enumerated partitions of {1, 2, 3, 4, 5} are 124/35, 134/25, 14/235, 13/245, 1245/3, 145/23, 125/34. - Augustine O. Munagi, Apr 11 2005

Numbers for which only one Fibonacci bit-representation is possible and for which the maximal and minimal Fibonacci bit-representations (A104326 and A014417) are equal. For example, a(12) = 10101 because 8 + 3 + 1 = 12. - Casey Mongoven, Mar 19 2006

Beginning with a(2), the "Recamán transform" (see A005132) of the Fibonacci numbers (A000045). - Nick Hobson, Mar 01 2007

Starting with nonzero terms, a(n) gives the row sums of triangle A158950. - Gary W. Adamson, Mar 31 2009

a(n+2) is the minimum number of elements in an AVL tree of height n. - Lennert Buytenhek (buytenh(AT)wantstofly.org), May 31 2010

a(n) is the number of branch nodes in the Fibonacci tree of order n-1. A Fibonacci tree of order n (n >= 2) is a complete binary tree whose left subtree is the Fibonacci tree of order n-1 and whose right subtree is the Fibonacci tree of order n-2; each of the Fibonacci trees of order 0 and 1 is defined as a single node (see the Knuth reference, p. 417). - Emeric Deutsch, Jun 14 2010

a(n+3) is the number of distinct three-strand positive braids of length n (cf. Burckel). - Maxime Bourrigan, Apr 04 2011

a(n+1) is the number of compositions of n with maximal part 2. - Joerg Arndt, May 21 2013

a(n+2) is the number of leafs of great-grandparent DAG (directed acyclic graph) of height n. A great-grandparent DAG of height n is a single node for n = 1; for n > 1 each leaf of ggpDAG(n-1) has two child nodes where pairs of adjacent new nodes are merged into single node if and only if they have disjoint grandparents and same great-grandparent. Consequence: a(n) = 2*a(n-1) - a(n-3). - Hermann Stamm-Wilbrandt, Jul 06 2014

2 and 7 are the only prime numbers in this sequence. - Emmanuel Vantieghem, Oct 01 2014

From Russell Jay Hendel, Mar 15 2015: (Start)

We can establish Gerald McGarvey's conjecture mentioned in the Formula section, however we require n > 4. We need the following 4 prerequisites.

(1) a(n) = F(n) - 1, with {F(n)}A000045.%20(2)%20(Binet%20form)%20F(n)%20=%20(d%5En%20-%20e%5En)/sqrt(5)%20with%20d%20=%20phi%20and%20e%20=%201%20-%20phi,%20de%20=%20-1%20and%20d%20+%20e%20=%201.%20It%20follows%20that%20a(n)%20=%20(d(n)%20-%20e(n))/sqrt(5)%20-%201.%20(3)%20To%20prove%20floor(x)%20=%20y%20is%20equivalent%20to%20proving%20that%20x%20-%20y%20lies%20in%20the%20half-open%20interval%20%5B0,%201).%20(4)%20The%20series%20%7Bs(n)%20=%20c1%20x%5En%20+%20c2%7D">{n >= 1} the Fibonacci numbers A000045. (2) (Binet form) F(n) = (d^n - e^n)/sqrt(5) with d = phi and e = 1 - phi, de = -1 and d + e = 1. It follows that a(n) = (d(n) - e(n))/sqrt(5) - 1. (3) To prove floor(x) = y is equivalent to proving that x - y lies in the half-open interval [0, 1). (4) The series {s(n) = c1 x^n + c2}{n >= 1}, with -1 < x < 0, and c1 and c2 positive constants, converges by oscillation with s(1) < s(3) < s(5) < ... < s(6) < s(4) < s(2). If follows that for any odd n, the open interval (s(n), s(n+1)) contains the subsequence {s(t)}_{t >= n + 2}. Using these prerequisites we can analyze the conjecture.

Using prerequisites (2) and (3) we see we must prove, for all n > 4, that d((d^(n-1) - e^(n-1))/sqrt(5) - 1) - (d^n - e^n)/sqrt(5) + 1 + c lies in the interval [0, 1). But de = -1, implying de^(n-1) = -e^(n-2). It follows that we must equivalently prove (for all n > 4) that E(n, c) = (e^(n-2) + e^n)/sqrt(5) + 1 - d + c = e^(n-2) (e^2 + 1)/sqrt(5) + e + c lies in [0, 1). Clearly, for any particular n, E(n, c) has extrema (maxima, minima) when c = 2*(1-d) and c = (1+d)*(1-d). Therefore, the proof is completed by using prerequisite (4). It suffices to verify E(5, 2*(1-d)) = 0, E(6, 2*(1-d)) = 0.236068, E(5, (1-d)*(1+d)) = 0.618034, E(6, (1-d)*(1+d)) = 0.854102, all lie in [0, 1).

(End)

a(n) can be shown to be the number of distinct nonempty matchings on a path with n vertices. (A matching is a collection of disjoint edges.) - Andrew Penland, Feb 14 2017

Also, for n > 3, the lexicographically earliest sequence of positive integers such that {phi*a(n)} is located strictly between {phi*a(n-1)} and {phi*a(n-2)}. - Ivan Neretin, Mar 23 2017

From Eric M. Schmidt, Jul 17 2017: (Start)

Number of sequences (e(1), ..., e(n-2)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i) != e(j) <= e(k). [Martinez and Savage, 2.5]

Number of sequences (e(1), ..., e(n-2)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i) >= e(j) <= e(k) and e(i) != e(k). [Martinez and Savage, 2.5]

(End)

Numbers whose Zeckendorf (A014417) and dual Zeckendorf (A104326) representations are the same: alternating digits of 1 and 0. - Amiram Eldar, Nov 01 2019

a(n+2) is the length of the longest array whose local maximum element can be found in at most n reveals. See link to the puzzle by Alexander S. Kulikov. - Dmitry Kamenetsky, Aug 08 2020

a(n+2) is the number of nonempty subsets of {1,2,...,n} that contain no consecutive elements. For example, the a(6)=7 subsets of {1,2,3,4} are {1}, {2}, {3}, {4}, {1,3}, {1,4} and {2,4}. - Muge Olucoglu, Mar 21 2021

a(n+3) is the number of allowed patterns of length n in the even shift (that is, a(n+3) is the number of binary words of length n in which there are an even number of 0s between any two occurrences of 1). For example, a(7)=12 and the 12 allowed patterns of length 4 in the even shift are 0000, 0001, 0010, 0011, 0100, 0110, 0111, 1000, 1001, 1100, 1110, 1111. - Zoran Sunic, Apr 06 2022

Conjecture: for k a positive odd integer, the sequence {a(k^n): n >= 1} is a strong divisibility sequence; that is, for n, m >= 1, gcd(a(k^n), a(k^m)) = a(k^gcd(n,m)). - Peter Bala, Dec 05 2022

In general, the sum of a second-order linear recurrence having signature (c,d) will be a third-order recurrence having a signature (c+1,d-c,-d). - Gary Detlefs, Jan 05 2023

a(n) is the number of binary strings of length n-2 whose longest run of 1's is of length 1, for n >= 3. - Félix Balado, Apr 03 2025

The 3x+1 or Collatz problem is as follows: start with any number n. If n is even, divide it by 2, otherwise multiply it by 3 and add 1. Do we always reach 1? This is a famous unsolved problem. It is conjectured that the answer is yes.

It seems that about half of the terms satisfy a(i) = a(i+1). For example, up to 10000000, 4964705 terms satisfy this condition.

n is an element of row a(n) in triangle A127824. - Reinhard Zumkeller, Oct 03 2012

The number of terms that satisfy a(i) = a(i+1) for i being a power of ten from 10^1 through 10^10 are: 0, 31, 365, 4161, 45022, 477245, 4964705, 51242281, 526051204, 5378743993. - John Mason, Mar 02 2018

5 seems to be the only number whose value matches its total number of steps (checked to n <= 10^9). - Peter Woodward, Feb 15 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

If b > 1 is a binary palindrome then both (2^(m+1) + 1)*b and 2^(m+1) + 2^m - b are also, where m = floor(log_2(b)). - Hieronymus Fischer, Feb 18 2012

Floor and ceiling: If d > 0 is any natural number, then A206913(d) is the greatest binary palindrome <= d and A206914(d) is the least binary palindrome >= d. - Hieronymus Fischer, Feb 18 2012

The greatest binary palindrome <= the n-th non-binary-palindrome is that binary palindrome with number A154809(n)-n+1. The corresponding formula identity is: A206913(A154809(n)) = A006995(A154809(n)-n+1). - Hieronymus Fischer, Mar 18 2012

From Hieronymus Fischer, Jan 23 2013: (Start)

The number of binary digits of a(n) is A070939(a(n)) = 1 + floor(log_2(n)) + floor(log_2(n/3)), for n > 1.

Also: A070939(a(n)) = A070939(n) + A070939(floor(n/3)) - 1, for n <> 2. (End)

Rajasekaran, Shallit, & Smith show that this is an additive basis of order 4. - Charles R Greathouse IV, Nov 06 2018

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

I conjecture that every number eventually appears - see A057167, A064227, A064228. - N. J. A. Sloane. That was written in 1991. Today I'm not so sure that every number appears. - N. J. A. Sloane, Feb 26 2017

As of Jan 25 2018, the first 13 missing numbers are 852655, 930058, 930557, 964420, 966052, 966727, 969194, 971330, 971626, 971866, 972275, 972827, 976367, ... For further information see the "Status Report" link. - Benjamin Chaffin, Jan 25 2018

From David W. Wilson, Jul 13 2009: (Start)

The sequence satisfies [1] a(n) >= 0, [2] |a(n)-a(n-1)| = n, and tries to avoid repeats by greedy choice of a(n) = a(n-1) -+ n.

This "wants" to be an injection on N = {0, 1, 2, ...}, as it attempts to avoid repeats by choice of a(n) = a(n-1) + n when a(n) = a(n-1) - n is a repeat.

Clearly, there are injections satisfying [1] and [2], e.g., a(n) = n(n+1)/2.

Is there a lexicographically earliest injection satisfying [1] and [2]? (End)

Answer: Yes, of course: The set of injections satisfying [1] and [2] is not empty, so there's a lexicographically least element. More concretely, it starts with the same 23 terms a(0..22) which are known to be minimal, but after a(22) = 41 it has to go on with a(23) = 41 + 23 = 64, since choosing "-" here necessarily yields a non-injective sequence. See A171884. - M. F. Hasler, Apr 01 2019

It appears that a(n) is also the value of "x" and "y" of the endpoint of the L-toothpick structure mentioned in A210606 after n-th stage. - Omar E. Pol, Mar 24 2012

Of course this is not a permutation of the integers: the first repeated term is 42 = a(24) = a(20). - M. F. Hasler, Nov 03 2014. Also 43 = a(18) = a(26). - Jon Perry, Nov 06 2014

Of all the sequences in the OEIS, this one is my favorite to listen to. Click the "listen" button at the top, set the instrument to "103. FX 7 (Echoes)", click "Save", and open the MIDI file with a program like QuickTime Player 7. - N. J. A. Sloane, Aug 08 2017

This sequence cycles clockwise around the OEIS logo. - Ryan Brooks, May 09 2020

For n > 3, gcd(a(n), a(n-1)) = 1 and gcd(a(n), a(n-2)) > 1. (This is just a restatement of the definition.)

This is now known to be a permutation of the natural numbers: see the 2015 article by Applegate, Havermann, Selcoe, Shevelev, Sloane, and Zumkeller.

From N. J. A. Sloane, Nov 28 2014: (Start)

Some of the known properties (but see the above-mentioned article for a fuller treatment):

1. The sequence is infinite. Proof: We can always take a(n) = a(n-2)*p, where p is a prime that is larger than any prime dividing a(1), ..., a(n-1). QED

2. At least one-third of the terms are composite. Proof: The sequence cannot contain three consecutive primes. So at least one term in three is composite. QED

3. For any prime p, there is a term that is divisible by p. Proof: Suppose not. (i) No prime q > p can divide any term. For if a(n)=kq is the first multiple of q to appear, then we could have used kp < kq instead, a contradiction. So every term a(n) is a product of primes < p. (ii) Choose N such that a(n) > p^2 for all n > N. For n > N, let a(n)=bg, a(n+1)=c, a(n+2)=dg, where g=gcd(a(n),a(n+2)). Let q be the largest prime factor of g. We know q < p, so qp < p^2 < dg, so we could have used qp instead of dg, a contradiction. QED

3a. Let a(n_p) be the first term that is divisible by p (this is A251541). Then a(n_p) = q*p where q is a prime less than p. If p < r are primes then n_p < n_r. Proof: Immediate consequences of the definition.

4. (From David Applegate, Nov 27 2014) There are infinitely many even terms. Proof:

Suppose not. Then let 2x be the maximum even entry. Because the sequence is infinite, there exists an N such that for any n > N, a(n) is odd, and a(n) > x^2.

In addition, there must be some n > N such that a(n) < a(n+2). For that n, let g = gcd(a(n),a(n+2)), a(n) = bg, a(n+1)=c, a(n+2)=dg, with all of b,c,d,g relatively prime, and odd.

Since dg > bg, d > b >= 1, so d >= 3. Also, g >= 3.

Since a(n) = bg > x^2, one of b or g is > x.

Case 1: b > x. Then 2b > 2x, so 2b has not yet occurred in the sequence. And gcd(bg,2b)=b > x > 1, gcd(2b,c)=1, and since g >= 3, 2b < bg < dg. So a(n+2) should have been 2b instead of dg.

Case 2: g > x. Then 2g > 2x, so 2g has not yet occurred in the sequence. And gcd(bg,2g)=g > 1, gcd(2g,c)=1, and since d >= 3, 2g < dg. So a(n+2) should have been 2g instead of dg.

In either case, we derive a contradiction. QED

Conjectures:

5. For any prime p > 97, the first time we see p, it is in the subsequence a(n) = 2b, a(n+2) = 2p, a(n+4) = p for some n, b, where n is about 2.14*p and gcd(b,p)=1.

6. The value of |{k=1,..,n: a(k)<=k}|/n tends to 1/2. - Jon Perry, Nov 22 2014 [Comment edited by N. J. A. Sloane, Nov 23 2014 and Dec 26 2014]

7. Based on the first 250000 terms, I conjectured on Nov 30 2014 that a(n)/n <= (Pi/2)*log n.

8. The primes in the sequence appear in their natural order. This conjecture is very plausible but as yet there is no proof. - N. J. A. Sloane, Jan 29 2015

(End)

The only fixed points seem to be {1, 2, 3, 4, 12, 50, 86} - see A251411. Checked up to n=10^4. - L. Edson Jeffery, Nov 30 2014. No further terms up to 10^5 - M. F. Hasler, Dec 01 2014; up to 250000 - Reinhard Zumkeller; up to 300000 (see graph) - Hans Havermann, Dec 01 2014; up to 10^6 - Chai Wah Wu, Dec 06 2014; up to 10^8 - David Applegate, Dec 08 2014.

From N. J. A. Sloane, Dec 04 2014: (Start)

The first 250000 points lie on about 8 roughly straight lines, whose slopes are approximately 0.467, 0.957, 1.15, 1.43, 2.40, 3.38, 5.25 and 6.20.

The first six lines seem well-established, but the two lines with highest slope at present are rather sparse. Presumably as the number of points increases, there will be more and more lines of ever-increasing slopes.

These lines can be seen in the Havermann link. See the "slopes" link for a list of the first 250000 terms sorted according to slope (the four columns in the table give n, a(n), the slope a(n)/n, and the number of divisors of a(n), respectively).

The primes (with two divisors) all lie on the lowest line, and the lines of slopes 1.43 and higher essentially consist of the products of two primes (with four divisors).

(End)

The eight roughly straight lines mentioned above are actually curves. A good fit for the "line" with slope ~= 1.15 is a(n)~=n(1+1.0/log(n/24.2)), and a good fit for the other "lines" is a(n)~= (c/2)*n(1-0.5/log(n/3.67)), for c = 1,2,3,5,7,11,13. The first of these curves consists of most of the odd terms in the sequence. The second family consists of the primes (c=1), even terms (c=2), and c*prime (c=3,5,7,11,13,...). This functional form for the fit is motivated by the observed pattern (after the first 204 terms) of alternating even and odd terms, except for the sequence pattern 2*p, odd, p, even, q*p when reaching a prime (with q a prime < p). - Jon E. Schoenfield and David Applegate, Dec 15 2014

For a generalization, see the sequence of monomials of primes in the comment in A247225. - Vladimir Shevelev, Jan 19 2015

From Vladimir Shevelev, Feb 24 2015: (Start)

Let P be prime. Denote by S_P*P the first multiple of P appearing in the sequence. Then

1) For P >= 5, S_P is prime.

Indeed, let

a(n-2)=v, a(n-1)=w, a(n)=S_P*P. (*)

Note that gcd(v,P)=1. Therefore, by the definition of the sequence, S_P*P should be the smallest number such that gcd(v,S_P) > 1.

So S_P is the smallest prime factor of v.

2) The first multiples of all primes appear in the natural order.

Suppose not. Then there is a pair of primes P < Q such that S_Q*Q appears earlier than S_P*P. Let

a(m-2)=v_1, a(m-1)=w_1, a(m)=S_Q*Q. (**)

Then, as in (*), S_Q is the smallest prime factor of v_1. But this does not depend on Q. So S_Q*P is a smaller candidate in (**), a contradiction.

3) S_P < P.

Indeed, from (*) it follows that the first multiple of S_P appears earlier than the first multiple of P. So, by 2), S_P < P.

(End)

For any given set S of primes, the subsequence consisting of numbers whose prime factors are exactly the primes in S appears in increasing order. For example, if S = {2,3}, 6 appears first, in due course followed by 12, 18, 24, 36, 48, 54, 72, etc. The smallest numbers in each subsequence (i.e., those that appear first) are the squarefree numbers A005117(n), n > 1. - Bob Selcoe, Mar 06 2015

As with decimal reversal, initial zeros are ignored; otherwise, the reverse of 1 would be 1000000... ad infinitum.

Numerators of the binary van der Corput sequence. - Eric Rowland, Feb 12 2008

It seems that in most cases A030101(x) = A000265(x) and that if A030101(x) <> A000265(x), the next time A030101(y) = A000265(x), A030101(x) = A000265(y). Also, it seems that if a pair of values exist at one index, they will exist at any index where one of them exist. It also seems like the greater of the pair always shows up on A000265 first. - Dylan Hamilton, Aug 04 2010

The number of occasions A030101(n) = A000265(n) before n = 2^k is A053599(k) + 1. For n = 0..2^19, the sequences match less than 1% of the time. - Andrew Woods, May 19 2012

For n > 0: a(a(n)) = n if and only if n is odd; a(A006995(n)) = A006995(n). - Juli Mallett, Nov 11 2010, corrected: Reinhard Zumkeller, Oct 21 2011

n is binary palindromic if and only if a(n) = n. - Reinhard Zumkeller, corrected: Jan 17 2012, thanks to Hieronymus Fischer, who pointed this out; Oct 21 2011

Given any n > 1, the set of numbers A030109(i) = (A030101(i) - 1)/2 for indexes i ranging from 2^n to 2^(n + 1) - 1 is a permutation of the set of consecutive integers {0, 1, 2, ..., 2^n - 1}. This is important in the standard FFT algorithms (starting or ending bit-reversal permutation). - Stanislav Sykora, Mar 15 2012

Row n of A030308 gives the binary digits of a(n), prepended with zero at even positions. - Reinhard Zumkeller, Jun 17 2012

The binary van der Corput sequence is the infinite sequence of fractions { A030101(n)/A062383(n), n = 0, 1, 2, 3, ... }, and begins 0, 1/2, 1/4, 3/4, 1/8, 5/8, 3/8, 7/8, 1/16, 9/16, 5/16, 13/16, 3/16, 11/16, 7/16, 15/16, 1/32, 17/32, 9/32, 25/32, 5/32, 21/32, 13/32, 29/32, 3/32, 19/32, 11/32, 27/32, 7/32, 23/32, 15/32, 31/32, 1/64, 33/64, 17/64, 49/64, ... - N. J. A. Sloane, Dec 01 2019

Record highs occur at n = A209492(m) (for n>=1) with values a(n) = A224195(m) (for n>=3). - Bill McEachen, Aug 02 2023

Essentially same as A091257 but computed starting from offset 0 instead of 1.

Each polynomial in GF(2)[X] is encoded as the number whose binary representation is given by the coefficients of the polynomial, e.g., 13 = 2^3 + 2^2 + 2^0 = 1101_2 encodes 1*X^3 + 1*X^2 + 0*X^1 + 1*X^0 = X^3 + X^2 + X^0. - Antti Karttunen and Peter Munn, Jan 22 2021

To listen to this sequence, I find instrument 99 (crystal) works well with the other parameters defaulted. - Peter Munn, Nov 01 2022