cp's OEIS Frontend

In other words, the n-th row gives the minimal partition of n into terms of A023758 (runs of consecutive 1's in binary).

Restricted growth sequence transform of the ordered pair [A069010(n), A324386(n)].

Restricted growth sequence transform of the ordered pair [A069010(n), A324055(n)].

In other words, the n-th row gives the numerical values of the runs of 1's in the binary expansion of n.

From Peter C. Heinig (algorithms(AT)gmx.de), Mar 08 2008: (Start)

This is also the number of maximal ideals of the ring (Z/nZ,+,*). Since every finite integral domain must be a field, every prime ideal of Z/nZ is a maximal ideal and since in general each maximal ideal is prime, there are just as many prime ideals as maximal ones in Z/nZ, so the sequence gives the number of prime ideals of Z/nZ as well.

The reason why this number is given by the sequence is that the ideals of Z/nZ are precisely the subgroups of (Z/nZ,+). Hence for an ideal to be maximal it has form a maximal subgroup of (Z/nZ,+) and this is equivalent to having prime index in (Z/nZ) and this is equivalent to being generated by a single prime divisor of n.

Finally, all the groups arising in this way have different orders, hence are different, so the number of maximal ideals equals the number of distinct primes dividing n. (End)

Equals double inverse Mobius transform of A143519, where A051731 = the inverse Mobius transform. - Gary W. Adamson, Aug 22 2008

a(n) is the number of unitary prime power divisors of n (not including 1). - Jaroslav Krizek, May 04 2009 [corrected by Ilya Gutkovskiy, Oct 09 2019]

Sum_{d|n} 2^(-A001221(d) - A001222(n/d)) = Sum_{d|n} 2^(-A001222(d) - A001221(n/d)) = 1 (see Dressler and van de Lune link). - Michel Marcus, Dec 18 2012

Up to 2*3*5*7*11*13*17*19*23*29 - 1 = 6469693230 - 1, also the decimal expansion of the constant 0.01111211... = Sum_{k>=0} 1/(10 ^ A000040(k) - 1) (see A073668). - Eric Desbiaux, Jan 20 2014

The average order of a(n): Sum_{k=1..n} a(k) ~ Sum_{k=1..n} log log k. - Daniel Forgues, Aug 13-16 2015

From Peter Luschny, Jul 13 2023: (Start)

We can use A001221 and A001222 to classify the positive integers as follows.

A001222(n) = A001221(n) = 0 singles out {1}.

Restricting to n > 1:

A001222(n)^A001221(n) = 1: A000040, prime numbers.

A001221(n)^A001222(n) = 1: A246655, prime powers.

A001222(n)^A001221(n) > 1: A002808, the composite numbers.

A001221(n)^A001222(n) > 1: A024619, complement of A246655.

n^(A001222(n) - A001221(n)) = 1: A144338, products of distinct primes. (End)

Inverse Möbius transform of the characteristic function of primes (A010051). - Wesley Ivan Hurt, Jun 22 2024

Dirichlet convolution of A010051(n) and 1. - Wesley Ivan Hurt, Jul 15 2025

The primes become the powers of 2 (2 -> 1, 3 -> 2, 5 -> 4, 7 -> 8); the composite numbers are formed by taking the values for the factors in the increasing order, multiplying them by the consecutive powers of 2, and summing. See the Example section.

From Antti Karttunen, Jun 27 2014: (Start)

The odd bisection (containing even terms) halved gives A244153.

The even bisection (containing odd terms), when one is subtracted from each and halved, gives this sequence back.

(End)

Question: Are there any other solutions that would satisfy the recurrence r(1) = 0; and for n > 1, r(n) = Sum_{d|n, d>1} 2^A033265(r(d)), apart from simple variants 2^k * A156552(n)? See also A297112, A297113. - Antti Karttunen, Dec 30 2017

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

This sequence can be used for filtering certain base-2 related sequences, because it matches only with any such sequence b that can be computed as b(n) = f(A005940(n+1)), where f(n) is any function that depends only on the prime signature of n (some of these are listed under the index entry for "sequences computed from exponents in ...").

Matching in this context means that the sequence a matches with the sequence b iff for all i, j: a(i) = a(j) => b(i) = b(j). In other words, iff the sequence b partitions the natural numbers to the same or coarser equivalence classes (as/than the sequence a) by the distinct values it obtains.

Because the Doudna map n -> A005940(1+n) is an isomorphism from "unary-binary encoding of factorization" (see A156552) to the ordinary representation of the prime factorization of n, it follows that the equivalence classes of this sequence match with any such sequence b, where b(n) is computed from the lengths of 1-runs in the binary representation of n and the order of those 1-runs does not matter. Particularly, this holds for any sequence that is obtained as a "Run Length Transform", i.e., where b(n) = Product S(i), for some function S, where i runs through the lengths of runs of 1's in the binary expansion of n. See for example A227349.

However, this sequence itself is not a run length transform of any sequence (which can be seen for example from the fact that A046523 is not multiplicative).

Furthermore, this matches not only with sequences involving products of S(i), but with any sequence obtained with any commutative function applied cumulatively, like e.g., A000120 (binary weight, obtained in this case as Sum identity(i)), and A069010 (number of runs of 1's in binary representation of n, obtained as Sum signum(i)).

A finite sequence is aperiodic if all of its cyclic rotations are distinct. See A000740 or A027375 for details.

Also numbers k such that the k-th composition in standard order is aperiodic. The k-th composition in standard order (graded reverse-lexicographic, A066099) 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 gives a bijective correspondence between nonnegative integers and integer compositions. - Gus Wiseman, Apr 28 2020