cp's OEIS Frontend

In other words, we partition n into pairs of fibbinary numbers whose binary expansions have no common 1's and list the corresponding fibbinary numbers to get the n-th row.

If n = Product p_i^a_i, d = Product p_i^c_i is a unitary divisor of n if each c_i is 0 or a_i.

Also the number of squarefree divisors of n. - Labos Elemer

Also number of divisors of the squarefree kernel of n: a(n) = A000005(A007947(n)). - Reinhard Zumkeller, Jul 19 2002

Also shadow transform of pronic numbers A002378.

For n >= 1 define an n X n (0,1) matrix A by A[i,j] = 1 if lcm(i,j) = n, A[i,j] = 0 if lcm(i,j) <> n for 1 <= i,j <= n. a(n) is the rank of A. - Yuval Dekel (dekelyuval(AT)hotmail.com), Aug 11 2003

a(n) is also the number of solutions to x^2 - x == 0 (mod n). - Yuval Dekel (dekelyuval(AT)hotmail.com), Sep 21 2003

a(n) is the number of squarefree divisors of n, but in general the set of unitary divisors of n is not the set of squarefree divisors (compare the rows of A077610 and A206778). - Jaroslav Krizek, May 04 2009

Row lengths of the triangles in A077610 and in A206778. - Reinhard Zumkeller, Feb 12 2012

a(n) is also the number of distinct residues of k^phi(n) (mod n), k=0..n-1. - Michel Lagneau, Nov 15 2012

a(n) is the number of irreducible fractions y/x that satisfy x*y=n (and gcd(x,y)=1), x and y positive integers. - Luc Rousseau, Jul 09 2017

a(n) is the number of (x,y) lattice points satisfying both x*y=n and (x,y) is visible from (0,0), x and y positive integers. - Luc Rousseau, Jul 10 2017

Conjecture: For any nonnegative integer k and positive integer n, the sum of the k-th powers of the unitary divisors of n is divisible by the sum of the k-th powers of the odd unitary divisors of n (note that this sequence lists the sum of the 0th powers of the unitary divisors of n). - Ivan N. Ianakiev, Feb 18 2018

a(n) is the number of one-digit numbers, k, when written in base n such that k and k^2 end in the same digit. - Matthew Scroggs, Jun 01 2018

Dirichlet convolution of A271102 and A000005. - Vaclav Kotesovec, Apr 08 2019

Conjecture: Let b(i; n), n > 0, be multiplicative sequences for some fixed integer i >= 0 with b(i; p^e) = (Sum_{k=1..i+1} A164652(i, k) * e^(k-1)) * (i+2) / (i!) for prime p and e > 0. Then we have Dirichlet generating functions: Sum_{n > 0} b(i; n) / n^s = (zeta(s))^(i+2) / zeta((i+2) * s). Examples for i=0 this sequence, for i=1 A226602, and for i=2 A286779. - Werner Schulte, Feb 17 2022

The smallest integer with 2^m unitary divisors, or equivalently, the smallest integer with 2^m squarefree divisors, is A002110(m). - Bernard Schott, Oct 04 2022

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

This sequence can be used for filtering certain Stern polynomial (see A125184, A260443) related sequences, because it matches only with any such sequence b that can be computed as b(n) = f(A260443(n)), 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.

Some of these are listed on the last line ("Sequences that partition N into ...") of Crossrefs section.

To compute a(n): replace every other bit with zero (starting with the second bit) in each run of consecutive 1's in the binary expansion of n.

From Gus Wiseman, Jul 11 2025: (Start)

This is the greatest binary rank of a sparse subset of the binary indices of n, where:

1. The binary indices of a nonnegative integer are the positions of 1 in its reversed binary expansion.

2. A set is sparse iff 1 is not a first difference.

3. The binary rank of a set {S_1,S_2,...} is Sum_i 2^(S_i-1).

(End)