cp's OEIS Frontend

Every number can be represented uniquely as a product of numbers of the form p^(2^k), sequence A050376. This sequence is a binary representation of this factorization, with a(p^(2^k)) = 2^(i-1), where i is the index (A302778) of p^(2^k) in A050376. Additive with a(p^e) = sum a(p^(2^e_k)) where e = sum(2^e_k) is the binary representation of e and a(p^(2^k)) is as described above. - Franklin T. Adams-Watters, Oct 25 2005 - Index offset corrected by Antti Karttunen, Apr 17 2018

With n having a unique factorization as A050376(i) * A050376(j) * ... * A050376(k), with i, j, ..., k all distinct, a(n) = A050376(1+i) * A050376(1+j) * ... * A050376(1+k).

Multiplicative because for coprime m and n the Fermi-Dirac factorizations of m and n are disjoint and their union is the Fermi-Dirac factorization of m * n. - Andrew Howroyd, Aug 02 2018

With n having a unique factorization as fdp(i) * fdp(j) * ... * fdp(k), with i, j, ..., k all distinct, a(n) = fdp(i-1) * fdp(j-1) * ... * fdp(k-1), where fdp(0) = 1 and fdp(n) = A050376(n) for n >= 1.

Multiplicative because for coprime m and n the Fermi-Dirac factorizations of m and n are disjoint and their union is the Fermi-Dirac factorization of m * n. - Andrew Howroyd, Aug 02 2018

Shares with A064736, A207901, A302781, A302350, etc. a property that a(n) is always either a divisor or a multiple of a(n+1). However, because multiple bits may change simultaneously when moving from A006068(n) to A006068(n+1) [with the restriction that the changing bits are all either toggled on or all toggled off], it means that also here the terms might be divided or multiplied by more than just a single Fermi-Dirac prime (A050376). E.g. a(3) = 3, while a(4) = A050376(1) * A050376(3) * 3 = 2*4*3 = 24. See also comments in A284003.

Consider A019565. Imagine that it is an automatic piano that "plays sequences" when an appropriate punched "tape" is fed to it (as its input), i.e., when it is composed from the right with an appropriate sequence p, as A019565(p(n)). The 1-bits in the binary expansion of each p(n) are the "holes" in the tape, and they determine which "tunes" are present on beat n. The "tunes" are actually primes that are multiplied together. Of course only "squarefree music" (sequences containing only squarefree numbers, A005117) is possible to generate this way, thus we call A019565 a "squarefree piano".

There is a more sophisticated instrument, called "Fermi-Dirac piano" (A052330), with which it is possible to produce sequences that may contain any numbers.

If the tape is constructed in such a way that between the successive beats (when moving from p(n) to p(n+1)), either a subset of 0-bits are toggled on (changed to 1's), or a subset of 1-bits are toggled off (changed to 0's), but no both kind of changes occur simultaneously, then when fed as an input to either of these pianos, the resulting sequence "s" (the output) is guaranteed to satisfy the condition that s(n+1) is either a multiple or a divisor of s(n). For example, Gray code A003188 and its inverse A006068 are examples of such tapes, and they produce sequences A302033, A207901 and A284003, A302783.

This divisor-or-multiple permutation is obtained by playing "Fermi-Dirac piano" with the same tape which yields A303760 when "squarefree piano" is played with it. Note that A303760 is not a subsequence of this sequence, as its terms occur in different order than the squarefree terms here.

See also Peter Munn's Apr 11 2018 message on SeqFan-mailing list and comments in A304537.

Each a(n) is always either a divisor or a multiple of a(n+1).

A052330 is not exactly a permutation because it has offset 0 but doesn't contain 0. A052331 is its exact inverse, which has offset 1 and contains 0. - David Wasserman, Feb 24 2005

For all i, j: a(i) = a(j) => A000120(i) = A000120(j).

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