cp's OEIS Frontend

From David A. Corneth, Dec 29 2019: (Start)

Each term is a perfect square. Proof: A050377(n) is multiplicative with a(p^e) = A018819(e) and A018819(2k) = A018819(2k+1) and this sequence considers just records so we only need exponents of the form 2k; i.e., terms are squares.

Furthermore, the exponent 2 occurs at most once in the prime factorization of a(n) as A018819(2)^2 = A018819(4) = 4. So if the last two exponents in the prime factorization of m are 2's then setting the first of those two exponents to 4 and the other to 0 gives the same A050377(m).

Example of an application of this proof: we have 3600 = 2^4 * 3^2 * 5^2. We see the last two exponents are 2's so we can set the first of those two to 4 and the second to 0. This gives 2^4 * 3^4 = 1296 and, indeed, A050377(1296) = A050377(3600) = 16.

It seems that most exponents of a(n) are divisible by 4.

More specifically: Let S(n) be the list, possibly with duplicates, of exponents occurring in the prime factorizations of terms with the sum of exponents in the prime factorization <= n.

Let R(n) = |{x : x==4, S(n)}| / |S(n)|.

For example, S(8) is found from the following terms: 4, 16, 64, 144, 256, 576 and 1296 as the exponents in the prime factorization are (2), (4), (6), (4, 2), (8), (6, 2), (4, 4). The sums of each of these exponents per term is <= 8. There are 10 exponents listed. Of these 10 there are 5 that are divisible by 4. Therefore R(8) = 5/10.

Then it seems that R(n) tends to some value > 0.8 as n grows. (End)

Like terms in A330687, also these are all found in A025487.

Every number n is a product of a unique subset of these numbers. This is sometimes called the Fermi-Dirac factorization of n (see A182979). Proof: In the prime factorization n = Product_{j>=1} p(j)^e(j) expand every exponent e(j) as binary number and pick the terms of this sequence corresponding to the positions of the ones in binary (it is clear that both n and n^2 have the same number of factors in this sequence, and that each factor appears with exponent 1 or 0).

Or, a(1) = 2; for n>1, a(n) = smallest number which cannot be obtained as the product of previous terms. This is evident from the unique factorization theorem and the fact that every number can be expressed as the sum of powers of 2. - Amarnath Murthy, Jan 09 2002

Except for the first term, same as A084400. - David Wasserman, Dec 22 2004

The least number having 2^n divisors (=A037992(n)) is the product of the first n terms of this sequence according to Ramanujan.

According to the Bose-Einstein distribution of particles, an unlimited number of particles may occupy the same state. On the other hand, according to the Fermi-Dirac distribution, no two particles can occupy the same state (by the Pauli exclusion principle). Unique factorizations of the positive integers by primes (A000040) and over terms of A050376 one can compare with two these distributions in physics of particles. In the correspondence with this, the factorizations over primes one can call "Bose-Einstein factorizations", while the factorizations over distinct terms of A050376 one can call "Fermi-Dirac factorizations". - Vladimir Shevelev, Apr 16 2010

The numbers of the form p^(2^k), where p is prime and k >= 0, might thus be called the "Fermi-Dirac primes", while the classic primes might be called the "Bose-Einstein primes". - Daniel Forgues, Feb 11 2011

In the theory of infinitary divisors, the most natural name of the terms is "infinitary primes" or "i-primes". Indeed, n is in the sequence, if and only if it has only two infinitary divisors. Since 1 and n are always infinitary divisors of n>1, an i-prime has no other infinitary divisors. - Vladimir Shevelev, Feb 28 2011

{a(n)} is the minimal set including all primes and closed with respect to squaring. In connection with this, note that n and n^2 have the same number of factors in their Fermi-Dirac representations. - Vladimir Shevelev, Mar 16 2012

In connection with this sequence, call an integer compact if the factors in its Fermi-Dirac factorization are pairwise coprime. The density of such integers equals (6/Pi^2)*Product_{prime p} (1+(Sum_{i>=1} p^(-(2^i-1))/(p+1))) = 0.872497... It is interesting that there exist only 7 compact factorials listed in A169661. - Vladimir Shevelev, Mar 17 2012

The first k terms of the sequence solve the following optimization problem:

Let x_1, x_2,..., x_k be integers with the restrictions: 2<=x_1A064547(Product{i=1..k} x_i) >= k. Let the goal function be Product_{i=1..k} x_i. Then the minimal value of the goal function is Product_{i=1..k} a(i). - Vladimir Shevelev, Apr 01 2012

From Joerg Arndt, Mar 11 2013: (Start)

Similarly to the first comment, for the sequence "Numbers of the form p^(3^k) or p^(2*3^k) where p is prime and k >= 0" one obtains a factorization into distinct factors by using the ternary expansion of the exponents (here n and n^3 have the same number of such factors).

The generalization to base r would use "Numbers of the form p^(r^k), p^(2*r^k), p^(3*r^k), ..., p^((r-1)*r^k) where p is prime and k >= 0" (here n and n^r have the same number of (distinct) factors). (End)

The first appearance of this sequence as a multiplicative basis in number theory with some new notions, formulas and theorems may have been in my 1981 paper (see the Abramovich reference). - Vladimir Shevelev, Apr 27 2014

Numbers n for which A064547(n) = 1. - Antti Karttunen, Feb 10 2016

Lexicographically earliest sequence of distinct nonnegative integers such that no term is a product of 2 or more distinct terms. Removing the distinctness requirement, the sequence becomes A000040 (the prime numbers); and the equivalent sequence where the product is of 2 distinct terms is A026416 (without its initial term, 1). - Peter Munn, Mar 05 2019

The sequence was independently developed as a multiplicative number system in 1985-1986 (and first published in 1995, see the Uhlmann reference) using a proof method involving representations of positive integers as sums of powers of 2. This approach offers an arguably simpler and more flexible means for analyzing the sequence. - Jeffrey K. Uhlmann, Nov 09 2022

Length of row n equals A056170(n) if A056170(n) is positive, or 1 if A056170(n) = 0.

The multiset of exponents >=2 in the prime factorization of n completely determines a(n) for over 20 sequences in the database (see crossreferences). It also determines the fractions A034444(n)/A000005(n) and A037445(n)/A000005(n).

For squarefree numbers, this multiset is { } (the empty multiset). The use of 0 in the table to represent each n with no exponents >=2 in its prime factorization accords with the usual OEIS practice of using 0 to represent nonexistent elements when possible. In comments, the second signature of squarefree numbers will be represented as { }.

For each second signature {S}, there exist values of j and k such that, if the second signature of n is {S}, then A085082(n) is congruent to j modulo k. These values are nontrivial unless {S} = { }. Analogous (but not necessarily identical) values of j and k also exist for each second signature with respect to A088873 and A181796.

Each sequence of integers with a given second signature {S} has a positive density, unlike the analogous sequences for prime signatures. The highest of these densities is 6/Pi^2 = 0.607927... for A005117 ({S} = { }).

Apparently the (ordinary) Dirichlet inverse of A050377. - R. J. Mathar, Jul 15 2010

Also analog of Liouville's function (A008836) in Fermi-Dirac arithmetic, where the terms of A050376 play the role of primes (see examples). - Vladimir Shevelev, Oct 28 2013.

First differs from A335428 at n = 36. Differs from A050377, A344417 and A347437 at n = 1 and then at n = 36.

In the unique factorization of n in terms of distinct "Fermi-Dirac primes", n is represented as a product of prime powers (A246655) whose exponents are powers of 2 (A000079). a(n) is the maximal exponent of these prime powers (and not the maximal exponent of the exponents that are powers of 2). Thus, a(n) is a power of 2 for n >= 2.

First differs from A384913 at n = 64.