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After 1, contains only the least representatives of such prime signatures where the maximal exponent occurs more than once. However, here the terms are not in ascending order.

Meyers (see Guy reference) conjectures that for all r >= 1, the least odd number not in the set {a(i): i < prime(r)} is prime(r+1). - N. J. A. Sloane, Jan 08 2021

Meyers' conjecture would be refuted if and only if for some r there were such a large gap between prime(r) and prime(r+1) that there existed a composite c for which prime(r) < c < a(c) < prime(r+1), in which case (by Bertrand's postulate) c would necessarily be a term of A246281. - Antti Karttunen, Mar 29 2021

a(n) is odd for all n and for each odd m there exists a k with a(k) = m (see A064216). a(n) > n for n > 1: bijection between the odd and all numbers. - Reinhard Zumkeller, Sep 26 2001

a(n) and n have the same number of distinct primes with (A001222) and without multiplicity (A001221). - Michel Marcus, Jun 13 2014

From Antti Karttunen, Nov 01 2019: (Start)

More generally, a(n) has the same prime signature as n, A046523(a(n)) = A046523(n). Also A246277(a(n)) = A246277(n) and A287170(a(n)) = A287170(n).

Many permutations and other sequences that employ prime factorization of n to encode either polynomials, partitions (via Heinz numbers) or multisets in general can be easily defined by using this sequence as one of their constituent functions. See the last line in the Crossrefs section for examples.

(End)

All numbers of the form 2^k1*3^k2*...*p_n^k_n, where k1 >= k2 >= ... >= k_n, sorted.

A111059 is a subsequence. - Reinhard Zumkeller, Jul 05 2010

Choie et al. (2007) call these "Hardy-Ramanujan integers". - Jean-François Alcover, Aug 14 2014

The exponents k1, k2, ... can be read off Abramowitz & Stegun p. 831, column labeled "pi".

For all such sequences b for which it holds that b(n) = b(A046523(n)), the sequence which gives the indices of records in b is a subsequence of this sequence. For example, A002182 which gives the indices of records for A000005, A002110 which gives them for A001221 and A000079 which gives them for A001222. - Antti Karttunen, Jan 18 2019

The prime signature corresponding to a(n) is given in row n of A124832. - M. F. Hasler, Jul 17 2019

Numbers k such that there is no x < k such that A000005(x) = A000005(k). - Benoit Cloitre, Apr 28 2002

A047983(a(n)) = 0. - Reinhard Zumkeller, Nov 03 2015

Subsequence of A025487. If some m in A025487 is the first term in that sequence having its number of divisors, m is in this sequence. - David A. Corneth, Aug 31 2019

a(n) = Product of distinct primorials larger than one, obtained as Product_{i} A002110(1+i), where i ranges over the zero-based positions of the 1-bits present in the binary representation of n.

This sequence can be represented as a binary tree. Each child to the left is obtained as A283980(k), and each child to the right is obtained as 2*A283980(k), when their parent contains k:

1

|

...................2....................

6 12

30......../ \........60 180......../ \......360

/ \ / \ / \ / \

/ \ / \ / \ / \

/ \ / \ / \ / \

210 420 1260 2520 6300 12600 37800 75600

etc.

Numbers k for which A007814(A025487(k)) > A007949(A025487(k)), i.e., numbers k for which the 2-adic valuation of A025487(k) is larger than its 3-adic valuation.

Numbers k for which A181815(k) is even.

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)