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a(0) = 0 and for n > 0, if there are one or more k_i that are not already present in the sequence among terms a(0) .. a(n-1), and for which bitor(k_i,a(n-1)) = a(n-1), then a(n) = that k_i which gives minimum value of A019565(k_i) amongst them; otherwise, when no such k_i exists, a(n) = the least number not already present that can be obtained by cumulatively filling the successive vacant bits of a(n-1) from the least significant end of its binary representation (by toggling 0's to 1's, possibly also one or more leading zeros).

Shares with permutations like A003188, A006068, A300838, A302846, A303763, and A303767 the property that when moving from any a(n) to a(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 may occur at the same step.

A300829 gives the positions of records (terms of A000225 in ascending order), that for cases n=2..79 are followed by 2^(n-1).

The greedy algorithm which constructs this sequence can be understood also in terms of Heinz encodings of partitions (see A215366): Any term a(n) corresponds to a particular integer partition {s1+...+sk} via mapping a(n) = prime(s1) * ... * prime(sk), where s1 .. sk are the summands of an integer partition. The choices for constructing the next partition are: If by removing any parts from the partition we can find any smaller partitions that have not already occurred in the sequence, then we choose the one which has the smallest Heinz encoding value. On the other hand, if all partitions obtained by such removals have already occurred in the sequence, then we add to the current partition the least number of copies of the least positive integer that is not yet a part of the partition (A257993), until a partition is found which is not yet in the sequence.

From Antti Karttunen & David A. Corneth, May 01 - 04 2018: (Start)

No two successive descending terms, that is, a(n) > a(n+1) > a(n+2) never occurs.

For n > 1, if a(n) is odd then a(n-1) = 2^h * k * a(n) and a(n+1) = 2^j * a(n) for some h, k and j, that is, odd terms occur between two larger even numbers.

If a(n) < a(n+1) < a(n+2) then (a(n+1) / a(n)) is a divisor of a(n+2).

However, when a(n) < a(n+1) > a(n+2) then (a(n+1) / a(n)) might not be a divisor of a(n+2). The first such case occurs at n=64..66, as a(64) = 280 = 2^3 * 5 * 7, a(65) = 7560 = 2^3 * 3^3 * 5 * 7, and a(66) = 72 = 2^3 * 3^2. We have 7560/280 = 27, which is not a divisor of 72 (72/27 = 8/3).

In most cases, when a(n+1) < a(n) then gcd(a(n+1), a(n)/a(n+1)) = 1 (about 87% for the first 100000 descents). However, there are many exceptions to this, the first case occurring at a(65) = 7560 = 2^3 * 3^3 * 5 * 7 and a(66) = 72 = 2^3 * 3^2, with gcd(72,7560/72) = 3.

(End)

From David A. Corneth, May 04 2018: (Start)

The sequence can be partitioned into a tabf sequence with rows having the first element odd and the others even. It would give (1, 2, 6), (3, 12, 4, 36), (9, 18, 90), (5, 10, 30), (15, 60, 20, 180), (45, 360, 8, 24, 120, 40, 1080), (27, 54, 270), ...

It turns out that some rows are multiples of others; for example, the row (5, 10, 30) is five times the row (1, 2, 6). (End)

See also "observed scaling patterns" in the Formula section.

A303750 gives the positions of odd terms.

A282291 and A304531 are unitary divisor variants that satisfy the condition gcd(a(n+1), a(n)/a(n+1)) = 1, whenever a(n) > a(n+1).

The primes 2, 3, 5, 7, 11, 13, 19, 23 and 29 occur at positions 2, 4, 11, 42, 176, 1343, 8470, 57949, 302739, 1632898, thus after 7 and except for 13, a little earlier than they occur in variant A304531.

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

If a(n+1) > a(n), then A001222(a(n+1)) = 1 + A001222(a(n)).

From Antti Karttunen, May 23 2018: (Start)

For n >= 1, A006530(a(n)) = A000040(A070939(n)), thus the greatest prime dividing n, or equally, its index (A061395), is monotonic and follows the length of binary representation of n. This follows by induction on the size of the binary representation of n, and the fact that the "least possible unused divisor" part of a greedy rule can find all the unused divisors of A002110(k) before the next larger prime A000040(1+k) is needed as a factor.

For n >= 1, a((2^k)+1) = A000040(k+1), that is, after the first term with the next larger prime factor, which always occurs at 2^k, the next term is that prime itself, which is prime(k+1).

(A) For r in range 1 .. (2^(k-1)), a((2^k)+r) = A000040(k+1) * a(r-1), and prime A000040(k) is not present in the factorization. Because we cannot divide prime(k+1) out, as that would give a term already encountered, and because every term in this range has it as a largest prime factor, the relative magnitude-wise order of the terms in this range follows the relative magnitude-wise order of terms in a(0) .. a((2^(k-1))-1).

(B) For r in range (2^(k-1))+1 .. (2^k)-1, a((2^k)+r) = A000040(k+1) * a(r-1), and prime A000040(k) is present in the factorization.

Now it might be case that prime(k) > a product m of some subset of primes prime(k-1) .. prime(1). Even though the algorithm in those cases "would like" to divide by prime(k) instead of dividing by that product m, because then the divisor would be smaller, it cannot, because dividing by prime(k) (or by any other divisor containing it) would give an already used term.

(End)

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

The construction is otherwise like that of A303760, except here we choose the largest divisor instead of the smallest one. In contrast to A303760, this sequence is NOT permutation of A005117: 70 = A019565(13) is the first missing squarefree number. See also comments in A303769, A303749 and A302775.

Index of greatest prime factor of a(n) is monotonic and increments at n = {0, 1, 2, 4, 8, 15, 31, 50, 102, 157, 317, 480, 964, 1451, 2907, 4366, 8738, 13113, 26233, 39356, ...} - Michael De Vlieger, May 22 2018