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For n > 0, the least k such that for at least n-1 iterations of map x -> A003961(x), starting from x=k, x stays nondeficient. In other words, from each a(n) starts a chain of at least n nondeficient numbers (A023196) obtained by successive prime shifts, e.g, for a(3) we have: 19399380 -> 334639305 -> 5391411025, where -> stands for applying A003961, the prime shift towards larger primes.

After 1 all other terms here are even, because if an odd number k is nondeficient, then A064989(k) is nondeficient also, where A064989 is the prime shift towards smaller primes. Moreover, because A047802 is defined for every n >= 0, also this sequence is.

From Peter Munn, Aug 13 2020 (Start)

Upper bounds for a(4) and a(5) are:

a(4) <= 195534950863140268380 = A064989(A064989(A064989(20169691981106018776756331))) = A337202(3).

a(5) <= 538938984694949877040715541221415046162838700 = A064989^4((A047802(4)*17*19)/137).

(End)

From David A. Corneth, Aug 21 2020: (Start)

Subsequence of A025487.

Let prime(n)# be the n-th primorial number, A002110(n) = A034386(prime(n)). Then:

a(6) <= 191# * 7#;

a(7) <= 311# * 5#;

a(8) <= 457# * 5#.

(End)

That each term occurs in A025487 follows because (1), the abundancy index of prime(i)^e is larger than that of prime(i+1)^e, that is, sigma(prime(i)^e)/prime(i)^e > sigma(prime(i+1)^e)/prime(i+1)^e, and (2) because the abundancy index of p^(e+d) * q^e is larger than that of p^e * q^(e+d), where p and q are distinct primes, p < q, and e, d > 0. Thus, for any n, we can first find a "prime-factorization compressed version" of it, A071364(n), and then sort the exponents to the non-ascending order with A046523 (and actually, A046523(A071364(n)) = A046523(n), so we need to apply just A046523), to get a term x of A025487, that certainly have the abundancy index >= n [and this inequivalence stays same for their successive prime shifts as well, the abundancy index of A003961(x) being at least that of A003961(n), etc.], and as A046523(n) <= n for all n, it is guaranteed that the least k for which A336835(k) >= n are found from A025487, which is the range of A046523.

a(n) is the product of elements of the multiset that covers an initial interval of positive integers with multiplicities equal to the parts of the n-th composition in standard order (graded reverse-lexicographic, A066099). This composition is obtained by taking the set of positions of 1's in the reversed binary expansion of n, prepending 0, taking first differences, and reversing again. For example, the 13th composition is (1,2,1) giving the multiset {1,2,2,3} with product 12, so a(13) = 12. - Gus Wiseman, Apr 26 2020

Restricted growth sequence transform of function f(n) = A046523(A329044(n)).

For all i, j:

A305800(i) = A305800(j) => a(i) = a(j),

a(i) = a(j) => A324888(i) = A324888(j),

a(i) = a(j) => A329046(i) = A329046(j).

Restricted growth sequence transform of function f(n) = A246277(A329044(n)).

For all i, j:

A305800(i) = A305800(j) => a(i) = a(j),

a(i) = a(j) => A329045(i) = A329045(j),

a(i) = a(j) => A329343(i) = A329343(j),

a(i) = a(j) => A329348(i) = A329348(j),

a(i) = a(j) => A329349(i) = A329349(j).

From Gus Wiseman, Aug 05 2021 and Antti Karttunen, Oct 13 2021: (Start)

Also the product of primes at even positions in the weakly decreasing list (with multiplicity) of prime factors of n. For example, the prime factors of 108 are (3,3,3,2,2), with even bisection (3,2), with product 6, so a(108) = 6.

Proof: A108951(n) gives a number with the same largest prime factor (A006530) and its exponent (A071178) as in n, and with each smaller prime p = 2, 3, 5, 7, ... < A006530(n) having as its exponent the partial sum of the exponents of all prime factors >= p present in n (with primes not present in n having the exponent 0). Then applying A000188 replaces each such "partial sum exponent" k with floor(k/2). Finally, A319626 replaces those halved exponents with their first differences (here the exponent of the largest prime present stays intact, because the next larger prime's exponent is 0 in n). It should be easy to see that if prime q is not present in n (i.e., does not divide it), then neither it is present in a(n). Moreover, if the partial sum exponent of q is odd and only one larger than the partial sum exponent of the next larger prime factor of n, then q will not be present in a(n), while in all other cases q is present in a(n). See also the last example.

(End)

The primorial inflation of n, A108951(n), divided by its largest squarefree divisor, which is also its largest primorial divisor.

A permutation of A181818.

A squarefree analog of A302783. Each term is either a divisor or a multiple of the next one. In contrast to A302033 at each step the previous term can be multiplied (or divided), not just by a single prime, but possibly by a product of several distinct ones, A019565(A000975(k)). E.g., a(3) = 3, a(4) = 2*5*a(3) = 30. - Antti Karttunen, Apr 17 2018

a(n) = 2 for nonsquare n. - David A. Corneth, Jul 24 2017