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If, for some prime p, A045948(p) > p^2, then all members of the sequence are less than A003418(p). (Let p_(n) be a prime for which the inequality is satisfied, and let p_(n+1) be the smallest prime > (p_(n))^2. No number smaller than A003418(p_(n+1)) can belong to this sequence. However, for any p_(n) that satisfies the inequality, so does p_(n+1), leading to an endless cycle.) This inequality is first satisfied at p=53, as A045948(53)=5040 > 53^2=2809.

Proof: It follows from the definitions of p_(n) and p_(n+1), and from Bertrand's Postulate, that 2(A045948(p_(n))) > 2((p_(n))^2) > p_(n+1). Therefore 2((A045948(p_(n)))^2 > (p_(n+1))^2.

Since any prime that divides A003418(p_(n)) divides A003418(p_(n+1)) at least twice as often, A045948(p_(n+1)) cannot be less than the product of (A045948(p_n))^2 and A034386(p_(n)). (The latter term greatly exceeds 2 for any actual p_(n).)

Therefore A045948(p_(n+1)) > 2((A045948(p_n))^2 > (p_(n+1))^2, and p_(n+1) satisfies the inequality, implying that no number smaller than A003418(p_(n+2)) can belong to this sequence.

From Antti Karttunen, Feb 27 2014: (Start)

For all n >= 4, a(n) mod 10 = 2 (as A003418(5) = 60, the first multiple of ten in that sequence).

For all n >= 24, a(n) mod 100 = 62 (as A003418(25) = 26771144400, the first multiple of one hundred in that sequence).

Cf. also A236856.

a(n-1) gives the position of the first element of row n in irregular tables like A238280.

(End)

a(0)=0, as its only partition is an empty partition {}, and by convention lcm()=1, thus it takes no steps to reach from 1 to A003418(0)=1.

The records occur at positions 0, 2, 3, 5, 9, 11, 13, 19, 27, 31, 38, 43, 47, 61, 73, 81, ... and they seem to occur in order, i.e., as A001477. Thus the record-positions probably also give the left inverse function for this sequence. It also seems that each integer occurs only finite times in this sequence, so there should be a right inverse function as well.

It is known that this sequence is nonnegative for n >= 7. This can be established using the methods used to show A059794 is nonnegative. - Carl Pomerance, Bell Labs, Feb 16 2002

Also the intersection of A003418 and A007416.

Equivalently, for n >= 1, the least number > 1 of objects that can be arranged as a k-cube (k-dimensional hypercube) for all 1 <= k <= n.

a(7) = 2^420 contains 127 decimal digits.

From Jianing Song, Jul 20 2021: (Start)

Let F_q be the finite field with q elements, then F_a(n) is the smallest extension field of F_2 such that every polynomial of degree at most n splits into linear factors.

Union_{n>=0} F_a(n) is the algebraic clousre of F_2, which is the unique algebraically closed field with characteristic 2 and transcendence degree 0 (note that an algebraically closed field is uniquely determined by its characteristic and transcendence degree). Union_{n>=0} F_(2^(n!)) = Union_{n>=0} F_A050923(n) gives the same field.

Obviously, here 2 can be replaced by any prime p provided that {a(n)} is defined as a(n) = p^A003418(n). (End)