A175391 -id:A175391 - cp's OEIS frontend

A297105 Smallest square k > 1 such that iterating the map (k -> number of divisors(k)) n times yields a square on each iteration.

4, 36, 1587600, 296110298950956428271894663224760734752656000000

Offset: 0

Jon E. Schoenfield, Dec 25 2017

nonn

Conjecture:
For n > 0, let F be the number of prime factors of a(n-1), counted with multiplicity. For j = 1..F, let p(j) be the j-th largest of the F prime factors, and let q(j) be the successor prime of p(j). Then a(n) = Product_{j=1..F} prime(j)^(q(j)-1).
E.g., a(2) = 1587600 = 7*7*5*5*3*3*3*3*2*2*2*2, so q(1)=q(2)=11, q(3)=q(4)=7, q(5)=q(6)=q(7)=q(8)=5, q(9)=q(10)=q(11)=q(12)=3; then a(3) = 2^10 * 3^10 * 5^6 * 7^6 * 11^4 * 13^4 * 17^4 * 19^4 * 23^2 * 29^2 * 31^2 * 37^2 = 296110298950956428271894663224760734752656000000.

			The smallest square larger than 1 is 4, so a(0) = 4.
The first several squares larger than 1 are k = 4, 9, 16, 25, 36, ..., for which the number of divisors tau(k) is 3, 3, 5, 3, 9, ..., respectively; the first of these numbers of divisors that is a square is tau(36) = 9 = 3^2, so a(1) = 36.
a(2) = 1587600 = 2^4 * 3^4 * 5^2 * 7^2, so tau(1587600) = (4+1)*(4+1)*(2+1)*(2+1) = 225 = 3^2 * 5^2, so tau(225) = (2+1)*(2+1) = 9 = 3^2.
a(3) = 296110298950956428271894663224760734752656000000; tau(tau(tau(296110298950956428271894663224760734752656000000))) = tau(tau(300155625)) = tau(225) = 9 = 3^2.

Cf. A000005, A175391.

cp's OEIS Frontend

A297105 Smallest square k > 1 such that iterating the map (k -> number of divisors(k)) n times yields a square on each iteration.

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