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## Sequence of the Day for April 19

A092287: $\prod_{j = 1}^n \prod_{k = 1}^n \gcd(j, k),\ n \ge 0.$

 { 1, 1, 2, 6, 96, 480, 414720, 2903040, ... }

Peter Bala conjectures that the order of the primes in the prime factorization of a(n) is given by the formula

$\operatorname{ord}_p\ a(n) = \sum_{k = 1}^{\lfloor \log_p(n) \rfloor} \left\lfloor\frac{n}{p^k}\right\rfloor ^2 = \left\lfloor \frac{n}{p} \right\rfloor ^2 + \left\lfloor \frac{n}{p^2} \right\rfloor ^2 + \left\lfloor \frac{n}{p^3} \right\rfloor ^2 + \cdots .$

Charles R Greathouse IV proved Bala's conjecture very recently.

Comparing this with the de Polignac–Legendre formula for the prime factorization of n!, i.e.

$\operatorname{ord}_p\ n! = \sum_{k = 1}^{\lfloor \log_p(n) \rfloor} \left\lfloor\frac{n}{p^k}\right\rfloor = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \cdots ,$

this suggests that a(n) can be considered as a generalization of the factorial numbers (the product between braces is obviously 1 if n is noncomposite)

$\frac{a(n)}{n!} = \left( \prod_{k = 1}^{n-1} \gcd(n, k) \right)^2 \frac{a(n-1)}{(n-1)!},\quad n \ge 1.$

Recurrence:

$a(0) := 1;\ a(n) := n \left( \prod_{k = 1}^{n-1} \gcd(n, k) \right)^2 a(n-1),\quad n \ge 1.$

Formula:

$a(n) = n! \left( \prod_{j = 1}^{n} \prod_{k = 1}^{j-1} \gcd(j, k) \right) ^2,\quad n \ge 0.$