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A few preliminary pages (most of these need to be updated!)
Sequence of the Day for April 19
A092287:
.

{ 1, 1, 2, 6, 96, 480, 414720, 2903040, ... } 
Peter Bala conjectures that the order of the
primes in the
prime factorization of
is given by the formula
 ${\begin{array}{l} \operatorname {ord} _{p}\ a(n)=\sum _{k=1}^{\lfloor \log _{p}(n)\rfloor }\left\lfloor {\frac {n}{p^{k}}}\right\rfloor ^{2}=\left\lfloor {\frac {n}{p}}\right\rfloor ^{2}+\left\lfloor {\frac {n}{p^{2}}}\right\rfloor ^{2}+\left\lfloor {\frac {n}{p^{3}}}\right\rfloor ^{2}+\cdots .}\end{array}}$
Charles R Greathouse IV proved Bala's conjecture very recently.
Comparing this with the
de Polignac–Legendre formula for the prime factorization of
, i.e.
 ${\begin{array}{l} \operatorname {ord} _{p}\ n!=\sum _{k=1}^{\lfloor \log _{p}(n)\rfloor }\left\lfloor {\frac {n}{p^{k}}}\right\rfloor =\left\lfloor {\frac {n}{p}}\right\rfloor +\left\lfloor {\frac {n}{p^{2}}}\right\rfloor +\left\lfloor {\frac {n}{p^{3}}}\right\rfloor +\cdots ,}\end{array}}$
this suggests that
can be considered as a "generalization of the
factorial numbers" * (the product between braces is obviously
1 if
is
noncomposite)
 ${\begin{array}{l} {\frac {a(n)}{n!}}=\left(\,\prod _{k=1}^{n1}\gcd(n,k)\right)^{2}{\frac {a(n1)}{(n1)!}},\quad n\geq 1.}\end{array}}$
Recurrence:
 ${\begin{array}{l} a(0):=1;\ a(n):=n\left(\,\prod _{k=1}^{n1}\gcd(n,k)\right)^{2}a(n1),\quad n\geq 1.}\end{array}}$
Formula:
 ${\begin{array}{l} a(n)=n!\left(\,\prod _{j=1}^{n}\prod _{k=1}^{j1}\gcd(j,k)\right)^{2},\quad n\geq 0.}\end{array}}$
A??????
( gcd ( j, k ))^{ 2}, n ≥ 0 
.

{1, 1, 1, 1, 4, 4, 576, 576, 147456, 11943936, 76441190400, 76441190400,...}
A224479 .

{1, 1, 1, 1, 2, 2, 24, 24, 384, 3456, 276480, 276480, 955514880, 955514880, ...}

* See GCD matrix generalization of the factorial.