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## Sequence of the Day for April 19

A092287:
 n

 j  = 1
 n

 k  = 1
gcd (  j, k ), n   ≥   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
${\displaystyle {\begin{array}{l}{\displaystyle \operatorname {ord} _{p}\ a(n)=\sum _{k=1}^{\lfloor \log _{p}(n)\rfloor }\left\lfloor {\frac {n}{p^{k}}}\right\rfloor ^{2}=\left\lfloor {\frac {n}{p}}\right\rfloor ^{2}+\left\lfloor {\frac {n}{p^{2}}}\right\rfloor ^{2}+\left\lfloor {\frac {n}{p^{3}}}\right\rfloor ^{2}+\cdots .}\end{array}}}$

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

Comparing this with the de Polignac–Legendre formula for the prime factorization of
 n!
, i.e.
${\displaystyle {\begin{array}{l}{\displaystyle \operatorname {ord} _{p}\ n!=\sum _{k=1}^{\lfloor \log _{p}(n)\rfloor }\left\lfloor {\frac {n}{p^{k}}}\right\rfloor =\left\lfloor {\frac {n}{p}}\right\rfloor +\left\lfloor {\frac {n}{p^{2}}}\right\rfloor +\left\lfloor {\frac {n}{p^{3}}}\right\rfloor +\cdots ,}\end{array}}}$
this suggests that
 a (n)
can be considered as a "generalization of the factorial numbers" * (the product between braces is obviously 1 if
 n
is noncomposite)
${\displaystyle {\begin{array}{l}{\displaystyle {\frac {a(n)}{n!}}=\left(\,\prod _{k=1}^{n-1}\gcd(n,k)\right)^{2}{\frac {a(n-1)}{(n-1)!}},\quad n\geq 1.}\end{array}}}$

Recurrence:

${\displaystyle {\begin{array}{l}{\displaystyle a(0):=1;\ a(n):=n\left(\,\prod _{k=1}^{n-1}\gcd(n,k)\right)^{2}a(n-1),\quad n\geq 1.}\end{array}}}$

Formula:

${\displaystyle {\begin{array}{l}{\displaystyle a(n)=n!\left(\,\prod _{j=1}^{n}\prod _{k=1}^{j-1}\gcd(j,k)\right)^{2},\quad n\geq 0.}\end{array}}}$
A??????
(
 n

 j  = 1
 j  − 1

 k  = 1
gcd (  j, k )) 2, n   ≥   0
.
{1, 1, 1, 1, 4, 4, 576, 576, 147456, 11943936, 76441190400, 76441190400,...}
A224479
 n

 j  = 1
 j  − 1

 k  = 1
gcd (  j, k ), n   ≥   0
.
{1, 1, 1, 1, 2, 2, 24, 24, 384, 3456, 276480, 276480, 955514880, 955514880, ...}

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