extend Oeis A152211 for more consecutive increasing terms A152211 is a very uneven sequence !
If you have powerful computations resources, find a longest consecutive list of increasing terms
What algorithm could we use ?
 A: This is a comment
rather than an answer.
The sequence is
$s(n)
=n\sigma_0(n)+\sigma_1(n)
$
where
$\sigma_k(n)
=\sum_{d|n} d^k
$,
so that
$s(n)
=\sum_{d|n} (n+d)
$.
Actually,
I think a more interesting sequence
would be
$t(n)
=n\sigma_0(n)-2\sigma_1(n)
=\sum_{d|n} (n-2d)
$
as I think its
fluctuations around zero
would be complicated.
Here are the first 15 values
according to Wolfy:
{{1, -1}, {2, -2}, {3, -2}, {4, -2}, {5, -2}, {6, 0}, {7, -2}, {8, 2}, {9, 1}, {10, 4}, {11, -2}, {12, 16}, {13, -2}, {14, 8}, {15, 12}}.
Note that
for prime $p$,
$s(p)
=p\sigma_0(p)+\sigma_1(p)
=2p+p+1
=3p+1
$
and
$t(p)
=2p-2(p+1)
=-2
$.
Note that
the average order
of
$\sigma_0(n)$
is
$\ln(n)
$
and
the average order
of
$\sigma_1(n)$
is
$n\pi^2/6
$
(see https://en.wikipedia.org/wiki/Average_order_of_an_arithmetic_function),
so,
if the two sums are correlated,
the average value of
$s(n)$
is
$n(\ln(n)+\pi^2/6)
$
and
the average value of
$t(n)$
is
$n(\ln(n)-\pi^2/3)
$.
From looking at
the plot produced
by Wolfy,
I would conjecture that
$t(n) \ge 0$
if $n$ is not a prime.
