Tempted to add to the "Clever" pool given that Yuuka is canonically one of Millennium's bigger math geniuses.

From bottom to top:
240 likes
100 RT's
1000 likes
480 RT's
9999 likes
As of the present time, 2.7k likes and 465 RTs so cut line is up to Yuuka's navel

From bottom to top:

1) 5!*log_2(4) = 120 * 2 = 240

2) a_n is an easy polynomial integral.
Get a_n=1/(n+2)-1/(n+1).
The sum of these a_n's is a telescoping series,
with a closed form -1/2 + 1/12 = -5/12.
So get -240*(-5/12) = 100

3) Expand the denominator, then divide both the
top, bottom by 6^n. This gives:
(1-(5/6)^n+(1/6)^n)/(1-(2/3)^n-(1/2)^n+(1/3)^n).
Take the limit and the exponentials all die out, so get 1/1=1 as limit. So, 1000*1=1000

4) This main integral is a bit tricky.

We proceed with the u-substitution technique.
With preprocessing, we'll integrate (xsin(x))/(4-cos^2(x)) from [0, pi].

With u=cos(x), we then integrate the u-function arccos(u)/(4-u^2) from [1, -1].

This u-function is nice: it's an odd function offset pi/2 units down; and odd functions are nice: they integrate to 0 on symmetric intervals; ala [1, -1].

So, add-subtracting this offset and exploiting the odd function trick, we obtain the integral of (pi/2)/(4-u^2) from [1, -1]; easily solved via partial fraction decomposition.

It evaluates as pi*ln(3)/4, thus giving 480 RT's.

5) Need to solve for the value of a.

Solving for 'a' boils down to finding the limit of (x^2+1)*G(x)/x as x approaches 0.

(G(x) is the integral of cos(pi*x^2) from [1, x+1])

A big challenge with this limit is that G(x) can't be evaluated as an elementary antiderivative, so it has no nice closed form.

However, a very clever person may instantly recognize that this doesn't matter, as we don't have to evaluate this integral at all.

Why? This integral's bounds of integration closes down to the single point [1, 1] as x approaches 1, and
hence, the integral has to evaluate to 0 as x approaches 0.

This means the entire limit of (x^2+1)G(x)/x becomes an indeterminate form 0/0,
meaning L'hopital's rule is applicable, and we can circumvent integration entirely.

The 0/0 form comes from the G(x)/x expression. Using L'Hopital's rule and the fundemental theorem of calculus, we can evaluate the limit as g(x+1)/1.

(We denote g(x)=cos(pi*x^2), and note G'(x)=g(x) here)

So, we'll evaluate the limit expression to be g(1) = -1, so a = -3.

So, we'll get that 3333*f(a) = 9999.

As of present time, there are 486 RT's and 2.8k likes. So, we'll likely never get to see nipples, but at least get to see underbooba