The answer is
. To prove that one can't do better, note that it suffices to find a construction for

that has no path with more than

circles. This is doable by placing the red cells "diagonally" such that no path intersects more than one red cell in rows
,
, 
through
, ...,

to
. An example with

is shown below:
Now we show that

is always achievable. To do this, for
, let

be the maximum number of red circles in any path starting from the
th circle in the
th row and going to the bottom of the triangle. We have the recurrence
![\[a_{i,j} = \max\{a_{i+1,j}, a_{i+1,j+1}\} + \begin{cases*} 1 & the $j$th circle in the $i$th row is colored \\ 0 & else \end{cases*}\]](https://latex.artofproblemsolving.com/5/f/6/5f6c7b6d27d146836d19a0ef40148ac705931e7d.png)
and we wish to show that
.
To do this, our main claim is that
![\[\sum_{j=1}^{i} 2^{a_{i,j}} \geq \sum_{j=1}^{i+1} 2^{a_{i+1,j}}\]](https://latex.artofproblemsolving.com/2/9/d/29d3a1bab4142258b0a4beaa421aeee1d63e8f62.png)
for all
. This will finish, since we can compute that
, and chaining together these inequalities shows that
.
To prove the inequality, pick a specific
. For a given
, the number of

with

is strictly greater than the number of

with
, unless the former quantity is zero. Therefore

On the other hand, if

is the location of the colored circle in row
,
![\[ \sum_{j=1}^{i} 2^{a_{i,j}} -\sum_{j=1}^{i} 2^{\max\{a_{i+1,j}, a_{i+1,j+1}\}} = 2^{\max\{a_{i+1,j^*}, a_{i+1,j^*+1}\}} \geq 2^{\min_j \{a_{i+1,j}\}}.\]](https://latex.artofproblemsolving.com/0/1/1/01179a79a325e623276e46215a671383705bb494.png)
Subtracting these two inequalities yields the claim.
Solution 2
The answer is
.
Proof that
is achievable.
Consider the following process. Start with an array composed of

zeroes. At any point, if the current array is
, replace it with the

element array
![\[[b_1, b_2, \dots, b_{k - 1}] = [\max(a_1, a_2), \max(a_2, a_3), \dots, \max(a_{k - 1}, a_k)]\]](https://latex.artofproblemsolving.com/c/4/b/c4b72537d42f1ec367e160615c87e62851fe52e8.png)
Finally, if the row with

circles has a red circle at position
, increase

by one. It is straightforward to verify that at any point our
-element array gives, for each circle in the row with

circles, the maximum number of red circles that can be visited starting from said circle. For any array consider the quantity
.
Main claim.
.
Proof. Let

be such that
. Then we have

So before adding

to any element, the value of

is at least
. After adding

to say,
, the value of

increases by
and therefore
![\[f(b_1, \dots, b_{k - 1}) \ge f(a_1, \dots, a_k) - 2^{a_i} + 2^{a_j} \ge f(a_1, \dots, a_k)\]](https://latex.artofproblemsolving.com/d/8/c/d8c94365c7dd0777b8dcba9c40423205c30762da.png)
Since

was chosen to be minimal. This proves the claim.
Initially we have

and at the end we have a single number
. Therefore

and

as desired.
Proof that
is the best possible.
It suffices to exhibit a construction with

rows in which at most

red circles can be achieved, then prefixes of the triangle provide constructions for other
. For each red circle consider drawing the two diagonals parallel to the non-base sides of the triangles and recording which circles on the bottom row they touch. These form an interval of length

where

is the number of the row counting from the bottom. For the example in the statement we get the intervals
. Notice that the intervals uniquely determine the circles, and that we can visit circle

after circle

if and only if the interval corresponding to

contains the interval corresponding to
, therefore it suffices to give a collection of intervals of sizes

such that no

of them are nested. Consider now the collection
![\[
\begin{matrix}
[1, 1] & [2, 3] & [3, 5] & \dots & [2^{k - 1}, 2^k - 1] \\
[1, 2^{k - 1} + 1] & [2, 2^{k - 1} + 3] & [3, 2^{k - 1} + 5] & \dots & [2^{k - 2}, 2^k - 1] \\
[1, 2^{k - 1} + 2^{k - 2} + 1] & [2, 2^{k - 1} + 2^{k - 2} + 3] & [3, 2^{k - 1} + 2^{k - 2} + 5] & \dots & [2^{k - 3}, 2^k - 1] \\
\vdots & \vdots & \vdots & & \vdots \\
[1, 2^{k - 1} + 2^{k - 2} + \dots + 2 + 1]
\end{matrix}
\]](https://latex.artofproblemsolving.com/d/d/b/ddb3152c2af25b99da879018bf35a5bb1fe6f5b5.png)
We can verify that this arrangement contains all lengths of intervals from

to

going from left to right, top to bottom. Moreover, no two intervals in the same row contain each other, and therefore any sequence of nested intervals contains at most

of them (one from each row), as desired.