Let's call this as the Mohanty's Triangle!

In any $n$ sided Mohanty's triangle, any number in the triangle is $4$ times of the number exactly above it. If there are $2015$ numbers in first row, the middle number is $1008$ . To obtain the number in the last row, we need to multiply the $1008$ by $4$ $1007$ times because there are $2014$ rows remaining out of which half the number have numbers exactly below $1008$ . Thus the number is $1008\times {4}^{1007}\\\huge\color{#3D99F6}{=\boxed{{2}^{2018}.{3}^{2}.{7}^{1}}}$

The bottom number in Mohanty's triangle starting with the numbers 1 through N on the top row comes out to be $2^{N-2}\cdot (N+1)$ . I originally saw this by looking at the leftmost entries of the kth row (which correspond to the bottom number of the triangle in the 1 through k case) and finding a pattern, then proving the general case via induction. The expression evaluates to $2^{2013} \cdot 2016$ $= 2^{2013} \cdot 2^5 \cdot 3^2 \cdot 7^1$ $= 2^{2018} \cdot 3^2 \cdot 7^1.$ This yields the correct answer of 2033.

Let's consider the sequence of integers from $1$ to $3$ .

If you write the triangle in terms of the elements in the first row, you will get this:

${ 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3 \\ \ \ \ \ \ \ 1+2 \ \ \ \ \ \ \ \ \ 2+3 \\ \ \ \ \ \ \ \ \ 1+2 \cdot 2+3 }$

I'm considering a small case but if you keep it going for sequences from $1$ to $4$ and from $1$ to $5$ for example, we can see that for a sequence from $1$ to $n$ , the bottom number follows the formula:

${ \sum\limits_{i=0}^{n - 1} {n-1 \choose i} \cdot (i + 1) }$

In this case, the bottom number is ${ {2 \choose 0} \cdot (1) + {2 \choose 1} \cdot (2) + {2 \choose 2} \cdot (3) = 8 }$ .

If we simply focus on just the first and the last numbers of each row downward, we find that:

Left-hand side: 1, 3, 8, 20,... Right-hand side: 2015, 4029, 8056, 16108,...

with each side defined by the following recursions (for $n \ge 2$ ):

LHS: $a_n = 2a_{n-1} + 2^{n-2}; a_1 = 1$

RHS: $b_n = 2b_{n-1} - 2^{n-2}; b_1 = 2015$

Let us now define a new sequence ${u_n}$ such that $u_n = a_n + b_n; u_1 = a_1 + b_1 = 2016.$ Solving the recursion $u_n = 2u_{n-1} \Rightarrow u_n = A\cdot2^n$ , which the initial condition yields:

$2016 = A\cdot2^{1} \Rightarrow A = 1008$

or $u_n = 1008\cdot2^{n}.$ The bottom-most vertex value occurs at $n = 2014$ , which will give the prime factorization:

$u_{2014} = 1008\cdot2^{2014} = (2^{4}3^{2}7^{1})\cdot2^{2014} = 2^{2018}3^{2}7^{1}.$

Hence, the required final sum is $2 + 2018 + 3 + 2 + 7 + 1 = \boxed{2,033}.$

Adding the first and last number of each row and writing them we get a g.p
(2016 , 4032 , 8064.........) with common ratio 2. Last term of it would give the lowest vertex number ( in the second last row we would have only two numbers, adding them would give the vertex no.) nth term of g.p = $a\times r^{n-1}$
so, 2014th term of g.p = $2016\times 2^{2013}$ = $(32 × 9 × 7)\times 2^{2013}$ = $2^{2018}\times 3^{2}\times 7$ now, 2+3+7+2018+2+1 = 2033