Up-side-down Triangle

Flip the triangle from left to right,(the first row becomes $2018,2017,2016,...3,2,1$ ),and add the two triangles together,it become $2019\quad2019\quad2019\quad2019\quad\cdots\quad2019\quad2019\quad2019\quad2019\\4038\quad4038\quad4038\quad\quad\cdots\quad\quad\quad4038\quad4038\quad4038\\8076\quad8076\quad\quad\quad\cdots\quad\quad\quad\quad8076\quad8076\\\vdots\quad\quad\\2n\quad\quad$ The $m$ th row is all $2019\times2^{m-1}$ ,and $2n$ is on the $2018$ th row,so $2n=2019\times2^{2018-1},n=2019\times2^{2016}$

If it was a Computer Science problem, the following would be one of the ways to find out $a$ and $b$ and / or $a+b$ ;

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# initially a=[1,2, ..., 2018]
a = [i+1 for i in range(2018)]

# replace [1,2, ..., 2018] with [1+2, ..., 2017+2018] and repeat until we have only one number
while (len(a) is not 1):
    a = [a[i]+a[i+1] for i in range(len(a)-1)]

a, b = a[0], 0 # a=n=a[0] and b=0 for now
# find the max b such that 2^b < n and remaining factor of n, i.e. a, would either be a prime number or 1
while(not a%2):
    a, b = a//2, b+1

print ("(a, b) = {} => a+b = {}".format((a, b), a+b))

(a, b) = (2019, 2016) => a+b = 4035

I wonder how algebra would help to find it out! (^,^)

We note that the first row of the table has 2018 terms, the second row has 2017, third row, 2016 and so on. Therefore, the $k$ th row has $2019-k$ terms and the 2018th row has only 1 term. Now if we add the first term and last term of each we get the following:

$\begin{array} {rl} \text{Row 1:} & 1 + 2018 = 2019 = 2019 \times 2^0 \\ \text{Row 2:} & 3 + 4035 = 4038 = 2019 \times 2^1 \\ \text{Row 3:} & 8 + 8068 = 8076 = 2019 \times 2^2 \\ \cdots & \qquad \qquad \qquad \cdots \\ \text{Row k:} & a_{k,1} + a_{k, 2019-k} = 2019 \times 2^{k-1} \\ \cdots & \qquad \qquad \qquad \cdots \\ \text{Row 2017:} & a_{2017,1} + a_{2017, 2} = 2019 \times 2^{2016} \end{array}$

We note that the 2017th row has two terms and the sum of the two terms equals to $n$ , the only term of the 2018th row. Therefore, $n = a_{2017,1} + a_{2017, 2} = 2019 \times 2^{2016}$ and $a+b = 2019+2016 = \boxed{4035}$ .

Nice problem.