Given $f(i, j) = f(i − 1, j − 1) + f(i, j − 1)$, find the value of $f(1009, 2019)$
I programmed the sequence to see if there is any pattern. Below is the $13 \times 13$ matrix including enough initial values to notice the pattern to be proven by induction after all:
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
[1, 2, 4, 7, 11, 16, 22, 29, 37, 46, 56, 67, 79]
[1, 2, 4, 8, 15, 26, 42, 64, 93, 130, 176, 232, 299]
[1, 2, 4, 8, 16, 31, 57, 99, 163, 256, 386, 562, 794]
[1, 2, 4, 8, 16, 32, 63, 120, 219, 382, 638, 1024, 1586]
[1, 2, 4, 8, 16, 32, 64, 127, 247, 466, 848, 1486, 2510]
[1, 2, 4, 8, 16, 32, 64, 128, 255, 502, 968, 1816, 3302]
[1, 2, 4, 8, 16, 32, 64, 128, 256, 511, 1013, 1981, 3797]
[1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1023, 2036, 4017]
[1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2047, 4083]
[1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4095]
[1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096]
So, the pattern I noticed is as follows: $$f(m,n) = \begin{cases} \displaystyle\sum\limits_{k=0}^{m} {n \choose k},\ \ m < n \\[1mm] 2^n, \qquad \ \ \ \ ~ m\ge n \end{cases}$$ Next, we should prove it by induction.
The case $m < n$ (which is more interesting since $1009 < 2019$):
$\square$ We have enough information for the base case (see the table above). Now assume that $$f(m-1,n-1) = \sum_{k=0}^{m-1}{n-1 \choose k} $$ and $$ f(m,n-1) = \sum_{k=0}^{m}{n-1 \choose k}$$ are true.
We should show that $f(m-1,n-1) + f(m,n-1) = f(m,n)$ is found by the same formula; that is, we are to prove $$\sum_{k=0}^{m-1}{n-1 \choose k} + \sum_{k=0}^{m}{n-1 \choose k} = \sum_{k=0}^{m} {n \choose k}$$
Indeed, $$ \begin{align} \sum_{k=0}^{m-1}{n-1 \choose k} + \sum_{k=0}^{m}{n-1 \choose k} &= \sum_{k=1}^{m}{n-1 \choose k-1} + {n-1\choose 0} + \sum_{k=1}^{m}{n-1 \choose k} \\[1mm] &= {n-1\choose 0} + \sum_{k=1}^{m}\left[{n-1 \choose k-1} +{n-1 \choose k}\right] \\[1mm] &\overset*= {n\choose 0} + \sum_{k=1}^{m}{n\choose k} \\[1mm] &= \sum_{k=0}^m {n\choose k} \end{align} $$ where (*) denotes the Pascal's formula. $\blacksquare$
The case $m \ge n$:
$\square$ The base case and the induction hypothesis are analogous and the induction step is trivially proven by noting that $$\underbrace{2^{n-1}}_{f(m-1,n-1)}+\underbrace{2^{n-1}}_{f(m,n-1)} = \underbrace{2^n}_{f(m,n)}. \ \ \blacksquare$$
Finally, we note that $$\boxed{f(1009, 2019) = \sum_{k=0}^{1009}{2019 \choose k}}$$
Here is a generating function approach. We consider the recurrence relation \begin{align*} f(i,j)&=f(i-1,j-1)+f(i,j-1)\qquad\qquad i,j\geq 1\tag{1}\\ f(i,0)&=f(0,j)=1\qquad\qquad\qquad\qquad\qquad\quad i,j\geq 0 \end{align*}
We set \begin{align*} \color{blue}{F(x,y)=\sum_{i=0}^\infty\sum_{j=0}^\infty f(i,j)x^iy^j} \end{align*} and use guided from (1) the Ansatz \begin{align*} \sum_{i=1}^\infty\sum_{j=1}^\infty f(i,j) x^iy^j=\sum_{i=1}^\infty\sum_{j=1}^\infty f(i-1,j-1) x^iy^j+\sum_{i=1}^\infty\sum_{j=1}^\infty f(i,j-1) x^iy^j\tag{2} \end{align*}
Left-hand side of (2):
We obtain \begin{align*} \sum_{i=1}^\infty&\sum_{j=1}^\infty f(i,j) x^iy^j\\ &=\sum_{i=0}^\infty\sum_{j=0}^\infty f(i,j) x^iy^j-\sum_{i=0}^\infty f(i,0)x^i-\sum_{j=0}^\infty f(0,j)y^j+f(0,0)\\ &=F(x,y)-\frac{1}{1-x}-\frac{1}{1-y}+1\tag{3} \end{align*}
Right-hand side of (2):
We obtain \begin{align*} \sum_{i=1}^\infty&\sum_{j=1}^\infty f(i-1,j-1) x^iy^j+\sum_{i=1}^\infty\sum_{j=1}^\infty f(i,j-1) x^iy^j\\ &=\sum_{i=0}^\infty\sum_{j=0}^\infty f(i,j) x^{i+1}y^{j+1}+\sum_{i=1}^\infty\sum_{j=0}^\infty f(i,j) x^{i}y^{j+1}\\ &=xyF(x,y)+yF(x,y)-\sum_{j=0}^\infty f(0,j)y^{y+1}\\ &=xyF(x,y)+yF(x,y)-\frac{y}{1-y}\tag{4} \end{align*}
From (3) and (4) we get:
\begin{align*} F(x,y)-\frac{1}{1-x}-\frac{1}{1-y}+1&=xyF(x,y)+yF(x,y)-\frac{y}{1-y}\\ F(x,y)(1-y(1+x))&=\frac{1}{1-x}\\ \color{blue}{F(x,y)}&\color{blue}{=\frac{1}{1-x}\,\frac{1}{1-y(1+x)}}\tag{5} \end{align*}
The final step is to extract $f(i,j)=[x^iy^j]F(x,y)$ where we use the coefficient of operator $[x^i]$ to denote the coefficient of $x^i$ of a series.
We obtain from (5) for $i,j\geq 0$: \begin{align*} \color{blue}{f(i,j)}&=[x^iy^j]\frac{1}{1-x}\,\frac{1}{1-y(1+x)}\\ &=[x^iy^j]\sum_{k=0}^\infty y^k(1+x)^k\,\frac{1}{1-x}\tag{6}\\ &=[x^i](1+x)^j\,\frac{1}{1-x}\tag{7}\\ &=[x^i](1+x)^j\sum_{k=0}^\infty x^k\tag{8}\\ &=\sum_{k=0}^i[x^{i-k}](1+x)^j\tag{9}\\ &=\sum_{k=0}^i\binom{j}{i-k}\tag{10}\\ &\color{blue}{=\begin{cases} \sum_{k=0}^i\binom{j}{k}&\qquad j>i\\ 2^j&\qquad j\leq i \end{cases}}\tag{11} \end{align*}
Comment:
In (6) we use the geometric series expansion.
In (7) we select the coefficient of $y^j$.
In (8) we expand $\frac{1}{1-x}$.
In (9) we use the linearity of the coefficient of operator and apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$. We also set the upper limit of the series to $i$ since other terms do not contribute.
In (10) we selet the coefficient of $x^{i-k}$.
In (11) we change the order of summation $k\to i-k$ and we use the binomial identity $\sum_{k=0}^j\binom{j}{k}=2^j$ as well as $\binom{j}{k}=0$ if $k>j$.
A slightly simpler solution based on VIVID's work (it's essentially the same).
If we utilize the generalized binomial coefficients where $$\binom{n}{k} = 0, n, k \in \mathbb Z, k > n \text { or } k<0$$ then VIVID's formula becomes $$f(m,n) = \sum_{k=0}^m \binom{n}{k} \tag 1$$
We also know the following recursive relation is true for these generalized coefficients as well: $$\binom{n}{k} = \binom{n-1}{k} + \binom{n-1}{k-1} \tag 2$$
Then, since $f$ is uniquely determined by $$f(0,x)=f(x,0)=1 \tag 3$$ and $$f(i, j) = f(i − 1, j − 1) + f(i, j − 1) \tag 4$$ we only need to show $(3)$ and $(4)$ are true when we plug in VIVID's formula $(1)$. Now, $(3)$ is obviously true. For $(4)$, indeed we have (via $(2)$)
$$f(i,j) = \sum_{k=0}^i \binom{j}{k} = \sum_{k=0}^i \binom{j-1}{k-1} + \sum_{k=0}^i \binom{j-1}{k} \\ = \sum_{k=1}^i \binom{j-1}{k-1} + f(i,j-1)\\ = \sum_{k=0}^{i-1} \binom{j-1}{k} + f(i,j-1)\\ = f(i-1,j-1) + f(i,j-1).\blacksquare $$