Note that the water is *not* transferred from container to container.

The additional water comes from an outside source.

Here is what I found . . .

The containers start like this:

Code:

| |
| |
| | | |
| | | |
| | | |
| | | |
t = 0 | | | |
| - - - | | | | |
|:::::::| | - - - | | |
|:::a:::| |:::b:::| | |
* - - - * * - - - * * - - - *
A B C

$\displaystyle b$ is added to container C.

Code:

| |
| |
| | | |
| | | |
| | | |
| | | |
t = 1 | | | |
| - - - | | | | |
|:::::::| | - - - | | - - - |
|:::a:::| |:::b:::| |:::b:::|
* - - - * * - - - * * - - - *
A B C

Then $\displaystyle b$ is added to container A.

Code:

| |
| |
| | | |
| | | |
| | | |
| | | |
t = 2 | - - - | | |
|:::::::| | | | |
|::a+b::| | - - - | | - - - |
|:::::::| |:::b:::| |:::b:::|
* - - - * * - - - * * - - - *
A B C

$\displaystyle b$ is added to Container C.

Code:

| |
| |
| | | |
| | | |
| | | |
| | | |
t = 3 | - - - | | - - - |
|:::::::| | | |:::::::|
|::a+b::| | - - - | |::2b:::|
|:::::::| |:::b:::| |:::::::|
* - - - * * - - - * * - - - *
A B C

Then $\displaystyle 2b$ is added to Container A.

Code:

| |
| |
| | | |
| | | |
| - - - | | |
|:::::::| | |
t = 4 |:::::::| | - - - |
|::a+3b:| | | |:::::::|
|:::::::| | - - - | |::2b:::|
|:::::::| |:::b:::| |:::::::|
* - - - * * - - - * * - - - *
A B C

$\displaystyle b$ is added to Container C.

Code:

| |
| |
| | | |
| | | |
| - - - | | |
|:::::::| | - - - |
t = 5 |:::::::| |:::::::|
|::a+3b:| | | |:::::::|
|:::::::| | - - - | |:::3b::|
|:::::::| |:::b:::| |:::::::|
* - - - * * - - - * * - - - *
A B C

Then $\displaystyle 3b$ is added to Container A.

Code:

| |
| |
| - - - | | |
|:::::::| | |
|:::::::| | |
|:::::::| | - - - |
t = 6 |::a+6b:| |:::::::|
|:::::::| | | |:::::::|
|:::::::| | - - - | |:::3b::|
|:::::::| |:::b:::| |:::::::|
* - - - * * - - - * * - - - *
A B C

We see a pattern developing.

. . $\displaystyle \begin{array}{cc}\text{Time} & \text{Container A} \\ \hline 0 & a \\ 2 & a+b \\ 4 & a+3b \\ 6 & a+6b \\ \vdots & \vdots \end{array}$

The coefficients of $\displaystyle b$ are triangular numbers: .$\displaystyle 0,1,3,6,10, 15, \hdots$

If time is an *even* number of seconds, $\displaystyle t\,=\,2k$

. . then: .$\displaystyle A \;=\;a + \frac{k(k+1)}{2}b$