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Let f(n) be the solution to this problem if there are n frogs. We have f(1) = 4 and f(2) = 11. Now assume we add one more frog and one more toad. We obtain the recursion
f(n+1) = 2f(n) + 3 - f(n-1).
It is possible to get a closed form equation for f(n), or we can just find f(2), f(3), etc.
| f(1) | 4 |
| f(2) | 11 |
| f(3) | 21 |
| f(4) | 34 |
| f(5) | 50 |
| f(6) | 69 |
| f(7) | 91 |
| f(8) | 116 |
| f(9) | 144 |
| f(10) | 175 |
| f(11) | 209 |
| f(12) | 246 |
| f(13) | 286 |
| f(14) | 329 |
| f(15) | 375 |
| f(16) | 424 |
| f(17) | 476 |
| f(18) | 531 |
| f(19) | 589 |
| f(20) | 650 |
| f(21) | 714 |
| f(22) | 781 |
| f(23) | 851 |
| f(24) | 924 |
| f(25) | 1000 |
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