## The challenge

As the name may already reveal, it works basically like a Fibonacci, but summing the last 3 (instead of 2) numbers of the sequence to generate the next.

So, if we are to start our Tribonacci sequence with `[1, 1, 1]` as a starting input (AKA signature), we have this sequence:

 ``````1 `````` ``````[1, 1 ,1, 3, 5, 9, 17, 31, ...] ``````

But what if we started with `[0, 0, 1]` as a signature? As starting with `[0, 1]` instead of `[1, 1]` basically shifts the common Fibonacci sequence by once place, you may be tempted to think that we would get the same sequence shifted by 2 places, but that is not the case and we would get:

 ``````1 `````` ``````[0, 0, 1, 1, 2, 4, 7, 13, 24, ...] ``````

## Test cases

 `````` 1 2 3 4 5 6 7 8 9 10 11 `````` ``````Test.describe("Basic tests") Test.assert_equals(tribonacci([1, 1, 1], 10), [1, 1, 1, 3, 5, 9, 17, 31, 57, 105]) Test.assert_equals(tribonacci([0, 0, 1], 10), [0, 0, 1, 1, 2, 4, 7, 13, 24, 44]) Test.assert_equals(tribonacci([0, 1, 1], 10), [0, 1, 1, 2, 4, 7, 13, 24, 44, 81]) Test.assert_equals(tribonacci([1, 0, 0], 10), [1, 0, 0, 1, 1, 2, 4, 7, 13, 24]) Test.assert_equals(tribonacci([0, 0, 0], 10), [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]) Test.assert_equals(tribonacci([1, 2, 3], 10), [1, 2, 3, 6, 11, 20, 37, 68, 125, 230]) Test.assert_equals(tribonacci([3, 2, 1], 10), [3, 2, 1, 6, 9, 16, 31, 56, 103, 190]) Test.assert_equals(tribonacci([1, 1, 1], 1), [1]) Test.assert_equals(tribonacci([300, 200, 100], 0), []) Test.assert_equals(tribonacci([0.5, 0.5, 0.5], 30), [0.5, 0.5, 0.5, 1.5, 2.5, 4.5, 8.5, 15.5, 28.5, 52.5, 96.5, 177.5, 326.5, 600.5, 1104.5, 2031.5, 3736.5, 6872.5, 12640.5, 23249.5, 42762.5, 78652.5, 144664.5, 266079.5, 489396.5, 900140.5, 1655616.5, 3045153.5, 5600910.5, 10301680.5]) ``````

## The solution using Python

 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 `````` ``````def tribonacci(signature, n): # if less than 1, return a blank list if n<1: return [] # if `n` is less than the signature, # return a list at the item's place if n

## A more elegant solution

 ``````1 2 3 4 `````` ``````def tribonacci(signature, n): res = signature[:n] for i in range(n - 3): res.append(sum(res[-3:])) return res ``````