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:

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[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:

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

Test cases

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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

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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<len(signature):
        return signature[0:n]
    
    # counter
    inc = 0
    
    # copy the signature list as a starting point
    seq = signature[:]
    
    # increment
    while inc<=n:
        # add up the last 3 items
        add = sum(seq[inc:inc+3])
        # add to the new list
        seq.append(add)
        # next!
        inc += 1
        
    # return the new list,
    # forcing to only the max size wanted
    return seq[0:n]

A more elegant solution

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