Digital Sum Fibonacci We are all familiar with the Fibonacci sequence:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765


However, instead of, f(n) = f(n-1) + f(n-2) we'll be taking digital sum of the previous 2 entries.  



The sequence should still start with 0, 1, after that the differences are rapidly apparent. This list is 0-indexed, you may use 1-indexed as well, state which you used.

f(0)  = 0
f(1)  = 1
f(2)  = 1   # 0 + 1
f(3)  = 2   # 1 + 1
f(4)  = 3   # 1 + 2
f(5)  = 5   # 2 + 3
f(6)  = 8   # 3 + 5
f(7)  = 13  # 8 + 5
f(8)  = 12  # 8 + 1 + 3
f(9)  = 7   # 1 + 3 + 1 + 2
f(10) = 10  # 1 + 2 + 7
f(11) = 8   # 7 + 1 + 0
f(12) = 9   # 1 + 0 + 8
f(13) = 17  # 8 + 9
f(14) = 17  # 9 + 1 + 7
f(15) = 16  # 1 + 7 + 1 + 7
f(16) = 15  # 1 + 7 + 1 + 6
f(17) = 13  # 1 + 6 + 1 + 5
f(18) = 10  # 1 + 5 + 1 + 3
f(19) = 5   # 1 + 3 + 1 + 0
f(20) = 6   # 1 + 0 + 5
f(21) = 11  # 5 + 6
f(22) = 8   # 6 + 1 + 1
f(23) = 10  # 1 + 1 + 8
f(24) = 9   # 8 + 1 + 0
f(25) = 10  # 1 + 0 + 9
f(26) = 10  # 9 + 1 + 0
f(27) = 2   # 1 + 0 + 1 + 0
(After this point it repeats at the 3rd term, 0-indexed)


Note: I didn't notice the repetition until I posted the challenge itself, and here I was thinking it'd be impossible to write another novel Fibonacci challenge.



Your task is, given a number n, output the nth digit of this sequence.

First 3 digits: [0,1,1],

24-digit repeated pattern: [2,3,5,8,13,12,7,10,8,9,17,17,16,15,13,10,5,6,11,8,10,9,10,10]

Hint: You may be able to exploit this repetition to your advantage.



This is code-golf, lowest byte-count is the winner.



BONUS: If you use the repetition in your answer, I will award the lowest byte-count answer that takes advantage of the repetition in the sequence a bounty of 100 points. This should be submitted as part of your original answer, after your original answer. See this post as an example of what I am talking about: https://codegolf.stackexchange.com/a/108972/59376

To qualify for this bonus your code must run in constant time (O(1)) with an explanation.

Bonus Winner: Dennis https://codegolf.stackexchange.com/a/108967/59376 &lt; Dennis won.

Most Unique Implementation: https://codegolf.stackexchange.com/a/108970/59376
(Also will receive 100 points, finalized after correct answer is chosen)
JavaScript (ES6), 45 bytes



f=(n,x=0,y=1)=&gt;n?f(n-1,y,(x%9||x)+(y%9||y)):x
&lt;input type=number min=0 oninput=o.textContent=f(this.value)&gt;&lt;pre id=o&gt;




x and y can't both be 9, since that would require the previous number to be 0, so their digital sum can't exceed 17. This means that the digital root for numbers greater than 9 is the same as the remainder modulo 9.