Implement the Fibonacci sequence... Shifted to the right

Question

Background

We all know the famous Fibonacci sequence, where each term is the sum of the previous two, with the first term being 0 and the second one being 1. The first few elements are:

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

What we are going to implement today is a similar sequence, starting at 0 and 1 as well, and each term is the sum of the previous two, but the sum's digits are shifted by one place to the right. What does "shifted" mean? – A digit shifted to the right is basically that digit + 1, unless it equals 9, which maps to 0 instead. Just as a reference, here is the full list of mappings:

0 -> 1;  1 -> 2;  2 -> 3;  3 -> 4;  4 -> 5;  5 -> 6;  6 -> 7;  7 -> 8;  8 -> 9;  9 -> 0

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With all that being said, these are the first terms of our Fibonacci-shifted sequence:

0, 1, 2, 4, 7, 22, 30, 63, 4, 78, 93, 282, 486, 879, 2476, 4466, 7053, 22620, 30784, 64515

Given an integer N (non-negative if 0-indexed and positive if 1-indexed), finds the Nth term of the Fibonacci-shifted sequence.

Rules

• The output may contain leading zero(es), in case 9 is the first (few) digit(s).

• Input and output can be handled using any default mean.

• Take note that Loopholes are forbidden.

• This is code-golf, hence your program will be scored in bytes, with less bytes being better.

Answer 1

Python 2, 78 72 bytes

f=lambda n:n>1and int(''.join(`-~int(i)%10`for i in`f(n-1)+f(n-2)`))or n

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Saved 6 bytes thanks to Halvard Hummel.

Answer 2

Jelly, 9 bytes

ð+D‘%⁵Ḍø¡

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-2 thanks to Dennis.

Explanation:

ð+D‘%⁵Ḍø¡ Full program. Takes input through STDIN.
ð         Start new dyadic chain
 +        Add the left (f(n - 1)) and right (f(n - 2)) arguments
  D       Get list of decimal digits
   ‘      Increment each
    %⁵    Modulo ⁵ (10)
      Ḍ   Convert from decimal to integer
       ø  Start new niladic chain
        ¡ Repeat previous chain z times (where z is taken from STDIN as there are no arguments)

Answer 3

JavaScript (ES6), 63 62 bytes

This is what I'd come up with before the challenge was deleted, don't have time to golf it further now.

Returns the nth number in the sequence, 0-indexed.

f=(n,x=0,y=1)=>n?f(--n,y,+[...x+y+""].map(z=>++z%10).join``):x

• 1 byte saved thanks to Conor.

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Alternative, 55 bytes

A quick port of VoteToReopen's excellent Python solution.

f=n=>n>1?+[...f(--n)+f(--n)+""].map(x=>++x%10).join``:n

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Answer 4

J, 40 bytes

[`(-&2(10|>:)&.(10&#.inv)@+&$:<:)@.(1&<)

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Highlighted following is the standard Fibonacci husk:

[`(-&2                    +&$:<:)@.(1&<)

Then, we compose the following function with the result of adding the previous two terms:

      (10|>:)&.(10&#.inv)@

This, in turn, is the function 10|>: atop 10&#.inv, which gets the digits of a number. That is to say, the digits are obtained, then the function 10|>: is applied, then the digits are combined to make a number again. 10|>: is a fork that is equivalent to (y + 1) % 10 in a conventional language for input y.