How can we make all ten numerals appear as the last digit in two consecutive Fibonacci numbers?

Question

In a previous question it was determined that only 1, 3, 7 and 9 could be the last digit in two consecutive Fibonacci numbers.

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It was assumed that the first two numbers were 1, 1 (or 0, 1).

More generally any sequence where the nth term is the sum of the previous terms is a Fibonacci sequence. The first two numbers can be any integer.

Name a starting pair for each numeral to be the last digit in two consecutive Fibonacci numbers. (No one pair can produce all numerals as last digit in two consecutive Fibonacci numbers.)

A good answer will tell all the numerals to fit the condition for that starting pair and why the other numerals are not the last digit in two consecutive Fibonacci numbers for that starting pair.

Bonus question: Find a starting pair that produce no numeral as the last digit in two consecutive Fibonacci numbers. Tell why.

Answer 1

The sequence starting with 2, 8 produces all even digits greater than zero doubled up:

2

8

10

18

28

46

74

120

194

314

508

822

1330

2152

3482

5634

9116

14750

23866

38616

All numbers in the sequence are even, so there are no odd terms, therefore an odd digit cannot be found at the end of a number once, let alone twice in a row.

As you have already found, the sequence starting 0, 1 produces all odd digits except 5 doubled up. From the last question you also have there the reason why it does not contain any even digits at the end twice in a row.

For a sequence to include a double 0, it must start 0, 0, and it will carry on with an infinite string of them.

For a sequence to include a double 5, it must start 5, 5 or 0, 5 or 5, 0:

5, 5, 10, 15, 25, 40...

0, 5, 5, 10, 15, 25...

5, 0, 5, 5, 10, 15...

The sequence 1, 3 does not have any zeroes in it, and therefore no doubled up digits. The last digits for this sequence repeat in the pattern:

[1, 3, 4, 7, 1, 8, 9, 7, 6, 3, 9, 2], [1, 3, 4, 7...

Answer 2

sorry. i have no answer for the real question. but for bonus Q, (2,1) could be one of the answers. for 60-recurrence of Fib's sequence, (2,1) sequence only have 12.

2,1,3,4,7,1,8,9,7,6,3,9 (12-recurrence).

it has no 5s or 0s, and without 0s, we cannot have twin consecutives.

answers :

(2,1) or (1,3) or (3,4) or (4,7) or (7,1) or (1,8) or (8,9) or (9,7) or (7,6) or (6,3) or (3,9) or (9,2).