TL;DR: The possible strings of magical operations are:
(first) 143 -> 484 -> 968 -> 1837 -> 9218 -> 17347 (sixth)
(first) 89 -> 187 -> 968 -> 1837 -> 9218 -> 17347 (sixth)
Rigorous proof follows...
Let's first make a few starting observations regarding the left and right numbers for a given number (result) and the number it is created from by a magical operation (operand):
1) If the left number of the result is 1 then the left and right numbers of the operand must either sum to 10 or more, or sum to 9 with a 1 carried from the addition of any middle numbers.
2) If the left number of the result is 1 more than the right number, then the left and right numbers of the operand must sum to the right number of the result, with a 1 carried from the addition of any middle numbers.
3) If the left and right numbers of the result are equal, then the left and right numbers of operand must sum to one of them, and there is no carried 1 from the addition of any middle numbers.
Any combination of left and right numbers for the result that don't meet the above criteria can't be formed by any operand.
Starting with the sixth number as
17347, then we can say that the fifth number must be...
a 4-digit number, since there is no 5-digit number that can produce the sixth number after a magical operation. The only two single-digit numbers that can produce 17 are 8 and 9, so the fifth number has to be
9ab8. The 9 has to be on the left so the numbers that sum to 8 on the right (from the prior magical operation on the fourth number) can be incremented by a carried 1. We can also determine that
b have to sum to 3, and that the carried one that creates the 9 will also be added to
a one larger than
b. The only solution is
This gives us the fifth number:
This then leads us to the fourth number...
which also has to be a 4-digit number
abcd. We know that
d have to add to 8, and
c have to add to 11, and are therefore different numbers. Let's first consider the possible values for
d. The combinations
7bc1 are immediately out since there are no combination of numbers in a 3-digit or 4-digit third number that could make those. If it's
4bc4, then a 4-digit third number would have to be either
3gh1, and there are no combinations of
h that could produce two different values for
c (since there can be no carried 1).
The only option left for the fourth number is
1bc7. This would be created by a third number
z would have to add to create a
7 on the right and a
1 on the left, and we know from above this would mean it's
9y8. The value for
b could be either
y+y+1 is less than 10, or
y+y+1 is greater than 10. Since
b+c must be 11, the only combination that works is
This gives us the fourth number:
As well as the third number:
Now onto the second number(s)...
which have to be a 3-digit number
a+c equals 8 and the middle number is 8 to produce a carried 1. As above, the options
781 are out since there are no 2-digit or 3-digit first numbers that can result in those. We're left with
This gives us two possibilities for a second number:
And now onto the first number(s)...
For a second number
484 the first number must be a 3-digit number
a+c is 4 and
a must be strictly less than (not equal to)
c=3 satisfy this.
For a second number
187, only a first number
89 works, since
98 has a left number greater than the right.
This gives us two possibilities for a first number: