Pairs of Bogotá Numbers

Partial answer, and other findings:

I'm going to call the number that generates a Bogotá number a Bogotá root.

i) Find the next five pairs of consecutive numbers consisting of two Bogotá numbers.

The first eight pairs are:

ii) Are there infinitely many such pairs?

Possible, but they seem fairly sparse.

iii) Can arbitrarily long sets of consecutive numbers be found all of which are Bogotá numbers?

The only sequence of more than two I can suggest twists the definition a bit:

Other observations

There are a number of Bogotá numbers with multiple Bogotá roots. There are 3905 numbers with multiple Bogotá roots where the roots are under 1000000. The first 10 are:

192 <- 24 32
648 <- 36 81
819 <- 91 117
1197 <- 133 171
1536 <- 48 64
4872 <- 87 174
4977 <- 79 711
5976 <- 166 332
7056 <- 98 441
9968 <- 178 712

and a few more with more Bogotá roots:

549504 <- 1696 2862 3392 3816
1798848 <- 6246 12492 33312
4193856 <- 19416 21843 29124
4804128 <- 4766 16681 21447
5827584 <- 8672 17344 182112
7426944 <- 7368 12894 14736
1578092544 <- 86976 97848 342468 913248

Some patterns of Bogotá numbers:

No Bogotá root contains a 0. These all generate 0, which violates the definition.

Any number composed of only the digit 1 is a Bogotá number. These are also their own Bogotá root.

Any number composed of any number of the digit 2 and one digit 4 is a Bogotá numbers. Similarly, for the digits 3 and 9, and for the digit 4 and digit sequence 56. Thus, the following are Bogotá numbers: 4, 9, 56, 222222422, 93333333333, 445644444444.

There are similar patterns for Bogotá roots composed of all but one digits being 1.

Edit:

I did some work generating odd Bogotá numbers trying to find consecutive odd Bogotá numbers. Based on the definition, all the digits in the root must be odd, which made the search space somewhat smaller. I found three: (9,11), (8197,8199), and (11977,11979). None of these is part of a consecutive pair. This was for Bogotá roots up to 10 billion, except for 56 that overflowed a 64 bit integer.

Since getting 4 or more consecutive Bogotá numbers requires two consecutive odd ones, I think it unlikely to find longer runs.