Runs of rootable numbers

[Keiji]

Oh yea the red and blue A's in my signature are there to represent a mathematical pattern that is similar to the pattern of prime numbers in the sense that there is no (known) natural process that can specifically produce that pattern of numbers. The blue A's are to represent the numbers that can be rooted by another other than one whole number to get a whole number. The pattern includes the numbers 1, 4, 8, 9, 16, 25, 27, 32, and so on. Basically if you take a number and try rooting it by some of the lesser whole numbers and when one or more of it's whole number roots equals another whole number it will be represented by one of the blue A's in my signature so long as it's lesser than the total number of A's in my signature.

This inspired me to think a bit...

2^3 is 8, and 3^2 is 9. That's a run of two rootable numbers ("blue A's") in a row.
Does a run of three ever occur?
If so, does a run of four, five, six... ever occur?
Is it possible to find, somewhere, a run of any positive integer length?

I'd google this, but it's one of those things where you really don't know what to search for without already knowing the answer.

[Mrrl]

Question is - is a row "8,9" the last or there is at least one row of two after that?

[Keiji]

Yep - precisely what I was thinking...

[wendy]

In general, there are no runs of powers, separated by one. Certainly, there's no run of three powers.

It is known that there is no run of three powerful numbers, where a powerful number is one where if a prime divides it, so does its square. There are examples of two such numbers.

The run of two proper powers that differ by one, is generally considered unlikely. To see why, one needs to consider that the numbers adjacent to a proper power has algebraic factors, for (a^n)-1, one for each divisor of n, and for (a^n)+1, one for each divisor of 2n, that does not divide n. These are generally co-prime, except where a prime p divides both the factor for x and for a(x).

As an example, consider the pair 12167 and 12168. The first is 23^3, so we would find factors corresponding to 23A2 = 24, 23A6 = 507. A common divisor of 3 exists, because 3 divides 23A2, and hence 23A(2*3^n). The balance of factors give a cube (8) and a square (169). While 12168 is a powerful number (2^3*3^2*13^2), it's not a proper power.

One would look for instances where there is a large power in an algebraic root, for example, 7³ = 18²+18+1, but these are increasingly rare. A trawl for instances where p³ divides some algebraic root (any algebraic root), where p>b, suggests that apart from 113³ dividing 68A112, there is no other instance below 2,000,000 (confirmed: "bond.rex running over the output of '40 smallest sevenites'.)

It's a long call to suggest that large powers (one would suggest large prime powers, like 3^5 vs 5^2), /could/ exist, but one would find that very large primes dominate the picture.