Would this be an interesting sequence?

[Chris Clark]

So, straight off I am no mathematician but I am a passable, albeit very old school software guy. Started off doing 6502 assembler in school computer club, early 80's.

Recently I became interested in numbers that are multiply perfect powers or inversely put, numbers that have many integer roots. I set out to try and find numbers that have more than a handful of integer roots. Wrote some code, clocked up a fair bit of cpu time and noticed that numbers of the the form (some small prime)^n were good candidates for long chains of roots. 
Seeing this I changed my approach from testing n^2 , n^3,n^4... n^34 where n=1 to 10E34 to p^n, p^n+1, p^n+2 where  p=2,3,5,7,9,11,13,17,19 and n=1-1e6 (so far) This soon started yielding numbers that were moderately large and had chains of > 100 integer roots.

Searching OEIS for the series of actual integers seems a little daunting given the smallest has 15,172 decimal digits but searching the individual terms expressed as a power of 2 has yielded nothing resembling what I am doing here.

Should I continue working along these lines to submit a sequence?

[Jeffrey Shallit]

What is a number with "many integer roots" or a "chain of integer roots"?  This seems rather undefined to me.   Can you give some specific examples or a rigorous definition?   What are the "integer roots" of 3^5?

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[Chris Clark]

Perhaps an example is best: 

4096 = 2^12 = 4^6 = 8^4  = 16^3
or
1073741824  = 32768^2 = 1024^3 = 64^5 = 32^6 = 8^10 =4^15 = 2^30
or a slightly larger one
5444517870735015415413993718908291383296  =73786976294838206464^2 = 17592186044416^3 = 8589934592^4 = 4194304^6 = 4096^11 = 2048^12 = 64^22 = 16^33 = 8^44 = 4^66 = 2^132

Does that help?

C

[M F Hasler]

On Wed, Oct 29, 2025 at 5:40 PM Chris Clark <chri...@gmail.com> wrote:

Perhaps an example is best: 

4096 = 2^12 = 4^6 = 8^4  = 16^3
or
1073741824  = 32768^2 = 1024^3 = 64^5 = 32^6 = 8^10 =4^15 = 2^30
or a slightly larger one
5444517870735015415413993718908291383296  =73786976294838206464^2 = 17592186044416^3 = 8589934592^4 = 4194304^6 = 4096^11 = 2048^12 = 64^22 = 16^33 = 8^44 = 4^66 = 2^132

All these examples are lists of the form  [ ( 2^(n/d) ) ^ d  for d in divisors(n) ]

with different values of n  (here : 12 , 30 and 132).

The same can of course be done with other bases (e.g., 3 instead of 2).

So I think it boils down to a table which lists divisors for the (possibly just composite) integers n

(which is already in OEIS, for example A269065).

[The list of  exponents  n/d  is  just the list of divisors  d  reversed.]

Values of n with many divisors could be found among the highly composite numbers A002182.

- Maximilian

[Allan Wechsler]

Chris, one interesting question to ask yourself, when deciding whether to submit a sequence is, "Is there any 'simpler' sequence embodying a similar idea, that is also not in the OEIS yet?" (I put the word "simpler" in scare-quotes because the concept of simplicity is not easy to define -- but trust your feelings for the moment.)

Let me give a foolish example. Suppose neither the square numbers 0, 1, 4, 9, 16, ... nor the incremented square numbers 1, 2, 5, 10, 17, ... were in the OEIS already. You wouldn't add n^2 + 1 first; surely it is more urgent to add n^2. And sure enough, when we look up those two sequences in OEIS, we find that the squares are A000290 while the incremented squares are A002522.

Now look at your sequence: integers that are a power of a power of a power ... of a power of an integer, where there are 100 or more links in the power chain. Notice that there is an arbitrary 100 in the formulation. There must also be another sequence, also not in OEIS, where the power chain is of length at least 73 -- and isn't this a simpler sequence? When we crank the threshold all the way down to 2, we finally find a sequence that is in the OEIS: A001597, the "perfect powers" sequence. But I suspect that even if the chain was required to be of length merely 3 or greater, that sequence would not be in the OEIS. And surely it would be worthy of earlier consideration.

There is another possible conceptual simplification. If you do a little bit of algebra, you realize that your sequence is basically the perfect powers restricted to a certain highly divisible class of exponents. What are these exponents? They are integers with a divisor chain of length greater that 100 -- in other words, a multiple of a multiple of a multiple ... of a multiple of an integer, where there are at least 100 multiples in the chain. Is this sequence in the OEIS? I doubt it -- and yet, surely, a sequence should go into the OEIS before its "exponential". That is, we have a more urgent need for the odd numbers, than for the sequence of integers raised to odd powers.

Now, as you already noted, the fact that the first nontrivial element of your sequence is the gargantuan 2^(2^100), makes adding this sequence to the OEIS highly problematic. But you don't have to reach so far to come up with original sequences! I've sketched two ideas for much simpler sequences that we also don't have yet.

Thank you for sharing your idea!

-- Allan

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[Jeffrey Shallit]

OK, first the terminology.   You're not talking about "roots" per se but rather "nth roots".  

Secondly, you're not talking about "chains" but simply numbers for which many of their n'th roots are integers.  There are no chains involved.

Since you restricted yourself to powers of a prime p, say p^m, their n'th roots are going to be of the form p^{m/n}.  So now you want m that are divisible by many different n.

There are multiple sequences in the OEIS like this.  For example, you might be wanting "highly composite m"; those m for which their number of divisors exceeds all small numbers.  This is sequence https://oeis.org/A002182 and if you take a prime p and raise it to numbers in this sequence you will find many integer n'th roots.  No smaller power of p will work to get this many.

Another thing you might be wanting is to get n = 1,2,3,..., to work; that is p^{m/n} is an integer for n = 1, 2, .., k.  In this case you can't do better than to take m=lcm(1,2,...,k).  This is sequence https://oeis.org/A003418 and if you take a prime p and raise it to numbers in this sequence you will get integer n'th roots for n=1,2,..., k.

Hope that helps.

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[Michael]

my names satoshi bitch

Sent with Mixmax

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Book a meeting with me

[Michael]

VAREMAN If you put 2 and 2 together you get 4 — the arithmetic truth. vareman.com Powered by Mixmax

SATOSHI!!!!!!!!

Sent with Mixmax

On Wed, Oct 29, 2025 at 3:32 AM Chris Clark <chri...@gmail.com> wrote:

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[Alex Violette]

Hey Michael,
idk if you're hacked or not but if not then this language shouldn't be thrown around here.

-Alex V.

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[Bob Lyons]

Michael is no longer a member of SeqFan.

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Bob

[Chris Clark]

Thanks for all the replies. Sounds like my sequence is just a subset of an existing sequence. Still, it's been fun writing code to follow this particular rabbit hole.

Thanks again :)

Chris