The Heraclitus transform of {6n, 27, 32}

[Geoffrey Caveney]

I was not a member of this list when Rémy Sigrist published A377091 in October 2024, but I see that it has received a significant amount of attention and study, including many other sequences derived from it. (Just within the past week, Neil Sloane and Paolo Xausa have published A399181-A399183 based on it.) It has been named "the Heraclitus transform" of the squares.

In the comments of A383442 (the corresponding transform of the triangular numbers), Neil Sloane defines a Heraclitus transform:

"Heraclitus (circa 500 BCE) observed that no man can step in the same river twice.
"The Heraclitus transform H(S) of a sequence S is formed by starting at 0, and moving s steps to the left or right, where s is any element of S, never visiting any number twice, and moving as close to 0 as possible. In case of a tie, move to the positive term.
"The present sequence is the Heraclitus transform of the triangular numbers A000217. For the squares, see A377091. Conjecture: both H(A000217) and H(A000290) contain every (positive or negative) integer. In fact it appears that this property holds whenever S is a monotonically strictly increasing sequence starting with 1. It does not hold for H(A000012), which is A001477."

I suspect that the reason for including the condition "starting with 1" was to rule out sequences such as the even integers, for which it is obvious that H(S) contains no odd integers, or in general the multiples of a given n, for which H(S) contains only such multiples of n.

But I propose that the inclusion of 1 in S may not be a necessary condition for this property. Rather, I conjecture that H(S) contains every integer whenever S is an unbounded sequence and at least one pair of its elements are coprime. Naturally 1 is coprime with every integer, but let us consider examples where only one pair of elements are coprime. A simple example is S = {2, 3, 6n}, where the only pair of coprime elements is 2, 3. In this case H(S) begins 0, 2, -1, 1, -2, 4, 6, 3, -3, -5, 7, 5, -7, -4, -6, ..., with first differences 2, -3, 2, -3, 6, 2, -3, -6, -2, 12, -2, -12, 3, -2, ..., and I see no reason to believe that it does not contain every integer.

It is more interesting and challenging to test cases where the only pair of coprime elements are larger integers. One may describe a general family of such sequences as S = {p^x, q^y, pqn}, where p and q are prime.

I like the example S = {6n, 27, 32} as an illustrative case, because the coprime pair are large enough that the occurrence of e.g. the integer 1 in H(S) is not prima facie obvious, but they are small enough that one can observe the behavior of such a sequence, including the eventual occurrence of the integer 1, in a reasonable number of terms:

H(S): 

0, 6, -6, 12, -12, 15, 3, -3, 9, -9, -15, 17, -1, 5, -7, 11, -13, 14, 2, -4, 8, -10, -16, 16, -2, 4, -8, 10, -14, 13, 1, -5, 7, -11, -17, ...

First differences: 

6, -12, 18, -24, 27, -12, -6, 12, -18, -6, 32, -18, 6, -12, 18, -24, 27, -12, -6, 12, -18, -6, 32, -18, 6, -12, 18, -24, 27, -12, -6, 12, -18, -6, ...

It is interesting that H(S), after the initial terms 0, 6, -6, 12, -12, then fills in all values |a(n)| < 18 before it reaches 18 or -18.

It may be interesting to examine the position n where a(n) = 1 occurs in such sequences H(S) for various values of p^x and q^y in S.

Geoffrey

[Geoffrey Caveney]

Apologies for replying to my own post. But for the investigation of the main case A377091, the Heraclitus transform of the squares, it may also be useful to consider the related case of the Heraclitus transform of S = {6n^2, 27, 32} = {6, 24, 27, 32, 54, 96, 150, ...}:

H(S):

0, 6, 12, -12, -6, 18, -9, -3, 3, 9, 15, -17, 7, 1, -5, -11, 13, -14, -8, -2, 4, 10, 16, -16, 8, 2, -4, -10, 14, -13, -7, -1, 5, 11, 17, -15, -21, -27, 27, 21, -33, -39, -45, -18, ...

First differences:

6, 6, -24, 6, 24, -27, 6, 6, 6, 6, -32, 24, -6, -6, -6, 24, -27, 6, 6, 6, 6, 6, -32, 24, -6, -6, -6, 24, -27, 6, 6, 6, 6, 6, -32, -6, -6, 54, -6, -54, -6, -6, 27, ...

[Neil Sloane]

Geoffrey, you didn't say, I think - what's the answer? does your set
of steps 6n, 27,32 hit every number or not?
Best regards
Neil

Neil J. A. Sloane, Chairman, OEIS Foundation.
Also Visiting Scientist, Math. Dept., Rutgers University,
Email: njas...@gmail.com

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[Geoffrey Caveney]

I conjecture that both 6n, 27, 32 and 6n^2, 27, 32 hit every number. Nothing that I have observed in their behavior has given me reason to doubt this conjecture.

Geoffrey 

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[Neil Sloane]

Geoffrey,
Thanks for that answer. My experiences with possible counterexamples
was similar, which is what emboldened me to make the conjecture in the
first place.

But I still don't 100% believe it!

Best regards
Neil

Neil J. A. Sloane, Chairman, OEIS Foundation.
Also Visiting Scientist, Math. Dept., Rutgers University,
Email: njas...@gmail.com

On Thu, Sep 10, 2026 at 3:41 PM Geoffrey Caveney

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[Geoffrey Caveney]

The intuitive argument in favor of the conjecture is that some integer would have to be the smallest absolute value that doesn't occur. But the definition of the sequence privileges the unused integer with the smallest absolute value as the preferred next term if permitted. If such an integer does not occur, it would require an infinite set of additional integers not to occur as well, and each of them would in turn require additional infinite sets of integers not to occur, etc. That seems implausible.

Of course that doesn't mean that mathematics has the tools to prove the conjecture either. I'm also confident that there are infinitely many primes of the form n^2 + 1, but no one has been able to prove it.

I wonder if sequences without squares or exponents in the definition, such as my 6n, 27, 32, or the simpler 6n, 2, 3, could be easier to prove. The proof of the case 2^n  (including 2^0 = 1) (A379719) and possibly even of n! (A393434) may point the way. Consider, for example, that 2^n is a very small subset of {2n, 1}. Just some food for thought.

Geoffrey

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[Geoffrey Caveney]

Neil,

I have an observation that may possibly lead to progress on the conjecture for the basic sequence A377091 for the squares:

It appears that the occurrence in the sequence of all integers k such that |k| <= m^2 / 2 requires only the squares up to m^2.

That is, the use of 1^2 and 2^2 covers all integers |k| <= 2,

1^2, 2^2, 3^2 cover all integers |k| < 9/2,

1^2 to 4^2 cover all integers |k| <= 8,

1^2 to 5^2 cover all integers |k| < 25/2,

1^2 to 6^2 cover all integers |k| <= 18, etc.

If an inductive step could show that this must be true for all m^2, then the truth of the complete conjecture would follow directly from the infinitude of the squares: every integer k satisfies |k| <= m^2 / 2 for some sufficiently large m^2.

Geoffrey

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