A Heraclitus transform variant with a nontrivial missing integer

[Geoffrey Caveney]

See A377091 and A383442 for information about Rémy Sigrist's "Heraclitus transform" sequences.

A curious variant is to select the second least (in absolute value) unused eligible integer. Such a sequence logically begins with a(0) = 1, skipping the least integer 0. Of course 0 can never occur in such a sequence; the question is whether other integers do or do not occur. It is consistent to follow the convention of A377091 that the positive value x is considered "less" than the negative value -x in case of a choice between them.

Applying this restriction to the squares (each difference between successive terms is a square, as in A377091), it is more or less trivial or at least very simple to see that 0, -1, and -2 never occur: -1 is never the second least eligible integer because |x-0| and |x-(-1)| cannot both be squares for x != 0, and since -1 never occurs, it follows that -2 never occurs because |x-0| and |x-(-2)| are both squares only for x = -1.

To allow -1 and -2 to occur, I considered the variant where each difference between successive terms can be a square or a cube. Here -1 and -2 can occur because |8-0| is a cube and |8-(-1)| is a square. But the interesting and nontrivial result in this variant is that the integer 6 never occurs.

The sequence begins 1, 2, 3, 4, -4, -3, 5, 9, -7, -8, 8, -1, -2, -6, 10, 11, 7, -9, -5, .... (The corresponding unselected least unused eligible values are 0, 0, -2, -1, 0, 0, -2, 6, 0, -6, 0, 0, 0, 6, -5, 6, -5, 6, 0, ....)

Thus, through term a(18) = -5, all values |a(n)| <= 6 have been used except 0 and 6. Henceforth, 6 could only occur in the sequence as the second least unused eligible integer if |a(n) - 0| and |a(n) - 6| were both either a square or a cube.

But that is impossible. Proof:

A. It is easy to see that no two cubes can have a difference of 6, since 2^3 - 1^3 = 7 and larger cubes have greater differences.

B. Also no two squares can have a difference of 6: Every square is == 0 or 1 modulo 4, so no difference of two squares is == 2 modulo 4.

C. Crucially, the difference of a square and a cube is the well-studied topic of Mordell curves: y^2 = x^3 + n. It is known that y^2 = x^3 + 6 and y^2 = x^3 - 6 have no integer solutions. See A054504 for the list of positive n with no integer solutions and A081121 for the list of negative n with no integer solutions. Thus, the absolute value of the difference of a square and a cube cannot be 6.

It appears likely that all integer values except 0 and 6 occur in the sequence. For any larger integer value k, the restriction on the difference of two squares in point B. above will not apply, since k-0 and k-6 cannot both be == 2 modulo 4.

Geoffrey