Proof that the Heraclitus transform of the squares and the integer 2 contains every integer

[Geoffrey Caveney]

I have submitted as draft sequence A400196 the Heraclitus transform of the squares and 2: "a(0) = 0; thereafter a(n) is the least integer (in absolute value) not yet in the sequence such that the absolute difference between a(n-1) and a(n) is either a square or 2; in case of a tie, preference is given to the positive value."

This is a modification of Rémy Sigrist's sequence A377091, the Heraclitus transform of the squares, by simply including the integer 2 together with the square numbers as permitted differences between successive terms of the sequence.

The initial terms of the modified sequence are 0, 1, -1, -2, 2, 3, 4, 5, -4, -3, -5, -6, -7, -8, 8, 6, 7, 9, 10, 11, 12, 13, -12, -10, -9, -11, -13, -14, -15, -16, -17, -18, 18, 14, 15, 16, 17, 19, ….

The attached pdf file presents a proof that this modified sequence contains every positive and negative integer. For the record, I did not use any AI program in producing this proof.

The inclusion of the integer 2 simplifies the structure of certain blocks of the sequence in crucial ways that make it much more amenable to a proof. In particular, it simplifies the structure of what the attached proof labels as Block B of negative integers (e.g., -12, -10, -9, -11, -13) and Block D of positive integers (e.g., 32, 28, 26, 27, 29, 30, 31). This smoothing of the internal structure of these blocks also smooths the transitions between blocks of positive and negative integers.

Here are the most crucial steps in the proof of the structure of Block B (and, mutatis mutandis, of Block D):

"(2) Because differences of 2 between terms are permitted, the only consecutive integers that may possibly (but not necessarily) occur in this first part of Block B, with terms having decreasing absolute value, are the terms –[(2k)^2 / 2 + 2] and –[(2k)^2 / 2 + 1].

[...]

"Due to property (2) above of the first part of Block B, and because differences of 1 and 2 between terms are permitted, this second part of Block B, with terms having increasing absolute value, must fill in all negative integer values between -[((2k+1)^2 - 1) / 2] and – [(2k)^2 / 2 + 1] that were not used in the first part of Block B in order of their absolute value."

I believe that a proof that A377091 also contains every positive and negative integer is not out of reach, but it will require the consideration and analysis of multiple technical cases where the transitions between blocks of positive and negative integers are more convoluted than those in this modified sequence. Perhaps the careful use of an AI program may be of assistance in completing such technical cases and steps of a proof.

Geoffrey Caveney

[Geoffrey Caveney]

I also now have a proof that A377092, the Heraclitus transform of the Fibonacci numbers, contains every integer. The structure is very similar to the proof for the squares and the integer 2, but there is a structure of six blocks rather than four blocks. For small k, the terms a(0), ..., a(F_3k) comprise all the integers m such that |m| <= F_3k / 2, and a(F_3k + 1) = F_3k / 2 + 1. The inductive step, proceeding through six blocks of terms, shows that the assumption of these facts for k implies the occurrence of the same for k+1. It follows that the sequence contains every positive and negative integer. I will write up and post the proof on this list as soon as I can. I did not use any AI program to find the proof.

Geoffrey