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In 1971 Richard Guy sent a letter to Neil Sloane outlining some integer sequences. One of these sequences, A279196, was added to the OEIS by Neil only in December of 2016:

A279196: Number of polynomials $P(x,y)$ with nonnegative integer coefficients such that $P(x,y) \equiv 1 \text{ (mod } x+y-1)$ and $P(1,1) = n$.

The first few terms of the sequence are:

1, 1, 2, 5, 13, 36, 102, 295, 864

In a recent lecture (see 9:50), Neil said:

I don't even know what the [...] polynomials are for the first few values, so it might be interesting to look into this.

I tried to address this in an ad hoc manner, and I managed to find all of the examples for $n \leq 3$ and three examples for $n = 4$:

$a(1) = 1$ with:

  • $1$

$a(2) = 1$ with:

  • $x + y = (x + y - 1) + 1$

$a(3) = 2$ with:

  • $xy + x + y^2 = (y + 1) (x + y - 1) + 1$ and
  • $xy + y + x^2 = (x + 1) (x + y - 1) + 1$

And for $n = 4$:

  • $x^2 + y^2 = (x + y + 1)(x + y - 1) + 1$
  • $x^2y + x^2 + xy^2 + y = (xy + x + 1)(x + y - 1) + 1$
  • $x^2y + xy^2 + x + y^2 = (xy + y + 1)(x + y - 1) + 1$

What is an algorithm or technique to enumerate polynomials for some arbitrary $n$?

Could anyone provide an example of, say, the remaining two polynomials when $n=4$? All thirteen polynomials for $n=5$?

1 Answers 1

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I think I figured it out (but I would still love to see a proof that this technique is exhaustive!)

It seems that (at least the small) values can be built up recursively:

To create a list of polynomials for $a(n)$, for each term (on the right) and polynomial (on the left) in $a(n-1)$, take a term from the left-hand side and move it to the factor of the right hand side.

For example, to create a polynomial for $n = 4$, take the first term/factor pair for $n = 3$ (highlighted below in bold) and move the term to be inside of the factor:

n = 4 (example 1)

$\boldsymbol{xy} + x + y^2 = \boldsymbol{(y + 1)}(x + y - 1) + 1$

Leads to: $x^2y + xy^2 + x + y^2 = \boldsymbol{(xy + y + 1)}(x + y - 1) + 1$

Remaining examples:

n = 4 (example 2)

$xy + \boldsymbol{x} + y^2 = \boldsymbol{(y + 1)}(x + y - 1) + 1$ and

$xy + \boldsymbol{y} + x^2 = \boldsymbol{(x + 1)}(x + y - 1) + 1$

Leads to: $x^2 + y^2 = \boldsymbol{(x + y + 1)}(x + y - 1) + 1$

n = 4 (example 3)

$xy + x + \boldsymbol{y^2} = \boldsymbol{(y + 1)}(x + y - 1) + 1$

Leads to: $xy^2 + xy + x + y^3 = \boldsymbol{(y^2 + y + 1)}(x + y - 1) + 1$

n = 4 (example 4)

$\boldsymbol{xy} + y + x^2 = \boldsymbol{(x + 1)} (x + y - 1) + 1$

Leads to: $x^2y + xy^2 + x + y^2 = \boldsymbol{(xy + y + 1)}(x + y - 1) + 1$

n = 4 (example 5)

$xy + y + \boldsymbol{x^2} = \boldsymbol{(x + 1)} (x + y - 1) + 1$

Leads to: $ x^3 + x^2y + xy + y = \boldsymbol{(x^2 + x + 1)}(x + y - 1) + 1$