# [seqfan] A279196: a plausible meaning?

Luc Rousseau luc_rousseau at hotmail.com
Tue Jun 27 19:54:53 CEST 2023

```Dear SeqFans,

In 1971 Richard Guy sent a letter to Neil Sloane outlining some integer sequences; one of them is A279196,
"Number of polynomials P(x,y) with nonnegative integer coefficients such that P(x,y)==1 (mod x+y−1) and P(1,1)=n".

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

In 2017 in a Winter Fruits talk (https://vimeo.com/201218446), Neil said about A279196:
"I don't even know what the [...] polynomials are for the first few values, so it might be interesting to look into this."

= = = =

Unaware of this context, a few days ago, I imagined this:
let the initial configuration be one token at (0, 0).
A configuration can be "degraded": to do so, { choose a nonempty pile of tokens in it, say that at (i, j); remove one token from that pile; then add one token at (i + 1, j) and one token at (i, j + 1) }.
The process being nondeterministic, define a(n) as the number of distinct configurations one can possibly get after (n - 1) "degradations" of the initial configuration.

A brute force program that I coded gave:

1, 1, 2, 5, 13, 36, 102, 295, 864, 2557, 7624, 22868, 68920, 208527, 632987, 1926752.

Superseeker then showed me that this sequence was extending A279196's known terms. Are the two sequences the same? I discovered the above-mentionned context.

It appears I can prove a(n) <= A279196(n) (see below); and I am asking: has anyone any insights to prove or disprove the equality?

= = = =

Translating languages, from the piles of tokens to polynomials, is easy: C*x^i*y^j means that the height of the pile at (i,j) is C.
The mod(x+y−1) sounds like a bridge between the two definitions, because clearly:

degrade(C*x^i*y^j) = (C−1)*x^i*y^j + x^(i+1)*y^j + x^i*y^(j+1) = C*x^i*y^j + (x+y−1)*(x^i*y^j)

Now, write P = Sum_{i,j} C[i,j]*x^i*y^j, and degrade P. We have to choose an (i0,j0) to actually degrade; this gives:

degrade(P) = (Sum_{(i,j)!=(i0,j0)} C[i,j]x^i*y^j) + (C[i0,j0]*x^i0*y^j0 + (x+y−1)(x^i0*y^j0))

degrade(P) = P + (x+y−1)*(x^i0*y^j0)

Applying "degrade" several times lets the cofactor of (x+y−1) grow accordingly; we can write:

degrade^<n>(P) = P + (x+y−1) * Q_n

where Q_n keeps the record of where we took the tokens. The fact we took exactly n tokens can be transcripted Q_n(1,1)=n. This implies:

degrade^<n−1>(1)(1,1) = 1+(1+1−1)*(n−1) = n.

Thus the polynomials of the degradation problem satisfy Richard Guy's axioms and a(n) <= A279196(n).
But are there other polynomials that do? Was the degradation problem the meaning Richard Guy had in mind, was the definition by polynomials a way to make it succinct?

Best regards,
Luc

P.S.: for essentially the same speech in a more comfortable, LaTeX form, here is a Math StackExchange thread:
https://math.stackexchange.com/questions/2127914/polynomials-px-y-with-nonnegative-integer-coefficients-such-that-px-y-eq

```

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