Wythoff column residue pairs

Abstract

The first two Wythoff columns attain all residue pairs modulo every positive integer.

Write phi=(1+sqrt(5))/2. Rows and columns start at one. The array W has W(n,0)=floor(n phi), W(n,1)=floor(floor(n phi) phi), W(n,2)=floor(floor(n phi) phi^2), and W(n,k+2)=W(n,k+1)+W(n,k) for k>=1. Its first row is 1,2,3,5,… . Let R(m) be the image of positive row indices under n -> (W(n,1) mod m,W(n,2) mod m). Let w(m,n,k) denote W(n,k) reduced to ZMod(m).

Theorem 1.1 (All-modulus residue coverage).

$$\forall m\in \mathbb{N},1\le m\Rightarrow \left(ncard\left(R\left(m\right)\right)=m^2\wedge \left(\forall a\in ZMod\left(m\right),b\in ZMod\left(m\right),\exists n\in \mathbb{N},1\le n\wedge \left(w\left(m,n,1\right)=a\wedge w\left(m,n,2\right)=b\right)\right)\right)$$

Proof. Machine-checked in Lean as D5/S3/Arith/Wythoff/ColumnResiduePairs.result (✓ std3). ∎

Resolves. Problems/oeis-a035513-wythoff-column-residue-pairs (proved) by D5/S3/Arith/Wythoff/ColumnResiduePairs.result.

Source. Repository-derived.

Acknowledgement. Clark Kimberling (2025). OEIS A035513: Wythoff array read by falling antidiagonals. URL: https://oeis.org/A035513.

Commentary.

The two floor expressions simplify to floor(n phi)+n-1 and 2 floor(n phi)+n-1. Given residues a,b, choose r=2a-b and s=b-a. Along positive indices n=r+1+mk, irrational rotation on the circle of circumference m is dense. An open interval between s and s+1 therefore supplies floor(n phi)=s modulo m, and the linear formulas give a,b. The image is the full product, which has m^2 elements. The modulus-one case strengthens Kimberling’s m>=2 assertion.

References

• Truth anchor: D5/S3/Arith/Wythoff/ColumnResiduePairs.result