Spezia’s A320043 square-value conjecture is false

Abstract

A prime row and the Pell equation give an additional square value of OEIS A320043.

Definition 1.1 (The gcd-weighted row sum).

$$\forall n\in \mathbb{N},a\left(n\right)=\sum _{k\in \mathrm{Finset}.\mathrm{Icc}\left(1,n\right)}\mathrm{Nat}.\mathrm{div}\left((n+1-k{)}^2\cdot k,\mathrm{Nat}.\gcd {\left(n+1-k,k\right)}^3\right)$$

Formalization. D5/S3/ArithSums/SpeziaA320043SquareValuesRefutation.a (✓ std3).

Citation. Stefano Spezia; OEIS Foundation Inc. (2018). OEIS A320043: Row sums of the triangle A322550. URL: https://oeis.org/A320043/internal.

Commentary.

OEIS A320043, revision 43, first FORMULA (unpaginated): “a(n) = Sum_{k=1..n} (n + 1 - k)^2*k/gcd(n + 1 - k, k)^3.” Here n and k are natural numbers. Nat.div is natural-number division, and subtraction is natural subtraction. For 1 <= k <= n, both gcd arguments are positive and the denominator divides the numerator exactly. The value at n = 0 is the empty sum; the source starts at n = 1.

Definition 1.2 (The source conjecture).

$$claim=\left(\forall n\in \mathbb{N},(1\le n)\Rightarrow (\forall m\in \mathbb{N},(a\left(n\right)=m^2)\Rightarrow ((n=1)\vee ((n=6)\vee (n=96))))\right)$$

Formalization. D5/S3/ArithSums/SpeziaA320043SquareValuesRefutation.claim (✓ std3).

Citation. Stefano Spezia; OEIS Foundation Inc. (2018). OEIS A320043: Row sums of the triangle A322550. URL: https://oeis.org/A320043/internal.

Commentary.

OEIS A320043, revision 43, sole COMMENTS conjecture (unpaginated): “Conjecture: a(n) is not a perfect square except for n = 1, 6 and 96.” The positive index n and the square root m range over natural numbers. The OFFSET field is 1,2; its first entry supplies the positive-index domain.

Theorem 1.3 (An additional square refutes the conjecture).

$$\neg (\forall n\in \mathbb{N},(1\le n)\Rightarrow (\forall m\in \mathbb{N},(a\left(n\right)=m^2)\Rightarrow ((n=1)\vee ((n=6)\vee (n=96)))))$$

Proof. Machine-checked in Lean as D5/S3/ArithSums/SpeziaA320043SquareValuesRefutation.result (✓ std3). ∎

Resolves. Problems/a320043-square-values-refutation (refuted) by D5/S3/ArithSums/SpeziaA320043SquareValuesRefutation.result.

Source. Repository-derived.

Acknowledgement. Stefano Spezia; OEIS Foundation Inc. (2018). OEIS A320043: Row sums of the triangle A322550. URL: https://oeis.org/A320043/internal.

Commentary.

Take p = 708158977 and y = 204427888. Lucas’s criterion proves p prime, and p squared minus 12 times y squared equals 1. For prime p, gcd(p-k,k) equals 1 on 1 <= k < p. A polynomial telescope gives 12a(p-1) = p squared times (p squared minus 1), so the Pell equation gives a(p-1) = (py) squared. Thus a(708158976) equals 144767444036350576 squared. This positive index differs from 1, 6 and 96. Minimality, infinitely many prime Pell inputs, and classification of other square values remain open.

References

• Truth anchor: D5/S3/ArithSums/SpeziaA320043SquareValuesRefutation.a
• Truth anchor: D5/S3/ArithSums/SpeziaA320043SquareValuesRefutation.claim
• Truth anchor: D5/S3/ArithSums/SpeziaA320043SquareValuesRefutation.result