Hanna’s Gaussian shifted-square parity

Abstract

Symmetric coefficient pairs leave exactly the odd diagonal contribution.

Definition 1.1 (Rational summands).

$$\forall m:\mathbb{N},(\mathrm{summand}\left(m\right)=(X{)}^m\cdot ((X{)}^{m+1}+C\left(\mathrm{Complex}.I\right){)}^m\cdot ((1+C\left(\mathrm{Complex}.I\right)\cdot (X{)}^{m+1}{)}^{m+1}{)}^{-1})$$

Formalization. D5/S1/Recurrence/Algebraic/HannaGaussianShiftedSquareParity.summand (✓ std3).

Citation. Paul D. Hanna (2019). OEIS A324300: Gaussian generating function and square parity. URL: https://oeis.org/A324300.

Commentary.

OEIS A324300, NAME (unpaginated): “G.f.: Sum_{n>=0} x^n * (x^(n+1) + i)^n / (1 + i*x^(n+1))^(n+1).” Complex.I is the imaginary unit and X is the formal series variable. The inverse has unit constant denominator.

Definition 1.2 (The coefficient series).

$$A=\mathrm{PowerSeries}.\mathrm{mk}\left((N:\mathbb{N}\mapsto \sum _{m\in \mathrm{Finset}.\mathrm{range}\left(N+1\right)}(\mathrm{coeff}\left(N,\mathrm{summand}\left(m\right)\right)))\right)$$

Formalization. D5/S1/Recurrence/Algebraic/HannaGaussianShiftedSquareParity.A (✓ std3).

Citation. Paul D. Hanna (2019). OEIS A324300: Gaussian generating function and square parity. URL: https://oeis.org/A324300.

Commentary.

OEIS A324300, NAME (unpaginated): “G.f.: Sum_{n>=0} x^n * (x^(n+1) + i)^n / (1 + i*x^(n+1))^(n+1).” The leading X power makes each coefficient sum finite.

Definition 1.3 (The parity assertion).

$$claim=(\forall N:\mathbb{N},(\exists z:\mathbb{Z},((\mathrm{coeff}\left(N,A\right)=(z:\mathbb{C}))\wedge (\mathrm{Odd}\left(z\right)⟺(\exists k:\mathbb{N},(N=k\cdot \left(k+2\right)))))))$$

Formalization. D5/S1/Recurrence/Algebraic/HannaGaussianShiftedSquareParity.claim (✓ std3).

Citation. Paul D. Hanna (2019). OEIS A324300: Gaussian generating function and square parity. URL: https://oeis.org/A324300.

Commentary.

OEIS A324300, COMMENTS and FORMULA (unpaginated): “Odd terms occur only at n = k*(k+2) for k >= 0 (conjecture); a(n*(n+2)) = 1 (mod 2) for n >= 0.” The index N is natural, z is an ordinary integer embedded in the complex numbers, and degree zero is included.

Theorem 1.4 (Integral coefficients and exact odd support).

$$\forall N:\mathbb{N},(\exists z:\mathbb{Z},((\mathrm{coeff}\left(N,A\right)=(z:\mathbb{C}))\wedge (\mathrm{Odd}\left(z\right)⟺(\exists k:\mathbb{N},(N=k\cdot \left(k+2\right))))))$$

Proof. Machine-checked in Lean as D5/S1/Recurrence/Algebraic/HannaGaussianShiftedSquareParity.result (✓ std3). ∎

Resolves. Problems/hanna324-gaussparity (proved) by D5/S1/Recurrence/Algebraic/HannaGaussianShiftedSquareParity.result.

Source. Repository-derived.

Acknowledgement. Paul D. Hanna (2019). OEIS A324300: Gaussian generating function and square parity. URL: https://oeis.org/A324300.

Commentary.

The exponent equation is symmetric in the two indices and off-diagonal pairs are even integers. A diagonal term is present exactly when N = k*(k+2), and its weight is odd.

References

• Truth anchor: D5/S1/Recurrence/Algebraic/HannaGaussianShiftedSquareParity.A
• Truth anchor: D5/S1/Recurrence/Algebraic/HannaGaussianShiftedSquareParity.claim
• Truth anchor: D5/S1/Recurrence/Algebraic/HannaGaussianShiftedSquareParity.result
• Truth anchor: D5/S1/Recurrence/Algebraic/HannaGaussianShiftedSquareParity.summand
• Dependency: D5/S1/Recurrence/Algebraic/HannaGaussianKernelParity