Conjectural divisibility for the non-split recurrence example

Determine whether, for the sequence defined by f_{n+1}=\frac{1}{n^2+1}\sum_{k=0}^{n}f_kf_{n-k} with f_0=1, the product \prod_{k=0}^{\lfloor n/2\rfloor}(k^2+1) divides 2^n d_n for every n≥0, where d_n is the denominator of f_n.

Background

The paper studies a differentially algebraic power series whose coefficient recurrence has the non-split denominator polynomial n2+1. Numerical experiments up to n=2000 suggest unusually large denominator growth and do not support a positive resolution of Mahler’s denominator conjecture in this non-split setting.

The same computations suggest a specific divisibility property involving the product of the values k2+1. If established, it would imply that no fixed expression of the form \delta{n+1}(\nu n+\mu)!s can always divide d_n, thereby reinforcing the obstruction associated with irreducible quadratic denominator factors.

References

Moreover, it seems that ∏⌊n/2⌋k=0 (k2 + 1) divides 2ndn for all n ≥ 0. If true, this conjecturally implies that there exist no integers δ, ν, μ, s ≥ 0 such that dn divides δn+1(νn + μ)!sfor all n ≥ 0.

Arithmetic properties of the Taylor coefficients of differentially algebraic power series  (2502.09259 - Krattenthaler et al., 13 Feb 2025) in Section 3.9, page 24