Page 343 of Ramanujan's Lost Notebook

θ = 2^1/5, a = 1/(θ−1) = 1+θ+θ²+θ³+θ⁴, b = √5/(1+θ²)^5/2 = 1−θ−θ²+θ⁴. Ramanujan claimed that a^mbⁿθ is extremely close to an integer p_m,n. Andrews–Berndt (Lost Notebook IV, §7.4) called the claims wrong; the integer is the trace and the error is a sum of Galois conjugates. self-test …

1. Calculator for (7.4.1)

m (0–3000)

n (0–500)

p_m,n = Tr(a^mbⁿθ) is computed exactly with integer arithmetic in Z[θ]. ε = a^mbⁿθ − p_m,n = −Σ_s=1..4 σ_s(a^mbⁿθ) is computed from the conjugates, so no big-float library is needed. Theorem: |ε| ≤ 2θ(|σ₁(a)|^m|σ₁(b)|ⁿ + |σ₂(a)|^m|σ₂(b)|ⁿ); the error tends to 0 exactly when m > 6.0114·n.

2. Why 5 and 7: the Pisot boundary

p = 5

a_p = 1/(2^1/p−1) is a unit. It is a Pisot number (all other conjugates inside the unit circle) exactly when 2^1/p > 2cos(2π/p), i.e. p ≤ 6. Ramanujan's two cases are p = 5 (last odd Pisot case) and p = 7 (first non-Pisot case, where he adds the units b and c).

Cite as: Kenjaev, O. U. (2026). Pisot units and the third manuscript on page 343 of Ramanujan's Lost Notebook. Zenodo. DOI 10.5281/zenodo.23003128 · ORCID 0009-0009-3566-9285

Source code and a Python verification script (verify_343.py, 15 checks; the note) accompany the note. Everything runs locally in your browser.