Ramanujan's hexagonal lattice sum

Completing identity (9.2.5) of Ramanujan's Lost Notebook: the two constants K and λ that Andrews and Berndt could not identify are K = μ−ν and λ = μ+ν. The completed identity turns a slowly convergent sum over the hexagonal lattice into a q-series that gives full double precision in about twenty terms.

The completed identity (s > 2)
σs−1(1)e−π√3 − σs−1(2)e−2π√3 + σs−1(3)e−3π√3 − ⋯
 = −2 Γ(s)ζ(s)/(2π)s · { cos(πs/6) + 2cos(π(s+1)/6)cos(π(s−1)/6) · R(s) },
R(s) = Σμ,ν≥1, (μ,ν)=1 cos( s·tan−1( (μ−ν) / ((μ+ν)√3) ) ) / (μ²+μν+ν²)s/2

Source: G. E. Andrews, B. C. Berndt, Ramanujan's Lost Notebook, Part IV (Springer 2013), §9.2–9.3, pp. 215–216 (lost notebook pp. 270–271). There K/(λ√3) appears with K, λ undefined. Proof: regroup the Andrews–Berndt form over the six units of the Eisenstein integers; three rotations of each primitive point fall in the half-plane μ ≥ 1 and combine into 4cos(π(s+1)/6)cos(π(s−1)/6)cos(sψ) with tan ψ = (μ−ν)/(√3(μ+ν)).

R(s) from the q-series
—
Left side L(s)

Check by direct summation

Adds the lattice sum term by term over all coprime μ, ν ≤ N. Near s = 2 the error only falls like N2−s, so millions of terms still leave percent-level error, while the q-series is exact to about 15 digits.

Npairsdirect sumrelative errortime

Exact values

Because E₄(ρ) = 0 and E₆(ρ) = 27Γ(1/3)¹⁸/(512π¹²) (Chowla–Selberg), the identity gives closed forms at s = 6 and s = 12. At s ≡ 2, 4 (mod 6) the coefficient of R(s) vanishes and the identity says nothing about R.

sclosed formvalueq-series − closed form

Cite as: Kenjaev, O. U. (2026). The constants K and λ in Ramanujan's identity (9.2.5) of the Lost Notebook. Zenodo. DOI 10.5281/zenodo.22987480 · ORCID 0009-0009-3566-9285 · Telegram @sheki · channel @nizomliy