OKOtakhon U. Kenjaev
Independent researcher · Khorezm, Uzbekistan

Otakhon U. Kenjaev

Working on the mathematics Ramanujan left behind — his Lost Notebook, special functions and the geometry of ellipses.

RamanujanSpecial functionsNumber theoryElliptic integrals
\(r(t) = (a\cos t,\ b\sin t,\ ct)\)

“An equation for me has no meaning unless it expresses a thought of God.” — attributed to S. Ramanujan

4papers with DOI
4open points from the Lost Notebook, Part IV
4interactive calculators
CC BYevery paper open access

Research

Each paper states one clear result, gives a proof, and comes with a calculator or code so the numbers can be checked.

Elliptic integrals · Geometry

Arc length of an elliptical helix via Ramanujan's second perimeter approximation, with explicit pitch thresholds

\[L_{\text{turn}} = P\!\left(\sqrt{a^2+c^2},\ \sqrt{b^2+c^2}\right)\]

One turn of the elliptical helix is exactly an ellipse perimeter — so Ramanujan's formula gives it in closed form.

  • \(c \ge 0.197254\,a \Rightarrow \text{rel. error} < 10^{-6}\)
  • \(c \ge 0.534032\,a \Rightarrow \text{rel. error} < 10^{-9}\)

Zenodo preprint · 2026 · doi:10.5281/zenodo.22969890

Lost Notebook · Lattice sums

The constants K and λ in Ramanujan's identity (9.2.5) of the Lost Notebook, with a hexagonal lattice sum calculator

\[K = \mu-\nu,\qquad \lambda = \mu+\nu \qquad (\gcd(\mu,\nu)=1)\]

Andrews and Berndt could not identify K and λ. Six-fold symmetry of the Eisenstein integers settles them.

  • \(R(6k) = \tfrac{(-1)^k}{3}\left(c_k\,E_6(\rho)^k - 3\right)\)
  • \(c_1 = 1,\ c_2 = \tfrac{250}{691},\ c_3 = \tfrac{5500}{43867}\)

Zenodo preprint · 2026 · doi:10.5281/zenodo.22987480

Lost Notebook · Algebraic number theory

Pisot units and the third manuscript on page 343 of Ramanujan's Lost Notebook, with a verification script and calculator

\[\frac{1}{2^{1/5}-1} = 1+\theta+\theta^2+\theta^3+\theta^4,\qquad \theta = 2^{1/5}\]

Claims that Andrews and Berndt called wrong become a theorem about units of \(\mathbb{Z}[2^{1/5}]\).

  • \(b = \frac{\sqrt5}{(1+4^{1/5})^{5/2}} = 1-\theta-\theta^2+\theta^4\)
  • \(\tfrac{1}{2^{1/p}-1}\ \text{is Pisot} \iff p \le 6\)

Zenodo preprint · 2026 · doi:10.5281/zenodo.23003128

Lost Notebook · Dirichlet series

On two open points in Part IV of Ramanujan's Lost Notebook: the domain of Entry 12.2.3 and a summation reading of Entry 19.2.2

\[\sigma_c\!\left(\sum_{n\ge1} \chi(n)\,d(n)\,n^{-s}\right) \in \left[\tfrac14,\ \tfrac13\right]\]

The expected domain of Entry 12.2.3 fails for r = 0: the series diverges for 0 < Re s < 1/4.

  • \(\chi = \text{non-principal character mod } 4\)
  • \(L(s,\chi)\,L(s-r,\chi) = \sum \chi(n)\,\sigma_r(n)\,n^{-s}\)

Zenodo preprint · 2026 · doi:10.5281/zenodo.23003573

About

I am an independent researcher from Khorezm, Uzbekistan. My work starts from places where Ramanujan wrote a formula and later readers could not say what it meant — and tries to make it precise, prove it, and turn it into something anyone can compute.

Every result is published openly with a DOI, and every paper comes with a calculator or code, so that the numbers can be checked by anyone, not taken on trust.

Methods: computations and drafting were assisted by AI tools; each paper states this in its declaration.

Profiles & identifiers