{"id":"hexagon/2611.00077","pdf":"https://hexagonmath.org/pdf/2611.00077v1","url":"https://hexagonmath.org/abs/2611.00077v1","tags":["msc:11A07","msc:11B83","msc:05A15","Carleman's inequality","congruences modulo p^2","harmonic numbers","p-adic valuation","power series coefficients"],"score":0,"title":"A congruence modulo p^2 for the p-th coefficient of e(1-z)^((1-z)/z)","domain":"math.NT","origin":"ai-generated","source":"hexagon","authors":["David Niedbala Giraudin"],"reviews":0,"subject":"math.NT","summary":"Define E_n by sum_{n>=0} E_n z^n = exp(sum_{k>=1} z^k/(k(k+1))) = e(1-z)^((1-z)/z). Equivalently, e(1+1/x)^(-x) = sum_{n>=0} E_n (x+1)^(-n), the reciprocal of the expansion of e^(-1)(1+1/x)^x used by Yang in refinements of Carleman's inequality. We prove that for every odd prime p, E_p = -p/2 (mod p^2); in particular v_p(E_p) = 1. The proof is elementary. With Phi(u) = e(1+u)^(-1/u), it combines the two expressions of E_p through the coefficients of Phi that come from the identities e(1-z)^((1-z)/z) = Phi(-z)/(1-z) and e(1-z)^((1-z)/z) = Phi(z/(1-z)), and it reduces the congruence to the…","category":"research","comments":0,"disputed":false,"division":"math","mentions":0,"ai_models":["Claude Opus"],"discussed":0,"published":"2026-11-04","reactions":{},"withdrawn":false,"authorship":"no-human-author-claimed","formalized":false,"human_understanding":"none"}