If Sn = X1 + . . . + Xn, where Xi are independent and identically distributed (i.i.d.) standard normal, then E|Sn| ≡ √2n/π, n ≧ 0. We show that no other symmetric law has exactly these “moments”; the general case remains (stubbornly) open. If X is standard two-sided exponential, then E|Sn| = 2n2-2n(2n/n). We show the latter moments are obtained exactly for all n also for Xi ~ B(2;0.5), the sum of two standard (± 1-valued) Bernoulli’s as well as for many other laws including unsymmetrical ones: Xi ~ G - 1, where G is geometric with mean 1, is one example. Our interest in this delicate nonlinear inverse problem (which was initiated by Klebanov, cf. [12]) of inverting the moments to recover the law was also drawn by the fact that it gives a way to study positive definite functions through the formula E|Sn| = (2/π) ∫0∞Re(1 - φn(1 / u))du, n ≧ 0, expressing E|Sn| in terms of the moments of φ, where φ is the characteristic function of X, φ(u) = Eexp(iuX). We show that if for some b > 0, ψb (u) = φ (btan (u / b)) is a positive definite function then the distributions corresponding to φ and ψb have the same E|Sn| moments for all n. We show that if X is Bernoulli with zero mean and values ±1 then the moments characterize the distribution uniquely even among nonsymmetric laws. In general however we expect that the moments do not characterize the law, and this may well be the only nontrivial case of uniqueness. We extend some of our results to the case of pth moments, p different from an even integer.