If [tex] \Omega_R [/tex] is a very small number,then couldn't there be many combinations of [tex] \Omega_{matter} [/tex] and [tex] \Omega_\Lambda [/tex] that give the same result?

Given the current value of the Hubble parameter H = 71, there are other models that give the same result. For example [itex]\Omega_m[/itex] = 0.1, [itex]\Omega_{\Lambda}[/itex] = 0.4. This is a degeneracy as any other. For example, mantaining [itex]\Omega_m[/itex] = 0.3 you get the same age with [itex]\Omega_{\Lambda}[/itex] = 0 and H = 60, which might not be a very unrealistic model (or at least it was not some years ago).

However, there is only one model that provides the same age for every value of H. This is the Milne model. I have not done any calculation but there must be some mathematical reason for this that should become clear when calculating the age as a function of H in the current model.

Yes, where [itex] \Omega_R [/itex] is the curvature component, the deviation of the total density parameter from unity.

However, lensing of distant quasars is observed to place [itex] \Omega_{matter} [/itex] in the 0.3 range and the standard WMAP concordance model puts it at 0.23 (DM) + 0.04 (baryon) with [itex] \Omega_{Dark Energy} = 0.76 [/itex].

The point of this paper is to point out that the result of this complicated standard theory results in the same age as the very simplest models, the linearly expanding one. Coincidence??