Eviews gives the same p-value. I think there is no "problem" here as the asymptotic distribution is highly asymmetric around zero including a band of positive ADF values but with very small probability area (and it is an one-sided test). Actually you need a test statistic of 2.8 ( urcpval(2.8, 0, 1, 3) ) to reach an asymptotic p-value of exactly 1 according to the code (and the response surface estimated by MacKinnon (1996)). P-value here (in the ADF test) is interpreted as the integral of the distribution to the left of the ADF statistic. If you get an estimate of ADF=0 the probability to the left is 99.62%, that is under the null hypothesis you will observe an ADF statistic of 0 or smaller (<= than 0) with 99.62 percent probability. Why should this be 1 or why should be 1 if the ADF statistic is positive? The estimated parameter could be higher than one - although "slightly" higher in relevant economic applications. For example you estimated AR=1.00390523 so the asymptotic p-value is 0.99749368 or 99.74% (is higher than 99.62 etc). Under the null that AR=1, the probability that you will estimate (with constant trend etc that is with all specification "correct") AR<=1.00390523 is 99.74%. Now in the left tail assume that you get ADF=-3.5 so that eval urcpval(-3.5, 0, 1, 3) gives 0.039304454 or under the null the probability to "draw" an ADF less than or equal to -3.5 is "only" 3.93%. At 5% (singificance) you reject the null etc. Yiannis