How to calculate autocorrelation manually

Next to the points about starting values raised in the comments, there is another issue related to the "manual" formula for a deterministic difference sequence of the type you use that, irrespective of the starting value, will give an autocorrelation of 1 when computed from the standard correlation formula:

Denote by $\bar{y}_{lag}$ the mean of the lagged series starting at $i=1$, and by $\bar{y}_{lead}$ the mean of the lead series starting at $i=2$. The correlation formula is

$$ corr=\frac{\sum_{i=2}^n(y_{i-1}-\bar{y}_{lag})(y_i-\bar{y}_{lead})}{\sqrt{\sum_{i=1}^{n-1}(y_i-\bar{y}_{lag})^2\sum_{i=2}^{n}(y_i-\bar{y}_{lead})^2}} $$ The difference sequence is such that $y_i=\rho y_{i-1}$ or, recursively, $y_i=\rho^{i-1}y_1$, where $y_1$ is the initial value in the series. Then, $$ \bar{y}_{lag}=\frac{1}{n-1}\sum_{i=1}^{n-1}y_i=\frac{y_1}{n-1}\sum_{i=0}^{n-2}\rho^i=y_1\frac{1-\rho^{n-1}}{(n-1)(1-\rho)}\equiv y_1\tilde M $$ Similarly, we may obtain $$ \bar{y}_{lead}=\rho y_1\tilde M $$ and the variations $$ \sum_{i=1}^{n-1}(y_i-\bar{y}_{lag})^2=y_1^2\sum_{i=1}^{n-1}(\rho^{i-1}-\tilde M)^2\equiv y_1^2D $$ and \begin{eqnarray*} \sum_{i=2}^{n}(y_i-\bar{y}_{lead})^2&=&\sum_{i=2}^{n}(\rho^{i-1}y_1-\rho y_1\tilde M)^2\\ &=&\rho^2y_1^2\sum_{i=2}^{n}(\rho^{i-2}-\tilde M)^2\\ &=&\rho^2y_1^2\sum_{i=1}^{n-1}(\rho^{i-1}-\tilde M)^2\\ &=&\rho^2y_1^2D \end{eqnarray*} For the numerator of the correlation, a similar calculation shows \begin{eqnarray*} \sum_{i=2}^n(y_{i-1}-\bar{y}_{lag})(y_i-\bar{y}_{lead}) &=&\sum_{i=2}^n(y_1\rho^{i-2}-y_1\tilde M)(y_1\rho^{i-1}-\rho y_1\tilde M)\\ &=&\rho y_1^2\sum_{i=2}^n(\rho^{i-2}-\tilde M)(\rho^{i-2}-\tilde M)\\ &=&\rho y_1^2\sum_{i=2}^n(\rho^{i-2}-\tilde M)^2\\ &=&\rho y_1^2\sum_{i=1}^{n-1}(\rho^{i-1}-\tilde M)^2=\rho y_1^2D\\ \end{eqnarray*} Thus, $$ corr=\frac{\rho y_1^2D}{\sqrt{\rho^2y_1^2Dy_1^2D}}=1 $$ The ACF, in turn, does not divide by the product of standard deviations, but, assuming stationarity, a common overall variance estimate. An analogous exercise will reveal that the cancellation then do not happen as with the standard correlation formula.

In particular, the full sample mean is now $$ \bar y=\frac{1}{n}\sum_{i=1}^ny_i=y_1\frac{1-\rho^n}{n(1-\rho)}\equiv y_1\breve M $$ such that the full variance computed around this full mean and the numerator in the correlation formula will not be proportional in the same way as the product of square roots of variations and numerator were. A bit more specifically, $$ \breve M=\tilde M+\frac{\rho^{n-1}}{n-1} $$ Also, in the variance formula, we now have $n$ squared deviations, such that $var(y)=y_1^2\cdot\breve D/(n-1)$, where $$ \breve D:=\sum_{i=1}^n(\rho^{i-1}-\breve M)^2 $$ Thus, the ratio of covariation and variance is $$ acf=\rho\frac{D}{\breve D}, $$ which is not equal to one, but rather close to, but also not equal to $\rho$. Note however that this expression is independent of the starting value $y_1$. (I haven't been able to further simplify the ratio, see also here.)