Password entropy is based on the number of possible combinations.
For a pattern to reduce entropy, the pattern needs to be known to the attacker.
If the pattern isn't fixed and known, the reduction in complexity will be specific to the attacker's algorithm. It will be different for different attackers. E.g., with a dictionary attack, effective complexity for dictionary words is dictionary size, for words out of the dictionary full brute force complexity.
I don't know whether there is or can be a mathematically optimal algorithm for letter repetitions. The simplest solution I can think of is to treat symbol repetitions, up to a N repetitions, as extra graphemes. In that case, the password's strength will be the number of graphemes it contains.
Against such an algorithm, the added strength of each consecutive symbol from the 2nd to Nth symbols is 0, and equivalent to another grapheme up to 2N. But such an algorithm itself will be slower against random passwords due to a larger effective character set. For instance, checking for 2-long repetitions for every symbol will drop its speed by a factor of 2^(length-1). But against a password that is all double symbols, its speed will improve to only a square root of above complexity.
If it's imperative that brute force efficiency is not compromised, a free tweak is always starting to pick the last grapheme as being equal to the previous one. In that case the added strength from the first (for variable length) or all repeat symbols at the end (for fixed length) is 0.
Doing so in the middle isn't a free tweak anymore. Neither is trying for extra repeats in a variable length password, although it's so cheap as to be nearly free.
In short, against an algorithm optimized to pick passwords with repeated characters, a password's strength can be estimated by dropping all consecutive digits past the first to get effective length L'. Theoretical strength would be approximately (charset*N)^L', where N is the maximum number of repetitions the attacker is testing for.
Against an algorithm optimized for brute-force efficiency, only the consecutive digits at the end should be dropped. Theoretical strength with a naive algorithm would be charset^(L'-1)*(charset+N). Any practical algorithm will be testing for a lot of suffixes already, though (passwords often end in "1" or "1!" to bypass complexity rules).
It's all a matter of what algorithm the attacker uses. Dictionaries, including leaked password lists, will generally be tried first.