Change the Code, Change the Sequence: Robbers This is a cops-and-robbers challenge. For the cops thread, go here.

This challenge involves two OEIS sequences chosen by the cops -- S1, S2 -- and how well those sequences can be golfed and obfuscated.

The cops are constructing code A that produces S1 and giving a number X that they claim is the best Levenshtein distance possible (in characters) to create B that produces S2.

The Robbers' Challenge

To crack a particular cop's submission, robbers must come up with a program C in the same language (and version) as that answer that produces S2(n) and is Y character changes away from A (with Y &lt;= X). Robbers do not necessarily need to find the exact same B code that the cop (secretly) produced. The robbers' submissions must adhere to the same 0-index or 1-index as specified by the cop's submission.

If you manage this, post an answer with the solution, linking to the cop's answer, and leave a comment on the cop's answer linking back to yours.

Each cop answer can only be cracked once, and of course, you're not allowed to crack your own answer. If the cop's answer turns out to be invalid before or after being cracked, it is not counted towards the robber's score.

Winning and Scoring

Robbers are scored by (X - Y)*5 + 5 for each of their cracks, and the robber with the overall highest score wins.

Further Rules


You must not use any built-ins for hashing, encryption, or random number generation (even if you seed the random number generator to a fixed value).
Either programs or functions are allowed, but the code must not be a snippet and you must not assume a REPL environment.
You may take input and give output in any convenient format. The input/output methods must be the same for both sequences.
The definitive calculator for the Levenshtein distance for this challenge is this one on Planet Calc.
In addition to being a CnR challenge, this is code-golf so all usual golfing rules apply.
Pyke, Levenshtein distance of 1, A036487, A135628 - score 5
Crack of an entry by muddyfish
wX*e

Try it here!
The original code, X*e, squares the input, X, multiplies that by the input *, and halves then floors the result, e.
The trick is that 'X' is 56 in the base 96 representation of w, so wX yields 56, multiply that by the input then floor and halve and you get 28 times the input as needed.