Passing as arguments the first day (first_date), the number of weeks between events (k), the days of the week (days_of_the_week) and the number of next event requested (n) will answer with the n next appointment dates

Get calendar dates.py

import datetime

FORMAT = '%d/%m/%Y'


def next_weekday(d, weekday):
    days_ahead = weekday - d.weekday()
    if days_ahead <= 0:  # Target day already happened this week
        days_ahead += 7
    return d + datetime.timedelta(days_ahead)


def recurring_task(first_date, k, days_of_the_week, n):
    fd = datetime.datetime.strptime(first_date, FORMAT)
    weekdays = ["Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday", "Sunday"]
    # get the first day of each days_of_the_week
    out = []
    fd -= datetime.timedelta(days=1)
    for day in days_of_the_week:
        next_day = weekdays.index(day)
        next_fd = next_weekday(fd, next_day)
        # build a list of next events for each day of the week
        # getting enough elements for the next elements
        for i in range((n / len(days_of_the_week)) + 1):
            next_fd_with_k = next_fd + datetime.timedelta(i * k * 7)
            out.append(next_fd_with_k)
    # collect the generated information and sort it
    out.sort()
    return map(lambda x: x.strftime(FORMAT), out[:n])


def test():
    assert recurring_task('01/01/2015', 2, ["Monday", "Thursday"], 4) == ["01/01/2015", "05/01/2015", 
                                                                          "15/01/2015", "19/01/2015"]