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A334288 Number of tie-less rugby (union) games with n scoring events. 2
1, 6, 30, 180, 1002, 6012, 34224, 205344, 1180010, 7080060, 40911324, 245467944, 1423944024, 8543664144, 49710351720, 298262110320, 1739627237002, 10437763422012, 61002039226716, 366012235360296, 2142786218045748, 12856717308274488, 75380119335678608 (list; graph; refs; listen; history; text; internal format)



In rugby (union) a scoring event can give 3, 5 or 7 points.

In April 1992 the current scoring format was introduced: 3 points are awarded for kicks/penalties, 5 points for unconverted tries and 7 points for converted tries. A game is a list of members of {-7,-5,-3,3,5,7} with negative points for the away team, positive for the home team.

A tie-less game is one in which the teams never have the same score (except at the beginning, when no team has scored yet).


Cameron Ford, Table of n, a(n) for n = 0..1286


a(2)=30, because there are 6^2=36 sequences of length 2 from {3,5,7,-3,-5,-7}; the 6 sequences that correspond to games with ties are precisely those of the form {k,-k}.



def number_of_tieless_rugby_games(n):


    Returns the number of tieless rugby games with n scoring events.

    A scoring event is a number in (-7, -5, -3, 3, 5, 7) and a game is tieless

    if the score is never zero, apart from at the start.

    Negative points represent points for the away team, positive points

    represent points for the home team


    dictionary_of_scores = {0:1}

    # The keys of this dictionary represent possible scores.

    # The values represent the number of ways this score can be reached.

    scoring_events = (-7, -5, -3, 3, 5, 7)

    for i in range(n):

        # At each stage, we have the non-zero scores with i scoring events in

        # dictionary_of_scores. To find non-zero scores with i+1 scoring events

        # consider each non-zero score, and each possibility for the next

        # scoring event.

        old_dictionary = dictionary_of_scores

        dictionary_of_scores = {}

        for score, number_of_ways in old_dictionary.items():

            for scoring_event in scoring_events:

                new_score = score + scoring_event

                if new_score != 0:

                    dictionary_of_scores[new_score] =\

                    dictionary_of_scores.get(new_score, 0) + number_of_ways

    return sum(dictionary_of_scores.values())


Inspired by A137684.

Sequence in context: A089896 A057754 A001473 * A063888 A029571 A259276

Adjacent sequences:  A334285 A334286 A334287 * A334289 A334290 A334291




Cameron Ford, Jun 13 2020



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Last modified November 26 09:38 EST 2020. Contains 338639 sequences. (Running on oeis4.)