A334288 Number of tieless rugby (union) games with n scoring events.

1, 6, 30, 180, 1002, 6012, 34224, 205344, 1180010, 7080060, 40911324, 245467944, 1423944024, 8543664144, 49710351720, 298262110320, 1739627237002, 10437763422012, 61002039226716, 366012235360296, 2142786218045748, 12856717308274488, 75380119335678608

Offset: 0

Author

Cameron Ford, Jun 13 2020

Keywords

• nonn

Comments

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

Examples

			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}.
		

Links

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

Crossrefs

Inspired by A137684.

Programs

Python

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 nonzero scores with i scoring events in
        # dictionary_of_scores. To find nonzero scores with i+1 scoring events
        # consider each nonzero 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())