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A334288 Number of tie-less rugby (union) games with n scoring events. 2


%S 1,6,30,180,1002,6012,34224,205344,1180010,7080060,40911324,245467944,

%T 1423944024,8543664144,49710351720,298262110320,1739627237002,

%U 10437763422012,61002039226716,366012235360296,2142786218045748,12856717308274488,75380119335678608

%N Number of tie-less rugby (union) games with n scoring events.

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

%C 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.

%C 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).

%H Cameron Ford, <a href="/A334288/b334288.txt">Table of n, a(n) for n = 0..1286</a>

%e 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}.

%o (Python)

%o def number_of_tieless_rugby_games(n):

%o """

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

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

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

%o Negative points represent points for the away team, positive points

%o represent points for the home team

%o """

%o dictionary_of_scores = {0:1}

%o # The keys of this dictionary represent possible scores.

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

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

%o for i in range(n):

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

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

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

%o # scoring event.

%o old_dictionary = dictionary_of_scores

%o dictionary_of_scores = {}

%o for score, number_of_ways in old_dictionary.items():

%o for scoring_event in scoring_events:

%o new_score = score + scoring_event

%o if new_score != 0:

%o dictionary_of_scores[new_score] =\

%o dictionary_of_scores.get(new_score, 0) + number_of_ways

%o return sum(dictionary_of_scores.values())

%Y Inspired by A137684.

%K nonn

%O 0,2

%A _Cameron Ford_, Jun 13 2020

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Last modified September 23 19:41 EDT 2020. Contains 337315 sequences. (Running on oeis4.)