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%I #8 May 01 2013 21:02:10
%S 0,0,0,72,7200,907200,151351200,32818867200,9022953139200,
%T 3074644210176000,1273420678578048000,630456744366408960000,
%U 367858475127819067392000,249891669321208978871808000,195560183245820886392739840000,174686971253081560944736665600000,176671699588778004249333176401920000
%N Let k=binomial(n-1,2); a(n) = n*(n-1)*k!/(k-n+1)! for n >= 4, with a(1)=a(2)=a(3)=0.
%C Suggested by Question 8 on the Mathpath 2006 Qualifying Quiz, which says:
%C "You are given 5 dots arranged on a circle and told to draw segments between pairs of the points to connect all the dots. It is always possible to do this with 4 segments. However, suppose you are required to use 5 segments, that is, the dots should not all be connected until you draw your fifth segment. For instance, if the dots are numbered 1,2,3,4,5, one way to do this is to draw the following sequence of segments: 12, 34, 24, 13, 35. Another sequence would be 34, 13, 12, 24, 35; it uses the same segments but in a different order. (But careful: some other orders of these 5 segments do not count; why not?). Another sequence, using some different edges, is 23, 24, 25, 34, 15. Note that you may not draw the same segment twice. In other words, 12, 23, 23, 34, 45 uses only 4 segments, not 5.
%C How many sequences are there which take 5 segments to connect all 5 dots?"
%C The sequence arises if we replace "5" by "n".
%C However, "Carl L." (minesweeperman337(AT)yahoo.com) wrote on Dec 28 2009 to say that in fact 7200 was not the right answer to the Mathpath problem, the correct answer was 8640. So there may be no connection with the Mathpath problem! - _N. J. A. Sloane_, Jan 04 2010.
%H Mathpath 2006, <a href="http://www.mathpath.org/Quiz/quiz.2006.pdf">Qualifying Quiz</a>
%p f:=proc(n) local t1,t2,t3; t1:=binomial(n-1,2); t2:=t1!/(t1-n+1)!; t3:=n*(n-1)*t2; end;
%K nonn
%O 1,4
%A _N. J. A. Sloane_, Aug 01 2006, Jan 04 2010
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