%I
%S 0,0,3,4,15,81,168,540,2271,5365,16698,63229,159250,489048,1749933,
%T 4576140,13955895,48211389,129211818,392441049,1323741156,3609608838,
%U 10933915743,36252591813,100126350090,302737691646,990855646563
%N Number of words of length n+1 created with the letters a,b,c with more c's than b's and more b's than a's.
%C Also, if n+1 voters vote for one of the three candidates (A, B, or C) in an election, a(n) is the number of possible ballot results in which candidate C gets more votes than candidate B and candidate B gets more votes than candidate A. We note that the number of all possible ballot results is 3^(n+1). Hence, if all three candidates are equallylikely to get a random voter's vote, the probability of no ties among any of the candidates is 3!*a(n)/3^(n+1).  _Dennis P. Walsh_, Jun 19 2013
%H Alois P. Heinz, <a href="/A115752/b115752.txt">Table of n, a(n) for n = 0..1000</a>
%H Mike Zabrocki, <a href="http://garsia.math.yorku.ca/~zabrocki/math5020f05/OLEISexercises.pdf">Math 5020, York University</a>
%F E.g.f.: (t(1)^33*t(1)*t(2)+2*t(3))/6 where t(1)=hypergeom([],[],x), t(2)=hypergeom([],[1],x^2) and t(3)=hypergeom([],[1,1],x^3).  _Vladeta Jovovic_, Sep 22 2007
%F a(n) = sum(sum(n!/(i!j!(nij)!), j=i+1..floor((ni)/2)), j=0..floor((n2)/3)).  _Dennis P. Walsh_, Jun 19 2013
%e For n=4, a(4)=15 since there are 15 fiveletter words with more c's than b's and more b's than a's. Ten of the words use 3 c's and 2 b's, namely, cccbb, ccbcb, ccbbc, cbccb, cbcbc, cbbcc, bcccb, bccbc, bcbcc, and bbccc; and 5 of the words use 4 c's and 1 b, namely, ccccb, cccbc, ccbcc, cbccc, and bcccc.  _Dennis P. Walsh_, Jun 19 2013
%p seq(add(binomial(n+1,i)*add(binomial(n+1i,j), j=i+1..floor((ni)/2)), i=0..floor((n2)/3)), n=0..30); # _Dennis P. Walsh_, Jun 19 2013
%Y Cf. A092255.
%K nonn
%O 0,3
%A Kevin Smith (kjsmith(AT)yorku.ca), Mar 28 2006
