A115752 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.

0, 0, 3, 4, 15, 81, 168, 540, 2271, 5365, 16698, 63229, 159250, 489048, 1749933, 4576140, 13955895, 48211389, 129211818, 392441049, 1323741156, 3609608838, 10933915743, 36252591813, 100126350090, 302737691646, 990855646563

Offset: 0

Author

Kevin Smith (kjsmith(AT)yorku.ca), Mar 28 2006

Keywords

• nonn

Comments

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 equally-likely 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

Examples

			For n=4, a(4)=15 since there are 15 five-letter 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
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..1000
• Mike Zabrocki, Math 5020, York University

Crossrefs

Cf. A092255.

Programs

Maple

seq(add(binomial(n+1,i)*add(binomial(n+1-i,j), j=i+1..floor((n-i)/2)), i=0..floor((n-2)/3)), n=0..30); # Dennis P. Walsh, Jun 19 2013

Formula

E.g.f.: (t(1)^3-3*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

a(n) = sum(sum(n!/(i!j!(n-i-j)!), j=i+1..floor((n-i)/2)), j=0..floor((n-2)/3)). - Dennis P. Walsh, Jun 19 2013