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
Keywords
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
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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
Comments