OFFSET
1,3
COMMENTS
See "Example" below for the starting construction of S(n,k) and T(n,k).
To understand S(n,k) and Q, consider example S(5,k), i.e., f+g+h = 5, and S(n,k) are listed in reverse lexicographic order. So S(5,k) = {5,0,0}, {4,1,0}, {3,2,0}, {3,1,1}, {2,2,1} k=1..5, respectively. Q is the number of permutations of S(n,k). So Q=3 when S(n,k) = {5,0,0}, {3,1,1} and {2,2,1}; and Q=6 when S(n,k) = {4,1,0} and {3,2,0}.
In general, by definition: Q=1 when all members of S(n,k) are equal, Q=3 when S(n,k) contains a pair, and Q=6 when none of the members of S(n,k) is equal.
Suppose three equally-matched players are playing a tournament of n games; and for each game there is one winner and two losers. Then S(n,k) is the "overall win record" (where player order does not matter) after n games. Let p be the probability that any S(n,k) occurs after n games. Then p = T(n,k)/3^(n-1). (See also "Example" section.)
Generally, when S(n,k) is a z-member set {f,g,h,i..,y}, then Q is the number of permutations of S(n,k), T(n,k) = n!*Q/(z*f!*g!*h!..*y!) and p = T(n,k)/z^(n-1). So when z=2 we get A008314. (Observation prompted by query from Linda Rogers.)
For triangle T(n,k):
Row sums are 3^(n-1).
Row lengths are A001399(n).
Final terms in each row are A199127(n).
For n >= 3: T(n,2) = 2*n.
For n >= 5: T(n,3) = T(n,4) = A002378(n-1) (oblong numbers).
For n >= 6: T(n,6) = A007531(n).
For n >= 8: T(n,9) = A033487(n-3).
LINKS
Nicolas Behr, Pawel Sobocinski, Rule Algebras for Adhesive Categories, arXiv:1807.00785 [cs.LO], 2018, also LIPIcs 27th EACSL Annual Conference on Computer Science Logic (CSL 2018), Vol. 119, pp. 11:1-11:21.
EXAMPLE
Triangle T(n,k) begins:
n/k 1 2 3 4 5 6 7 8 9 10 11 12 13 14
1: 1
2: 1, 2
3: 1, 6, 2
4: 1, 8, 6, 12
5: 1, 10, 20, 20, 30
6: 1, 12, 30, 30, 20, 120, 30
7: 1, 14, 42, 42, 70, 210, 140, 210
8: 1, 16, 56, 56, 112, 336, 70, 560, 420, 560
9: 1, 18, 72, 72, 168, 504, 252, 1008, 756, 630, 2520, 560
10: 1, 20, 90, 90, 240, 720, 420, 1680, 1260, 252, 2520, 5040, 3150, 4200
Triangle S(n,k) begins:
n/k 1 2 3 4 5 6 7
1: {1,0,0}
2: {2,0,0} {1,1,0}
3: {3,0,0} {2,1,0} {1,1,1}
4: {4,0,0} {3,1,0} {2,2,0} {2,1,1}
5: {5,0,0} {4,1,0} {3,2,0} {3,1,1} {2,2,1}
6: {6,0,0} {5,1,0} {4,2,0} {4,1,1} {3,3,0} {3,2,1} {2,2,2}
T(4,3) = 6 because n=4 and S(4,3) = {2,2,0}; so Q=3 and 3*4!/(3*2!*2!*0!) = 6. Therefore p = 6/27 = 2/9 that the overall win record = {2,2,0} after playing 4 tournament games.
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Bob Selcoe, Apr 30 2017
STATUS
approved