A199127 -id:A199127 - cp's OEIS frontend

A286030 Irregular triangle T(n,k) read by rows: Let S be a 3-member set of integers {f,g,h} where f >= g >= h >= 0 and f+g+h = n. Let S(n,k) be an irregular triangle composed of all S listed in reverse lexicographic order by row n. Then T(n,k) = n!Q/(3f!g!h!), where Q is the number of permutations of S(n,k). (See "Comments" and "Examples" for additional explanation.)

1, 1, 2, 1, 6, 2, 1, 8, 6, 12, 1, 10, 20, 20, 30, 1, 12, 30, 30, 20, 120, 30, 1, 14, 42, 42, 70, 210, 140, 210, 1, 16, 56, 56, 112, 336, 70, 560, 420, 560, 1, 18, 72, 72, 168, 504, 252, 1008, 756, 630, 2520, 560

Offset: 1

Bob Selcoe, Apr 30 2017

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).

			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.

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.

Cf. A001399, A002378, A007531, A008314, A033487, A199127.

Cf. A000041 (partition numbers).

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