OFFSET
4,2
COMMENTS
The formula given below was conjectured by Martha Yip and proved by Robert Cori and Gábor Hetyei.
More generally one may consider genus-dependent Stirling numbers S(n, k, g) that count the partitions of genus g and k parts of the n-set.
Then T(n, k) = S(n, k, 1). See Robert Coquereaux and Jean-Bernard Zuber.
LINKS
Robert Coquereaux and Jean-Bernard Zuber, Counting partitions by genus: a compendium of results, Journal of Integer Sequences, Vol. 27 (2024), Article 24.2.6. See p. 9. See also arXiv:2305.01100, 2023.
Robert Cori and Gábor Hetyei, Counting genus one partitions and permutations, arXiv:1306.4628 [math.CO], 2013.
Robert Cori and Gábor Hetyei, Counting genus one partitions and permutations, Sémin. Lothar. Comb. 70, B70e, 30 p. (2014).
Martha Yip, Genus one partitions, Master Thesis, University of Waterloo, 2006.
FORMULA
T(n, k) = (1/6)*binomial(n, 2)*binomial(n-2, k)*binomial(n-2, k-2).
EXAMPLE
Triangle begins (see Table 3.1 in Yip's thesis):
1;
5, 5;
15, 40, 15;
35, 175, 175, 35;
70, 560, 1050, 560, 70;
126, 1470, 4410, 4410, 1470, 126;
MATHEMATICA
T[n_, k_] := (1/6) Binomial[n, 2] Binomial[n-2, k] Binomial[n-2, k-2];
Table[T[n, k], {n, 4, 12}, {k, 2, n-2}]//Flatten (* Stefano Spezia, Feb 14 2024 *)
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Robert Coquereaux, Feb 12 2024
STATUS
approved