OFFSET
3,2
LINKS
Alois P. Heinz, Table of n, a(n) for n = 3..10585
EXAMPLE
T(4,3) = 4, because 4 simple graphs on 4 labeled nodes with 3 edges contain a cycle subgraph:
..1-2...1-2...1.2...1.2..
..|/.....\|...|\...../|..
..3.4...3.4...3-4...3-4..
Triangle begins:
1;
4, 15, 6, 1;
10, 85, 252, 210, 120, 45, 10, 1;
20, 285, 1707, 5005, 6435, 6435, 5005, 3003, 1365, 455, 105, 15, 1;
MAPLE
B:= proc(n) option remember; if n=0 then 0 else B(n-1) +n^(n-1) *x^n/n! fi end: BB:= proc(n) option remember; expand (B(n) -B(n)^2/2) end: f:= proc(k) option remember; if k=0 then 1 else unapply (f(k-1)(x) +x^k/k!, x) fi end: A:= proc(n, k) option remember; series(f(k)(BB(n)), x, n+1) end: aa:= (n, k)-> coeff (A(n, k), x, n) *n!: b:= (n, k)-> if k>=n then 0 else aa(n, n-k) -aa(n, n-k-1) fi: T:= (n, k)-> product (n*(n-1)/2-j, j=0..k-1)/k! -b(n, k): seq (seq (T(n, k), k=3..n*(n-1)/2), n=3..8);
MATHEMATICA
(* t = A138464 *) t[0, 0] = 1; t[n_, k_] /; (0 <= k <= n-1) := t[n, k] = Sum[(i+1)^(i-1)*Binomial[n-1, i]*t[n-i-1, k-i], {i, 0, k}]; t[_, _] = 0; T[n_, k_] := Binomial[n*(n-1)/2, k]-t[n, k]; Table[Table[T[n, k], {k, 3, n*(n-1)/2}], {n, 3, 8}] // Flatten (* Jean-François Alcover, Feb 14 2014 *)
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Alois P. Heinz, Sep 04 2008
STATUS
approved