A367948
Triangular array read by rows. T(n,k) is the number of strongly connected binary relations on [n] (A186081) with period k, n >= 1, 1<=k<=n.
Original entry on oeis.org
1, 3, 1, 139, 3, 2, 25575, 103, 12, 6, 18077431, 4815, 230, 60, 24
Offset: 1
Triangle begins ...
1;
3, 1;
139, 3, 2;
25575, 103, 12, 6;
18077431, 4815, 230, 60, 24;
...
T(4,3) = 12. Let A be the strongly connected relation on [4] whose adjacency matrix is {{0,0,0,1},{0,0,0,1},{1,1,0,0},{0,0,1,0}}. It is easy to check that the period of A is 3. Also, G(A) contains two cycles of length 3 so that the GCD of its cycle length is 3. Also {A^i:i>=1} contains the equivalence relation corresponding to the set partition {1,2}{3}{4}. There are 12 relations in the same isomorphism class as A so that T(4,3) = 12.
- R. Brualdi and H. Ryser, Combinatorial Matrix Theory, Cambridge Univ. Press, 1991, pages 53-96.
- Ki Hang Kim, Boolean Matrix Theory and Applications, Marcel Dekker, 1982, pages 177-226.
A340264
T(n, k) = Sum_{j=0..k} binomial(n, k - j)*Stirling2(n - k + j, j). Triangle read by rows, 0 <= k <= n.
Original entry on oeis.org
1, 0, 2, 0, 1, 4, 0, 1, 6, 8, 0, 1, 11, 24, 16, 0, 1, 20, 70, 80, 32, 0, 1, 37, 195, 340, 240, 64, 0, 1, 70, 539, 1330, 1400, 672, 128, 0, 1, 135, 1498, 5033, 7280, 5152, 1792, 256, 0, 1, 264, 4204, 18816, 35826, 34272, 17472, 4608, 512
Offset: 0
[0] 1;
[1] 0, 2;
[2] 0, 1, 4;
[3] 0, 1, 6, 8;
[4] 0, 1, 11, 24, 16;
[5] 0, 1, 20, 70, 80, 32;
[6] 0, 1, 37, 195, 340, 240, 64;
[7] 0, 1, 70, 539, 1330, 1400, 672, 128;
[8] 0, 1, 135, 1498, 5033, 7280, 5152, 1792, 256;
[9] 0, 1, 264, 4204, 18816, 35826, 34272, 17472, 4608, 512;
Alternating sum of row(n) is
A109747(n).
-
egf := exp(t*(exp(-x) - x - 1));
ser := series(egf, x, 22):
p := n -> coeff(ser, x, n);
seq(seq((-1)^n*n!*coeff(p(n), t, k), k=0..n), n = 0..10);
# Alternative:
T := (n, k) -> add(binomial(n, k - j)*Stirling2(n - k + j, j), j=0..k):
seq(seq(T(n, k), k = 0..n), n=0..9); # Peter Luschny, Feb 09 2021
-
T[ n_, k_] := Sum[ Binomial[n, k-j] StirlingS2[n-k+j, j], {j, 0 ,k}]; (* Michael Somos, Jul 18 2021 *)
-
T(n, k) = sum(j=0, k, binomial(n, j)*stirling(n-j, k-j, 2)); /* Michael Somos, Jul 18 2021 */
Showing 1-2 of 2 results.
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