A348436 Triangle read by rows. T(n,k) is the number of labeled threshold graphs on n vertices with k components, for 1 <= k <= n.
1, 1, 1, 4, 3, 1, 23, 16, 6, 1, 166, 115, 40, 10, 1, 1437, 996, 345, 80, 15, 1, 14512, 10059, 3486, 805, 140, 21, 1, 167491, 116096, 40236, 9296, 1610, 224, 28, 1, 2174746, 1507419, 522432, 120708, 20916, 2898, 336, 36, 1, 31374953, 21747460, 7537095, 1741440, 301770, 41832, 4830, 480, 45, 1
Offset: 1
Examples
Triangle begins: 1; 1, 1; 4, 3, 1; 23, 16, 6, 1; 166, 115, 40, 10, 1; 1437, 996, 345, 80, 15, 1; 14512, 10059, 3486, 805, 140, 21, 1; 167491, 116096, 40236, 9296, 1610, 224, 28, 1; 2174746, 1507419, 522432, 120708, 20916, 2898, 336, 36, 1; 31374953, 21747460, 7537095, 1741440, 301770, 41832, 4830, 480, 45, 1; ...
Links
- D. Galvin, G. Wesley and B. Zacovic, Enumerating threshold graphs and some related graph classes, arXiv:2110.08953 [math.CO], 2021.
- Sam Spiro, Counting Threshold Graphs with Eulerian Numbers, arXiv:1909.06518 [math.CO], 2019.
Programs
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Maple
T := (n, k) -> `if`(n = k, 1, binomial(n, k-1)*A053525(n-k+1)): for n from 1 to 10 do seq(T(n, k), k=1..n) od; # Peter Luschny, Oct 24 2021
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Mathematica
eulerian[0, 0] := 1; eulerian[n_, m_] := eulerian[n, m] = Sum[((-1)^k)*Binomial[n + 1, k]*((m + 1 - k)^n), {k, 0, m + 1}]; (* t[n] counts the labeled threshold graphs on n vertices *) t[0] = 1; t[1] = 1; t[n_] := t[n] = Sum[(n - k)*eulerian[n - 1, k - 1]*(2^k), {k, 1, n - 1}]; T[1, 1] := 1; T[n_, 1] := T[n, 1] = (1/2)*t[n]; T[n_, n_] := T[n, n] = 1; T[n_, k_] := T[n, k] = Binomial[n, k - 1]*T[n - k + 1, 1]; Table[T[n, k], {n, 1, 10}, {k, 1, n}] // Flatten
Formula
T(1,1) = 1; for n >= 2, T(n,1) = A005840(n)/2; for n >= 3 and 2 <= k <= n-1, T(n,k) = binomial(n,k-1)*T(n-k+1,1); and for n >= 2, T(n,n)=1.
T(n, k) = binomial(n, k-1)*A053525(n - k + 1) if k != n, otherwise 1. - Peter Luschny, Oct 24 2021
Comments