A326323 A(n, k) = A_{n}(k) where A_{n}(x) are the Eulerian polynomials, square array read by ascending antidiagonals, for n >= 0 and k >= 0.
1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 6, 1, 1, 1, 4, 13, 24, 1, 1, 1, 5, 22, 75, 120, 1, 1, 1, 6, 33, 160, 541, 720, 1, 1, 1, 7, 46, 285, 1456, 4683, 5040, 1, 1, 1, 8, 61, 456, 3081, 15904, 47293, 40320, 1, 1, 1, 9, 78, 679, 5656, 40005, 202672, 545835, 362880, 1
Offset: 0
Examples
Array starts: k=0: 1, 1, 1, 1, 1, 1, 1, 1, 1, ... [A000012] k=1: 1, 1, 2, 6, 24, 120, 720, 5040, 40320, ... [A000142] k=2: 1, 1, 3, 13, 75, 541, 4683, 47293, 545835, ... [A000670] k=3: 1, 1, 4, 22, 160, 1456, 15904, 202672, 2951680, ... [A122704] k=4: 1, 1, 5, 33, 285, 3081, 40005, 606033, 10491885, ... [A255927] k=5: 1, 1, 6, 46, 456, 5656, 84336, 1467376, 29175936, ... [A326324] k=6: 1, 1, 7, 61, 679, 9445, 158095, 3088765, 68958295, ... [A384525] k=7: 1, 1, 8, 78, 960, 14736, 272448, 5881968, 145105920, ... [A384514] k=8: 1, 1, 9, 97, 1305, 21841, 440649, 10386817, 279768825, ... Seen as a triangle: [0], 1 [1], 1, 1 [2], 1, 1, 1 [3], 1, 1, 2, 1 [4], 1, 1, 3, 6, 1 [5], 1, 1, 4, 13, 24, 1 [6], 1, 1, 5, 22, 75, 120, 1 [7], 1, 1, 6, 33, 160, 541, 720, 1 [8], 1, 1, 7, 46, 285, 1456, 4683, 5040, 1 [9], 1, 1, 8, 61, 456, 3081, 15904, 47293, 40320, 1
Links
- OEIS Wiki, Eulerian polynomials.
Programs
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Maple
A := (n, k) -> add(combinat:-eulerian1(k, j)*n^j, j=0..k): seq(seq(A(n-k, k), k=0..n), n=0..10); # Alternative: egf := n -> `if`(n=1, 1/(1-x), (n-1)/(n - exp((n-1)*x))): ser := n -> series(egf(n), x, 21): for n from 0 to 6 do seq(k!*coeff(ser(n), x, k), k=0..9) od;
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Mathematica
a[n_, 0] := 1; a[n_, 1] := n!; a[n_, k_] := (k - 1)^(n + 1)/k HurwitzLerchPhi[1/k, -n, 0]; (* Alternative: *) a[n_, k_] := Sum[StirlingS2[n, j] (k - 1)^(n - j) j!, {j, 0, n}]; Table[Print[Table[a[n, k], {n, 0, 10}]], {k, 0, 8}]
Formula
A(n, k) = Sum_{j=0..k} a(k, j)*n^j where a(k, j) are the Eulerian numbers.
E.g.f.: (n - 1)/(n - exp((n-1)*x)) for n = 0 and n >= 2, 1/(1 - x) if n = 1.
A(n, 0) = 1; A(n, 1) = n!.
A(n, k) = (k - 1)^(n + 1)/k HurwitzLerchPhi(1/k, -n, 0) for k >= 2.
A(n, k) = Sum_{j=0..n} j! * Stirling2(n, j) * (k - 1)^(n - j) for k >= 2.