A356545
Triangle read by rows. T(n, k) are the coefficients of polynomials p_n(x) based on the Eulerian numbers of first order representing the Bernoulli numbers as B_n = p_n(1) / (n + 1)!.
Original entry on oeis.org
1, 1, 0, 2, -1, 0, 6, -8, 2, 0, 24, -66, 44, -6, 0, 120, -624, 792, -312, 24, 0, 720, -6840, 14496, -10872, 2736, -120, 0, 5040, -86400, 285840, -347904, 171504, -28800, 720, 0, 40320, -1244880, 6181920, -11245680, 8996544, -3090960, 355680, -5040, 0
Offset: 0
The table T(n, k) of the coefficients, sorted in ascending order, starts:
[0] 1;
[1] 1, 0;
[2] 2, -1, 0;
[3] 6, -8, 2, 0;
[4] 24, -66, 44, -6, 0;
[5] 120, -624, 792, -312, 24, 0;
[6] 720, -6840, 14496, -10872, 2736, -120, 0;
[7] 5040, -86400, 285840, -347904, 171504, -28800, 720, 0;
[8] 40320, -1244880, 6181920, -11245680, 8996544, -3090960, 355680, -5040, 0;
- R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, 2nd ed. Addison-Wesley, Reading, MA, 1994, p. 268. (Since the thirty-fourth printing, Jan. 2022, with B(1) = 1/2.)
- Ira Gessel, Eulerian number identity, MathOverflow, Apr 2019.
- Peter Luschny, How are the Eulerian numbers of the first-order related to the Eulerian numbers of the second-order?, MathOverflow, Feb. 2021.
- Peter Luschny, Eulerian polynomials.
- Oskar Schlömilch, Ueber die Bernoulli'sche Funktion und deren Gebrauch bei der Entwickelung halbconvergenter Reihen, Zeitschrift fuer Mathematik und Pysik, vol. 1 (1856), p. 193-211.
- Julius Worpitsky, Studien über die Bernoullischen und Eulerschen Zahlen, Journal für die reine und angewandte Mathematik (Crelle), 94 (1883), 203-232. See page 22, first formula.
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E1 := proc(n, k) combinat:-eulerian1(n, k) end:
p := (n, x) -> add(E1(n, k)*k!*(n - k)!*(-x)^k, k = 0..n):
seq(print(seq(coeff(p(n, x), x, k), k=0..n)), n = 0..8);
seq(p(n, 1)/(n + 1)!, n = 0..14); # check the Bernoulli representation
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T[n_, k_] := k! * (n-k)! * Sum[(-1)^(k-j) * (k-j+1)^n * Binomial[n+1, j], {j, 0, k}]; Table[T[n, k], {n, 0, 9}, {k, 0, n}] // TableForm
(* Diagonals: *)
d[n_, k_] := k! * (n - k)! * Sum[(-1)^(n-k-j)*(n - j - k + 1)^n * Binomial[n + 1, j], {j, 0, n - k}];
A342321
T(n, k) = A343277(n)*[x^k] p(n, x) where p(n, x) = (1/(n+1))*Sum_{k=0..n} (-1)^k*E1(n, k)*x^(n - k) / binomial(n, k), and E1(n, k) are the Eulerian numbers A123125. Triangle read by rows, for 0 <= k <= n.
Original entry on oeis.org
1, 0, 1, 0, -1, 2, 0, 1, -4, 3, 0, -3, 22, -33, 12, 0, 1, -13, 33, -26, 5, 0, -5, 114, -453, 604, -285, 30, 0, 5, -200, 1191, -2416, 1985, -600, 35, 0, -35, 2470, -21465, 62476, -78095, 42930, -8645, 280, 0, 14, -1757, 21912, -88234, 156190, -132351, 51128, -7028, 126
Offset: 0
Triangle starts:
[n] T(n, k) A343277(n)
----------------------------------------------------------
[0] 1; [1]
[1] 0, 1; [2]
[2] 0, -1, 2; [6]
[3] 0, 1, -4, 3; [12]
[4] 0, -3, 22, -33, 12; [60]
[5] 0, 1, -13, 33, -26, 5; [30]
[6] 0, -5, 114, -453, 604, -285, 30; [210]
[7] 0, 5, -200, 1191, -2416, 1985, -600, 35; [280]
.
The coefficients of the polynomials p(n, x) = (Sum_{k = 0..n} T(n, k) x^k) / A343277(n) for the first few n:
[0] 1;
[1] 0, 1/2;
[2] 0, -1/6, 1/3;
[3] 0, 1/12, -1/3, 1/4;
[4] 0, -1/20, 11/30, -11/20, 1/5;
[5] 0, 1/30, -13/30, 11/10, -13/15, 1/6.
Sequences of rational polynomials p(n, x) with p(n, 1) = Bernoulli(n, 1):
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CoeffList := p -> op(PolynomialTools:-CoefficientList(p,x)):
E1 := (n, k) -> combinat:-eulerian1(n, k):
poly := n -> (1/(n+1))*add((-1)^k*E1(n,k)*x^(n-k)/binomial(n,k), k=0..n):
Trow := n -> denom(poly(n))*CoeffList(poly(n)): seq(Trow(n), n = 0..9);
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Poly342321[n_, x_] := If[n == 0, 1, Sum[x^k*k!*Sum[(-1)^(n - j)*StirlingS2[n, j] /((k - j)!(n - j + 1) Binomial[n + 1, j]), {j, 0, k}], {k, 1, n}]];
Table[A343277[n] CoefficientList[Poly342321[n, x], x][[k+1]], {n, 0, 9}, {k, 0, n}] // Flatten
A356602
Triangle read by rows. T(n, k) = numerator(Integral_{z=0..1} Eulerian(n, k)*z^(k + 1)*(z - 1)^(n - k - 1) dz), where Eulerian(n, k) = A173018(n, k) for n >= 1, and T(0, 0) = 1.
Original entry on oeis.org
1, 1, 0, -1, 1, 0, 1, -1, 1, 0, -1, 11, -11, 1, 0, 1, -13, 11, -13, 1, 0, -1, 19, -151, 302, -19, 1, 0, 1, -5, 1191, -302, 397, -15, 1, 0, -1, 247, -477, 15619, -15619, 477, -247, 1, 0, 1, -251, 1826, -44117, 15619, -44117, 1826, -251, 1, 0
Offset: 0
Triangle T(n, k) starts:
[0] 1;
[1] 1, 0;
[2] -1, 1, 0;
[3] 1, -1, 1, 0;
[4] -1, 11, -11, 1, 0;
[5] 1, -13, 11, -13, 1, 0;
[6] -1, 19, -151, 302, -19, 1, 0;
[7] 1, -5, 1191, -302, 397, -15, 1, 0;
[8] -1, 247, -477, 15619, -15619, 477, -247, 1, 0;
[9] 1, -251, 1826, -44117, 15619, -44117, 1826, -251, 1, 0;
The Bernoulli numbers (with B(1) = 1/2) are the row sums of the fractions.
[0] 1 = 1;
[1] + 1/2 = 1/2;
[2] - 1/6 + 1/3 = 1/6;
[3] + 1/12 - 1/3 + 1/4 = 0;
[4] - 1/20 + 11/30 - 11/20 + 1/5 = -1/30;
[5] + 1/30 - 13/30 + 11/10 - 13/15 + 1/6 = 0;
[6] - 1/42 + 19/35 - 151/70 + 302/105 - 19/14 + 1/7 = 1/42;
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E1 := proc(n, k) combinat:-eulerian1(n, k) end:
Trow := proc(n, z) if n = 0 then return 1 fi;
seq(numer(int(E1(n, k)*z^(k + 1)*(z - 1)^(n - k - 1), z=0..1)), k=0..n) end:
for n from 0 to 9 do Trow(n, z) od;
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Unprotect[Power]; Power[0, 0] = 1;
E1[n_, k_] /; n == k = 0^k; E1[n_, k_] /; k < 0 || k > n = 0;
E1[n_, k_] := E1[n, k] = (k + 1)*E1[n - 1, k] + (n - k)*E1[n - 1, k - 1];
T[n_, k_] /; n == k = 0^k;
T[n_, k_] := (-1)^(k - n + 1)*E1[n, k]*Gamma[k + 2]*Gamma[n - k]/Gamma[n + 2];
Table[Numerator[T[n, k]], {n, 0, 8}, {k, 0, n}] // TableForm
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