A356545 -id:A356545 - cp's OEIS frontend

A356547 Triangle read by rows. T(n, k) are the coefficients of polynomials p_n(x) based on the Eulerian numbers of second order representing the Bernoulli numbers as B_n = p_n(1) / (2(2n - 1)!).

1, 1, 0, 6, -4, 0, 120, -192, 72, 0, 5040, -15840, 13920, -3456, 0, 362880, -2096640, 3306240, -1918080, 345600, 0, 39916800, -413683200, 1053803520, -1064448000, 448519680, -62208000, 0, 6227020800, -114960384000, 447866496000, -699342336000, 506348236800, -164428185600, 18289152000, 0

Offset: 0

Peter Luschny, Aug 12 2022

The Bernoulli numbers with B(1) = 1/2 can be represented as the weighted sum of Eulerian numbers of second order, where we use the definition as given by Graham et al., Eulerian2(n, k) = A201637(n, k). For n >= 1 we have
   B_(n) = (1/2)*Sum_{k=0..n} (-1)^k*Eulerian2(n, k) / binomial(2*n - 1, k).
Although this representation looks classical it was apparently first proved by Majer in 2010; later Fu and recently O'Sullivan gave an alternative proof (see links).
An analogous representation based on the Eulerian numbers of first order is given in A356545.

			The triangle T(n, k) of the coefficients, sorted in ascending order, starts:
[0]        1;
[1]        1,          0;
[2]        6,         -4,          0;
[3]      120,       -192,         72,           0;
[4]     5040,     -15840,      13920,       -3456,         0;
[5]   362880,   -2096640,    3306240,    -1918080,    345600,         0;
[6] 39916800, -413683200, 1053803520, -1064448000, 448519680, -62208000, 0;

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, 2nd ed. Addison-Wesley, Reading, MA, 1994, p. 270. (Since the thirty-fourth printing, Jan. 2022, with B(1) = 1/2.)

Amy M. Fu, Some Identities Related to the Second-Order Eulerian Numbers, arXiv:2104.09316 [math.CO], Apr. 2021.
Peter Luschny, How are the Eulerian numbers of the first-order related to the Eulerian numbers of the second-order?, MathOverflow, Feb. 2021.
Pietro Majer, Expressions involving Eulerian numbers of the second kind, MathOverflow, Nov 2010.
G. Rzadkowski, M. Urlinska, A Generalization of the Eulerian Numbers, arXiv:1612.06635 [math.CO], 2016
Cormac O'Sullivan, Stirling's approximation and a hidden link between two of Ramanujan's approximations, arXiv:2208.02898 [math.NT], Aug. 2022.

Cf. A201637 (Eulerian number 2nd order), A164555(n)/A027642(n) (Bernoulli numbers with B(1) = 1/2).

Cf. A356545.

E2 := proc(n, k) combinat:-eulerian2(n, k) end:
p := (n, x) -> `if`(n = 0, 1, add(E2(n, k)*k!*(2*n - k - 1)!*(-x)^k, k = 0..n)):
seq(print([n], seq(coeff(p(n, x), x, k), k = 0..n)), n = 0..7);
seq(`if`(n = 0, 1, p(n, 1)/(2*(2*n-1)!)), n = 0..14); # check Bernoulli numbers

Let p_n(x) = Sum_{k=0..n} Eulerian2(n, k)*k!*(2*n - k - 1)! * (-x)^k.
T(n, k) = [x^k] p_n(x).
T(n, k) = (-1)^k*Eulerian2(n, k)*k!*(2*n - k - 1)!.

A356601 Triangle read by rows. T(n, k) = denominator(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, 2, 1, 6, 3, 1, 12, 3, 4, 1, 20, 30, 20, 5, 1, 30, 30, 10, 15, 6, 1, 42, 35, 70, 105, 14, 7, 1, 56, 7, 280, 35, 56, 7, 8, 1, 72, 252, 56, 630, 504, 28, 72, 9, 1, 90, 180, 105, 630, 126, 420, 45, 45, 10, 1, 110, 495, 33, 1155, 1386, 1155, 165, 99, 110, 11, 1

Offset: 0

Peter Luschny, Aug 15 2022

			Triangle T(n, k) starts:
[0]  1;
[1]  2,   1;
[2]  6,   3,   1;
[3] 12,   3,   4,   1;
[4] 20,  30,  20,   5,   1;
[5] 30,  30,  10,  15,   6,   1;
[6] 42,  35,  70, 105,  14,   7,  1;
[7] 56,   7, 280,  35,  56,   7,  8,  1;
[8] 72, 252,  56, 630, 504,  28, 72,  9,  1;
[9] 90, 180, 105, 630, 126, 420, 45, 45, 10,  1;
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;

Grzegorz Rządkowski, Bernoulli numbers and solitons - revisited, Journal of Nonlinear Mathematical Physics, (2010) 17:1, 121-126.

Cf. A356602 (numerator), A173018, A278075, A356545, A356547.

E1 := proc(n, k) combinat:-eulerian1(n, k) end:
Trow := proc(n, z) if n = 0 then return 1 fi;
seq(denom(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;

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[Denominator[T[n, k]], {n, 0, 8}, {k, 0, n}] // TableForm

R(n, k) = (-1)^(k - n + 1)*Eulerian(n, k)*Gamma(k + 2)*Gamma(n - k)/Gamma(n + 2) for 0 <= k < n, and T(n, n) = 0^n.
Bernoulli(n) = Sum_{k=0..n} R(n, k), where Bernoulli(1) = 1/2.
T(n, k) = denominator(R(n, k)).

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

Peter Luschny, Aug 15 2022

			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;

Cf. A356601 (denominator), A173018, A278075, A356545, A356547.

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;

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

R(n, k) = (-1)^(k - n + 1)*Eulerian(n, k)*Gamma(k + 2)*Gamma(n - k)/Gamma(n + 2) for 0 <= k < n, and T(n, n) = 0^n.
Bernoulli(n) = Sum_{k=0..n} R(n, k), where Bernoulli(1) = 1/2.
T(n, k) = numerator(R(n, k)).

cp's OEIS Frontend

A356547 Triangle read by rows. T(n, k) are the coefficients of polynomials p_n(x) based on the Eulerian numbers of second order representing the Bernoulli numbers as B_n = p_n(1) / (2(2n - 1)!).

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A356601 Triangle read by rows. T(n, k) = denominator(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.

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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.

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cp's OEIS Frontend

A356547 Triangle read by rows. T(n, k) are the coefficients of polynomials p_n(x) based on the Eulerian numbers of second order representing the Bernoulli numbers as B_n = p_n(1) / (2*(2*n - 1)!).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Maple

Formula

A356601 Triangle read by rows. T(n, k) = denominator(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.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

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.

Views

Author

Keywords

Examples

Crossrefs

Programs

Maple

Mathematica

Formula

A356547 Triangle read by rows. T(n, k) are the coefficients of polynomials p_n(x) based on the Eulerian numbers of second order representing the Bernoulli numbers as B_n = p_n(1) / (2(2n - 1)!).

A356601 Triangle read by rows. T(n, k) = denominator(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.

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.