

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)!).


3



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
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OFFSET

0,4


COMMENTS

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.


REFERENCES

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


LINKS



FORMULA

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)!.


EXAMPLE

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;


MAPLE

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*n1)!)), n = 0..14); # check Bernoulli numbers


CROSSREFS



KEYWORD



AUTHOR



STATUS

approved



