A354042 Triangle read by rows. The Faulhaber numbers. F(0, k) = 1 and otherwise F(n, k) = (n + 1)!*(-1)^(k+1)*Sum_{j=0..floor((k-1)/2)} C(2*k-2*j, k+1)*C(2*n+1, 2*j+1) * Bernoulli(2*n-2*j) / (k - j).

1, 0, 1, 0, -1, 2, 0, 4, -8, 6, 0, -36, 72, -60, 24, 0, 600, -1200, 1020, -480, 120, 0, -16584, 33168, -28320, 13776, -4200, 720, 0, 705600, -1411200, 1206240, -591360, 188160, -40320, 5040, 0, -43751232, 87502464, -74813760, 36747648, -11813760, 2661120, -423360, 40320

Offset: 0

Author

Peter Luschny, May 17 2022

Keywords

• sign
• tabl

Comments

I. Gessel and X. Viennot call the rational numbers F(n, k)/(n + 1)! 'Faulhaber numbers'. However, for our purposes it is more convenient to define the integers F(n, k). For the Faulhaber polynomials see A335951/A335952.

Let S(r, m) = Sum_{k=0..m} k^r, with 0^0 = 1 and S(0, m) = m + 1. Faulhaber's theorem (the sums of powers formula) is:

S(2*n+1, m) = (1/(n+1)!)*(1/2)*Sum_{k=0..n} F(n, k)*(m*(m + 1))^(k + 1).

Gessel and Viennot give two combinatorial interpretations for the Faulhaber numbers, for this see A354043.

Examples

			Triangle starts:
0: 1
1: 0,         1
2: 0,        -1,        2
3: 0,         4,       -8,         6
4: 0,       -36,       72,       -60,       24
5: 0,       600,    -1200,      1020,     -480,       120
6: 0,    -16584,    33168,    -28320,    13776,     -4200,     720
7: 0,    705600, -1411200,   1206240,  -591360,    188160,  -40320,    5040
8: 0, -43751232, 87502464, -74813760, 36747648, -11813760, 2661120, -423360, 40320
.
Let n = 4 and m = 3, then S(2*n + 1, m) = S(9, 3) = 20196. Faulhaber's formula gives this as (0*12 + (-36)*144 + 72*1728 + (-60)*20736 + 24*248832) / (2*120).
		

Links

• I. M. Gessel and X. G. Viennot, Determinants, Paths, and Plane Partitions, 1989 preprint.

Crossrefs

Cf. A335951/A335952, A000367/A002445, A354043, A263445.

Programs

Maple

F := (n, k) -> ifelse(n = 0, 1, (n + 1)!*(-1)^(k + 1)*add(binomial(2*k - 2*j, k + 1)*binomial(2*n + 1, 2*j + 1)*bernoulli(2*n - 2*j) / (k - j), j = 0..(k - 1)/2)): for n from 0 to 8 do seq(F(n, k), k = 0..n) od;

Formula

F(n,1) = (2*n +1)*Bernoulli(2*n)*(n+1)! for n >= 1.

F(n,2) = -(4*n+2)*Bernoulli(2*n)*(n+1)! for n >= 2.

F(n,3) = ((10*n+5)*Bernoulli(2*n) + binomial(2*n+1,3)*Bernoulli(2*n-2)/2)*(n+1)! for n >= 3.