A201454 Triangle of denominators of dual coefficients of Faulhaber.

1, 3, 3, 5, 3, 15, 7, 5, 3, 105, 9, 21, 15, 9, 105, 11, 9, 21, 3, 9, 231, 13, 11, 3, 7, 5, 3, 15015, 15, 39, 165, 9, 15, 5, 45, 2145, 17, 5, 13, 55, 9, 15, 15, 45, 36465, 19, 17, 5, 13, 55, 3, 35, 1, 5, 969969, 21, 57, 17, 21, 13, 33, 63, 7, 5, 63, 4849845

Offset: 0

Author

Damir Yeliussizov, Jan 09 2013

Keywords

• nonn
• frac
• tabl
• easy

Comments

Sum((k*(k + 1))^(m), k=0..N-1)=Sum(F(m,i)*N^(2*m-2*i+1),i=0..m), m=0,1,2,...

The coefficients F(m,i) are dual to Faulhaber coefficients, because they are obtained from the inverse expression Sum((k*(k + 1))^(m), k=0..N-1) to Faulhaber's formula from Sum((k)^(2*m-1), k=0..N-1) and there holds the identity F(m+i-1,i)=(-1)^i Fe(-m,i), where Fe(-m,i)=A093558(-m,i)/A093559(-m,i) is a Faulhaber coefficient for the sums of even powers of the first N-1 integers (for details see the link, from p. 19).

Examples

			Triangle begins:
1;
3,  3;
5,  3,  15;
7,  5,  3,   10;
9,  21, 15,  9,  105;
11, 9,  21,  3,  9,   231;
13, 11, 3,   7,  5,   3,   15015;
15, 39, 165, 9,  15,  5,   45,    2145;
17, 5,  13,  55, 9,   15,  15,    45,   36465;
19, 17, 5,   13, 55,  3,   35,    1,    5,    969969;
21, 57, 17,  21, 13,  33,  63,    7,    5,    63,    4849845;
etc.
		

Links

• Askar Dzhumadil'daev, Damir Yeliussizov, Power sums of binomial coefficients, Journal of Integer Sequences, 16 (2013), Article 13.1.6

Crossrefs

Cf. A201453.

Programs

Magma

[Denominator((1/(2*m-2*k+1))*&+[Binomial(m,2*k-i)*Binomial(2*m-2*k+i, i)*BernoulliNumber(i): i in [0..2*k]]): k in [0..m], m in [0..10]]; // Bruno Berselli, Jan 21 2013

Mathematica

f[m_, k_] := (1/(2*m - 2*k + 1))* Sum[Binomial[m, 2*k - i]*Binomial[2*m - 2*k + i, i]*BernoulliB[i], {i, 0, 2 k}];
a[m_, k_] := f[m, k] // Denominator;
Table[a[m, k], {m, 0, 10}, {k, 0, m}] // Flatten (* Jean-François Alcover, Jan 18 2013 *)

Formula

a(m,k) = denominator(F(m,k)) with F(m,k) = (1/(2*m-2*k+1)) * sum(i=0..2*k, binomial(m,2*k-i) * binomial(2*m-2*k+i,i) * Bernoulli(i) ).

A recursion is given by F(x,0) = 1/(2*x+1) and 2*(m-k)*(2*m-2*k+1)*F(m,k)=2*m*(2*m-1)*F(m-1,k)+m*(m-1)*F(m-2,k-1).