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A201454
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Triangle of denominators of dual coefficients of Faulhaber.
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2
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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
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OFFSET
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0,2
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COMMENTS
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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).
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LINKS
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FORMULA
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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).
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EXAMPLE
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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.
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MATHEMATICA
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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;
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PROG
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(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
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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