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A027462
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a(n) is the numerator of (-1/6) * Integral_{x=0..1} x^n * log^3(1-x).
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3
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1, 15, 575, 5845, 874853, 336581, 129973303, 1149858589, 101622655189, 21945415349, 31276937512951, 33264031387717, 77287019174361937, 81347802723340093, 17055178843123409, 142531324182321979
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OFFSET
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1,2
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COMMENTS
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LINKS
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FORMULA
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Numerators of sequence a(1,n) in (a(i,j))^4 where a(i,j) = 1/i if j <= i, 0 if j > i.
Numerators of (H(n,1)^3 + 3*H(n,1)*H(n,2) + 2*H(n,3))/(6*n) = ((gamma+Psi(n+1))^3 + 3*(gamma+Psi(n+1))*(1/6*Pi^2 - Psi(1, n+1)) + 2*Zeta(3) + Psi(2, n+1))/(6*n), where H(n, m) = Sum_{i=1..n} 1/i^m are generalized harmonic numbers. - Vladeta Jovovic, Aug 10 2002
For n>=1, (H(n,1)^3 + 3*H(n,1)*H(n,2) + 2*H(n,3))/(6*n) = -(1/6)*Integral_{x=0..1} x^(n-1)*log^3(1-x) dx = (1/n)*Sum_{j=1..n} (H(n,1) - H(j-1,1))*H(j,1)/j.
For every k>=1 the first column of (a(i,j))^k is the binomial transform of (-1)^n/(n+1)^k.
Proof: the sequence S(n,k) = ((-1)^k/k!)*Integral_{x=0..1} x^(n-1)*log^k(1-x) dx gives the binomial transform of (-1)^n/(n+1)^(k+1) and can be evaluated by parts with S(n,k) = (1/n)*Sum_{j=1..k} S(j,k-1) according to the generation of the first column of (a(i,j))^k.
(End)
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MATHEMATICA
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a[n_] := -1/6 Integrate[x^(n-1) Log[1-x]^3, {x, 0, 1}] // Numerator;
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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