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A096663
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Numerator of b(n), where Sum_{k>=1} b(k)/k^r = 1/(Sum_{k>=1} H(k)/k^r). H(k) = Sum_{j=1..k} 1/j, the k-th harmonic number.
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1
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1, -3, -11, 1, -137, 61, -363, 11, 149, 9881, -83711, -3391, -1145993, 1631353, 1821257, 3397, -42142223, -1565387, -275295799, -20644219, 151619971, 59515289, -444316699, -203021927, 374167685, 7248582529, 950047851, -8741096671, -9227046511387, -22795769741183
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OFFSET
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1,2
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COMMENTS
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For r = integer >= 2, Sum_{k>=1} b(k)/k^r also equals 1/(zeta(r+1)(r/2 + 1) - (1/2)Sum_{j=2..r-1} zeta(j)zeta(r+1-j)), where zeta(n) is Sum_{k>=1} 1/k^n.
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LINKS
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FORMULA
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b(1)=1; for n>=2, b(n) = -Sum_{k|n, k>=2} (H(k) b(n/k)).
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EXAMPLE
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1, -3/2, -11/6, 1/6, -137/60, 61/20, -363/140, ...
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MAPLE
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with(numtheory): H:=n->sum(1/j, j=1..n): b[1]:=1: for n from 2 to 32 do div:=sort(convert(divisors(n), list)):b[n]:=-sum(H(div[i])*b[n/div[i]], i=2..nops(div)) od: seq(numer(b[n]), n=1..32); # Emeric Deutsch
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CROSSREFS
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KEYWORD
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frac,sign
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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