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A096663 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. 1

%I #14 Mar 28 2015 22:36:55

%S 1,-3,-11,1,-137,61,-363,11,149,9881,-83711,-3391,-1145993,1631353,

%T 1821257,3397,-42142223,-1565387,-275295799,-20644219,151619971,

%U 59515289,-444316699,-203021927,374167685,7248582529,950047851,-8741096671,-9227046511387,-22795769741183

%N 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.

%C 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.

%F b(1)=1; for n>=2, b(n) = -Sum_{k|n, k>=2} (H(k) b(n/k)).

%e 1, -3/2, -11/6, 1/6, -137/60, 61/20, -363/140, ...

%p 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_

%Y Cf. A097504.

%K frac,sign

%O 1,2

%A _Leroy Quet_, Aug 25 2004

%E More terms from _Emeric Deutsch_ and _Max Alekseyev_, Apr 13 2005

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Last modified April 25 01:35 EDT 2024. Contains 371964 sequences. (Running on oeis4.)