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a(n) = numerator of sum of reciprocals of the terms of the continued fraction for H(n) = Sum_{k=1..n} 1/k.
7

%I #12 Mar 23 2017 04:36:30

%S 1,3,11,7,71,7,17,152,2699,701,691,248,133,137,61933,809,20705,64896,

%T 3587,17449,445,61897,208,20663,1163,982,27281,1871,2466139,44339,

%U 21293609,13417971,6229,54238033,99737,3585191,33583,40756259,5956441

%N a(n) = numerator of sum of reciprocals of the terms of the continued fraction for H(n) = Sum_{k=1..n} 1/k.

%H G. C. Greubel, <a href="/A112286/b112286.txt">Table of n, a(n) for n = 1..1000</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/HarmonicNumber.html">Harmonic Number</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/ContinuedFraction.html">Continued Fraction</a>

%e 1 +1/2 +1/3 +1/4 +1/5 +1/6 = 49/20 = 2 + 1/(2 + 1/(4 + 1/2)).

%e So a(6) is 7, the numerator of 7/4 = 1/2 + 1/2 + 1/4 + 1/2.

%t f[n_] := Plus @@ (1/# &) /@ ContinuedFraction[Sum[1/k, {k, n}]]; Table[Numerator[f[n]], {n, 40}] (* _Ray Chandler_, Sep 06 2005 * )

%Y m-th harmonic number H(m) = A001008(m)/A002805(m).

%Y Cf. A055573, A058027, A100398, A110020, A112287.

%K nonn,frac

%O 1,2

%A _Leroy Quet_, Sep 01 2005

%E Extended by _Hans Havermann_ and _Ray Chandler_, Sep 06 2005