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_Zak Seidov (zakseidov(AT)yahoo.com), _, May 28 2005
a(10)=5 because 10th prime is 29, harmonic mean of 10 and 29 is 580/39, and continued fraction for 580/39 has terms {14,1,6,1,4} and length 5.
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Eric W. Weisstein, 's World of Mathematics, <a href="http://mathworld.wolfram.com/ContinuedFraction.html">Continued Fraction</a>.
Eric W. Weisstein, 's World of Mathematics, <a href="http://mathworld.wolfram.com/HarmonicMean.html">Harmonic Mean</a>.
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A107918[n_]:=Length[ContinuedFraction[HarmonicMean[{n, Prime[n]}]]]
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Length of continued fraction for the harmonic mean of n and n-th prime.
2, 3, 3, 2, 3, 4, 3, 4, 3, 5, 3, 3, 5, 3, 5, 8, 5, 5, 6, 6, 8, 5, 2, 5, 4, 8, 7, 6, 5, 7, 4, 8, 4, 7, 6, 4, 6, 8, 5, 6, 8, 7, 4, 7, 7, 8, 4, 3, 6, 5, 6, 7, 5, 5, 8, 7, 5, 6, 7, 7, 8, 5, 6, 6, 9, 6, 7, 4, 3, 7, 5, 5, 5, 3, 7, 7, 6, 9, 3, 8, 5, 9, 4, 9, 7, 5, 5, 10, 5, 9, 6, 9, 6, 8, 9, 7, 5, 8, 6, 6, 7, 9, 5, 4
1,1
Eric W. Weisstein, <a href="http://mathworld.wolfram.com/ContinuedFraction.html">Continued Fraction</a>.
Eric W. Weisstein, <a href="http://mathworld.wolfram.com/HarmonicMean.html">Harmonic Mean</a>.
a(10)=5 because 10th prime is 29, harmonic mean of 10 and 29 is 580/39, and continued fraction for 580/39 has terms {14,1,6,1,4} and length 5.
A107918[n_]:=Length[ContinuedFraction[HarmonicMean[{n, Prime[n]}]]]
Cf. A107919.
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Zak Seidov (zakseidov(AT)yahoo.com), May 28 2005
approved