editing
approved
editing
approved
Cf. A107918
a(10)=26 because the 10th prime is 29, the harmonic mean of 10 and 29 is 580/39 and the continued fraction for 580/39 has terms {14,1,6,1,4} and sum of terms 26.
approved
editing
_Zak Seidov (zakseidov(AT)yahoo.com), _, May 28 2005
a(10)=26 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 sum of terms 26.
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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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A107919[n_]:=PLus@@ContinuedFraction[HarmonicMean[{n, Prime[n]}]]
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Sum of terms of continued fraction for the harmonic mean of n and n-th prime.
4, 6, 7, 16, 14, 16, 21, 21, 28, 26, 25, 32, 29, 36, 33, 34, 37, 47, 41, 43, 41, 48, 89, 52, 58, 53, 53, 60, 57, 59, 68, 63, 66, 75, 69, 75, 74, 78, 75, 110, 78, 83, 88, 102, 85, 92, 100, 349, 111, 104, 97, 101, 103, 109, 104, 119, 115, 119, 111, 119, 112, 126, 127, 124
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Cf. A107918
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)=26 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 sum of terms 26.
A107919[n_]:=PLus@@ContinuedFraction[HarmonicMean[{n, Prime[n]}]]
Cf. A107918.
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Zak Seidov (zakseidov(AT)yahoo.com), May 28 2005
approved