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Search: id:A055573
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| A055573 |
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Number of terms in simple continued fraction for n-th harmonic number H_n = sum_{k=1 to n} [1/k]. |
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+0
19
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| 1, 2, 3, 2, 5, 4, 6, 7, 10, 8, 7, 10, 15, 9, 9, 17, 18, 11, 20, 16, 18, 18, 23, 19, 24, 25, 24, 26, 29, 21, 24, 23, 26, 25, 32, 34, 33, 26, 24, 31, 32, 31, 36, 36, 39, 32, 34, 42, 47, 44, 46, 35, 40, 48, 43, 47, 59, 50, 49, 39, 50, 66, 54, 44, 54, 49, 41, 64, 47, 46, 54, 71, 72
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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By "simple continued fraction" is meant a continued fraction whose terms are positive integers and the final term is >= 2.
Does any number appear infinitely often in this sequence?
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REFERENCES
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S. R. Finch, Mathematical Constants, Cambridge, 2003, pp. 156
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LINKS
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M. F. Hasler, Table of n, a(n) for n=1,...,500.
Eric Weisstein's World of Mathematics, Harmonic Number
Eric Weisstein's World of Mathematics, Continued Fraction
G. Xiao, Contfrac server, To evaluate H(m) and display its continued fraction expansion, operate on "sum(n=1, m, 1/n)"
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FORMULA
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It appears that lim n -> infinity a(n)/n = C = 0.84... - Benoit Cloitre (benoit7848c(AT)orange.fr), May 04 2002
Conjecture : limit n ->infty a(n)/n = 12*ln(2)/Pi^2 = 0.84..... = A089729 Levy's constant. (Benoit Cloitre), Jan 17 2004
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EXAMPLE
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Sum_{k=1 to 3} [1/k] = 11/6 = 1 + 1/(1 + 1/5), so the 3_rd term is 3 because the simple continued fraction for the 3_rd harmonic number has 3 terms.
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MATHEMATICA
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Table[ Length[ ContinuedFraction[ HarmonicNumber[n]]], {n, 1, 75}] (from Robert G. Wilson v Dec 22 2003)
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PROGRAM
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(PARI) c=0; h=0; for(n=1, 500, write("projects/b055573.txt", c++, " ", #contfrac(h+=1/n))) - M. F. Hasler (www.univ-ag.fr/~mhasler), May 31 2008
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CROSSREFS
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m-th harmonic number H(m) = A001008(m)/A002805(m).
Cf. A058027, A100398, A110020, A112286, A112287.
Cf. A139001 (partial sums).
Sequence in context: A089587 A067316 A127433 this_sequence A072969 A139712 A075365
Adjacent sequences: A055570 A055571 A055572 this_sequence A055574 A055575 A055576
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KEYWORD
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nonn
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AUTHOR
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Leroy Quet (qq-quet(AT)mindspring.com), Jul 10 2000
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