

A055573


Number of terms in simple continued fraction for nth harmonic number H_n = Sum_{k=1..n} (1/k).


19



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;
refs;
listen;
history;
text;
internal format)



OFFSET

1,2


COMMENTS

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?


REFERENCES

S. R. Finch, Mathematical Constants, Cambridge, 2003, pp. 156


LINKS

G. C. Greubel, Table of n, a(n) for n = 1..10000 (terms 1..500 from M. F. Hasler)
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)"


FORMULA

It appears that lim n > infinity a(n)/n = C = 0.84...  Benoit Cloitre, May 04 2002
Conjecture: limit n > infinity a(n)/n = 12*log(2)/Pi^2 = 0.84..... = A089729 Levy's constant.  Benoit Cloitre, Jan 17 2004


EXAMPLE

Sum_{k=1 to 3} [1/k] = 11/6 = 1 + 1/(1 + 1/5), so the 3rd term is 3 because the simple continued fraction for the 3rd harmonic number has 3 terms.


MATHEMATICA

Table[ Length[ ContinuedFraction[ HarmonicNumber[n]]], {n, 1, 75}] (* Robert G. Wilson v, Dec 22 2003 *)


PROG

(PARI) c=0; h=0; for(n=1, 500, write("projects/b055573.txt", c++, " ", #contfrac(h+=1/n))) \\ M. F. Hasler, May 31 2008


CROSSREFS

mth harmonic number H(m) = A001008(m)/A002805(m).
Cf. A058027, A100398, A110020, A112286, A112287.
Cf. A139001 (partial sums).
Sequence in context: A216475 A267807 A127433 * A238729 A182816 A195637
Adjacent sequences: A055570 A055571 A055572 * A055574 A055575 A055576


KEYWORD

nonn


AUTHOR

Leroy Quet, Jul 10 2000


STATUS

approved



