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A055573
Number of terms in simple continued fraction for n-th harmonic number H_n = Sum_{k=1..n} (1/k).
24
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
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
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
(Python)
from sympy import harmonic
from sympy.ntheory.continued_fraction import continued_fraction
def A055573(n): return len(continued_fraction(harmonic(n))) # Chai Wah Wu, Jun 27 2024
CROSSREFS
m-th harmonic number H(m) = A001008(m)/A002805(m).
Cf. A139001 (partial sums).
Sequence in context: A216475 A267807 A127433 * A238729 A182816 A367580
KEYWORD
nonn
AUTHOR
Leroy Quet, Jul 10 2000
STATUS
approved