A055573 Number of terms in simple continued fraction for n-th harmonic number H_n = Sum_{k=1..n} (1/k).

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

Author

Leroy Quet, Jul 10 2000

Keywords

• nonn

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?

Examples

			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.
		

References

• Steven 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)"

Crossrefs

m-th harmonic number H(m) = A001008(m)/A002805(m).

Cf. A058027, A100398, A110020, A112286, A112287.

Cf. A139001 (partial sums).

Programs

Mathematica

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

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

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