A055573 -id:A055573 - cp's OEIS frontend

A373893 a(n) is the length of the simple continued fraction for the n-th alternating harmonic number.

0, 1, 2, 4, 7, 7, 5, 9, 8, 12, 9, 12, 11, 12, 13, 12, 18, 12, 17, 15, 15, 15, 19, 21, 18, 13, 21, 23, 25, 23, 26, 32, 28, 25, 24, 24, 31, 32, 33, 36, 41, 38, 38, 37, 44, 41, 37, 39, 47, 48, 42, 43, 43, 44, 46, 42, 44, 51, 45, 49, 52, 53, 62, 50, 57, 48, 55, 60, 52, 58, 70, 58, 60, 73, 67

Offset: 1

Ilya Gutkovskiy, Jun 21 2024

nonn

By "simple continued fraction" is meant a continued fraction whose terms are positive integers and the final term is >= 2.

			Sum_{k=1..7} (-1)^(k+1)/k = 319/420 = 1/(1 + 1/(3 + 1/(6 + 1/(3 + 1/5)))), so a(7) = 5.

Cf. A055573, A058312, A058313.

Table[Length[ContinuedFraction[Sum[(-1)^(k + 1)/k, {k, 1, n}]]] - 1, {n, 1, 75}]

from fractions import Fraction
from sympy.ntheory.continued_fraction import continued_fraction
def A373893(n): return len(continued_fraction(sum(Fraction(1 if k&1 else -1,k) for k in range(1,n+1))))-1 # Chai Wah Wu, Jun 27 2024

A091590 Number of terms in the simple continued fraction for the 10^n-th harmonic number, H_n = sum_{k=1 to n} (1/k).

Original entry on oeis.org

1, 8, 68, 834, 8356, 84548, 841817, 8425934, 84277586

Offset: 0

Robert G. Wilson v, Jan 22 2004

Conjecture: lim n -> infinity, a(n)/10^n -> C = 12*log(2)/Pi^2 = 0.842... - Benoit Cloitre, May 04 2002

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

J. Sondow and E. W. Weisstein, MathWorld: Harmonic Number

Cf. A055573. n-th harmonic number H(m) = A001008(n)/A002805(n).

s = 0; k = 1; Do[ While[s = s + 1/k; k < 10^n, k++ ]; Print[ Length[ ContinuedFraction[s]]]; k++, {n, 0, 6}]
Table[Length[ContinuedFraction[HarmonicNumber[10^n]]], {n, 0, 7}] (* Harvey P. Dale, Aug 24 2015 *)

a(n) = A055573(10^n). - Andrew Howroyd, Aug 10 2024

Corrected and extended by Eric W. Weisstein, Jan 23 2004

A336089 k such that L(H(k,1)^2) = 2*L(H(k,1)) where L(x) is the number of terms in the continued fraction of x and H(k,r) = Sum_{u=1..k} 1/u^r.

Original entry on oeis.org

7, 10, 14, 275, 293, 359, 509, 518, 526, 531, 643, 671, 701, 710, 1081, 1158, 1318, 1798, 1836, 2368, 2441, 2507, 2591, 2990, 3477, 3589, 3818, 4096, 5582, 5851, 6968, 7180, 7523, 8718, 8745, 8782, 8817, 8844, 8946, 9772, 9905

Offset: 1

Benoit Cloitre, Oct 04 2020

nonn

Conjecture: this sequence is infinite. More generally, for any fixed integers a,b,c,d >=1, there are infinitely many k's such that c*d*L(H(k,a)^b) = a*b*L(H(k,c)^d) where L(x) is the number of terms in the continued fraction of x and H(k,r) = Sum_{u=1..k} 1/u^r. Here (a,b,c,d) =(1,2,1,1).

Cf. A055573, A336088.

c[n_, r_] := Length @ ContinuedFraction @ (HarmonicNumber[n]^r); Select[Range[10^4], c[#, 2] == 2 * c[#, 1] &] (* Amiram Eldar, Oct 04 2020*)

H1=1; for(n=2, 10000, H1=H1+1/n; if(length(contfrac(H1^2))==2*length(contfrac(H1)), print1(n, ", ")))

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A373893 a(n) is the length of the simple continued fraction for the n-th alternating harmonic number.

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A091590 Number of terms in the simple continued fraction for the 10^n-th harmonic number, H_n = sum_{k=1 to n} (1/k).

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A336089 k such that L(H(k,1)^2) = 2*L(H(k,1)) where L(x) is the number of terms in the continued fraction of x and H(k,r) = Sum_{u=1..k} 1/u^r.

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