A058027 -id:A058027 - cp's OEIS frontend

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

Leroy Quet, Jul 10 2000

nonn

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?

			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.

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

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)"

m-th harmonic number H(m) = A001008(m)/A002805(m).
Cf. A058027, A100398, A110020, A112286, A112287.
Cf. A139001 (partial sums).

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

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

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

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

A100398 Array where n-th row (of A055573(n) terms) is the continued fraction terms for the n-th harmonic number, sum{ k=1 to n} 1/k.

Original entry on oeis.org

1, 1, 2, 1, 1, 5, 2, 12, 2, 3, 1, 1, 8, 2, 2, 4, 2, 2, 1, 1, 2, 5, 5, 2, 1, 2, 1, 1, 5, 7, 2, 1, 4, 1, 5, 1, 1, 7, 1, 3, 2, 1, 13, 12, 1, 3, 1, 2, 3, 50, 3, 4, 6, 1, 5, 3, 9, 1, 2, 4, 1, 1, 1, 15, 4, 3, 5, 1, 1, 4, 2, 1, 3, 2, 1, 3, 1, 4, 1, 6, 3, 3, 1, 39, 3, 1, 13, 3, 13, 3, 3, 7, 43, 1, 1, 1, 17, 7, 3, 2

Offset: 1

Leroy Quet, Dec 30 2004

Terms corresponding to H(n) (i.e. the n-th row) end at index A139001(n)=sum(i=1..n,A055573(n)) - M. F. Hasler, May 31 2008

			Since the 3rd harmonic number is 11/6 = 1 +1/(1 +1/5), the 3rd row is 1,1,5.

M. F. Hasler, Table of n, a(n) for n=1..105013.
Eric Weisstein's World of Mathematics, Harmonic Number
Eric Weisstein's World of Mathematics, Continued Fraction

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

Cf. A055573, A058027, A110020, A112286, A112287.

Flatten[Table[ContinuedFraction[HarmonicNumber[n]], {n, 16}]] (* Ray Chandler, Sep 17 2005 *)

c=0;h=0;for(n=1,500,for(i=1,#t=contfrac(h+=1/n),write("b100398.txt",c++," ",t[i]))) \\ M. F. Hasler, May 31 2008

Extended by Ray Chandler, Sep 17 2005

A110020 Final term of the simple continued fraction for H(n), where H(n) = Sum_{k=1..n} 1/k.

Original entry on oeis.org

1, 2, 5, 12, 8, 2, 5, 7, 3, 2, 5, 4, 6, 13, 7, 2, 11, 15, 6, 2, 36, 13, 2, 6, 3, 7, 3, 4, 2, 9, 4, 2, 2, 2, 2, 2, 6, 5, 2, 3, 2, 2, 2, 2, 11, 3, 59, 8, 2, 4, 104, 103, 5, 6, 2, 2, 2, 59, 2, 2, 3, 9, 20, 4, 2, 3, 4, 3, 4, 2, 2, 2, 2, 2, 2, 4, 3, 4, 2, 3, 2, 37, 2, 49, 6, 2, 6, 10, 2, 4, 8, 15, 2, 2, 23, 2

Offset: 1

Leroy Quet, Sep 03 2005

			H(5) = 137/60 = 2 + 1/(3 + 1/(1 + 1/(1 + 1/8))); a(5) is the final term, 8.

G. C. Greubel, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Harmonic Number
Eric Weisstein's World of Mathematics, Continued Fraction

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

Cf. A055573, A058027, A100398, A112286, A112287.

Table[Last[ContinuedFraction[HarmonicNumber[n]]], {n, 100}] (* Ray Chandler, Sep 17 2005 *)

Extended by Ray Chandler, Sep 17 2005

A112286 a(n) = numerator of sum of reciprocals of the terms of the continued fraction for H(n) = Sum_{k=1..n} 1/k.

Original entry on oeis.org

1, 3, 11, 7, 71, 7, 17, 152, 2699, 701, 691, 248, 133, 137, 61933, 809, 20705, 64896, 3587, 17449, 445, 61897, 208, 20663, 1163, 982, 27281, 1871, 2466139, 44339, 21293609, 13417971, 6229, 54238033, 99737, 3585191, 33583, 40756259, 5956441

Offset: 1

Leroy Quet, Sep 01 2005

			1 +1/2 +1/3 +1/4 +1/5 +1/6 = 49/20 = 2 + 1/(2 + 1/(4 + 1/2)).
So a(6) is 7, the numerator of 7/4 = 1/2 + 1/2 + 1/4 + 1/2.

G. C. Greubel, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Harmonic Number
Eric Weisstein's World of Mathematics, Continued Fraction

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

Cf. A055573, A058027, A100398, A110020, A112287.

f[n_] := Plus @@ (1/# &) /@ ContinuedFraction[Sum[1/k, {k, n}]]; Table[Numerator[f[n]], {n, 40}] (* Ray Chandler, Sep 06 2005 *)

Extended by Hans Havermann and Ray Chandler, Sep 06 2005

A112287 a(n) = denominator of sum of reciprocals of the terms of the continued fraction for H(n) = Sum_{k=1..n} 1/k.

Original entry on oeis.org

1, 2, 5, 12, 24, 4, 5, 35, 420, 156, 300, 45, 15, 39, 15351, 72, 1848, 10675, 300, 2142, 36, 5460, 15, 1870, 90, 63, 2040, 120, 138600, 3960, 1750320, 1324895, 440, 3945480, 5220, 158340, 1680, 3341100, 498960, 48048, 1260, 69264, 1510, 1168200, 568260

Offset: 1

Leroy Quet, Sep 01 2005

			1 +1/2 +1/3 +1/4 +1/5 +1/6 = 49/20 = 2 + 1/(2 + 1/(4 + 1/2)).
So a(6) is 4, the denominator of 7/4 = 1/2 + 1/2 + 1/4 + 1/2.

G. C. Greubel, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Harmonic Number
Eric Weisstein's World of Mathematics, Continued Fraction

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

Cf. A055573, A058027, A100398, A110020, A112286.

f[n_] := Plus @@ (1/# &) /@ ContinuedFraction[Sum[1/k, {k, n}]]; Table[Denominator[f[n]], {n, 45}] (* Ray Chandler, Sep 06 2005 *)

Extended by Hans Havermann and Ray Chandler, Sep 06 2005

A139001 Partial sums of A055573 = number of terms in continued fraction of H(n)=sum(1/k,k=1..n).

Original entry on oeis.org

1, 3, 6, 8, 13, 17, 23, 30, 40, 48, 55, 65, 80, 89, 98, 115, 133, 144, 164, 180, 198, 216, 239, 258, 282, 307, 331, 357, 386, 407, 431, 454, 480, 505, 537, 571, 604, 630, 654, 685, 717, 748, 784, 820, 859, 891, 925, 967, 1014, 1058, 1104, 1139, 1179, 1227, 1270

Offset: 1

M. F. Hasler, May 31 2008

nonn

Sequence A100398 holds the array having as n-th row the continued frac. of H(n); a(n) is the last term of the n-th row and accordingly, a(n-1)+1 is the index where the n-th row starts.

M. F. Hasler, Table of n, a(n) for n = 1..500

Cf. A001008, A002805, A055573, A058027, A100398, A110020, A112286, A112287.

h=s=0;vector(100,n,s+=#contfrac(h+=1/n))

a(n) = sum_{k=1..n} A055573(k)

A332342 Table T(n, k) read by antidiagonals upwards: sum of the terms of the continued fraction for the fractional part of n/k (n>=1, k>=1).

Original entry on oeis.org

0, 0, 2, 0, 0, 3, 0, 2, 3, 4, 0, 0, 0, 2, 5, 0, 2, 3, 4, 4, 6, 0, 0, 3, 0, 4, 3, 7, 0, 2, 0, 4, 5, 2, 5, 8, 0, 0, 3, 2, 0, 3, 5, 4, 9, 0, 2, 3, 4, 5, 6, 5, 5, 6, 10, 0, 0, 0, 0, 4, 0, 5, 2, 3, 5, 11, 0, 2, 3, 4, 4, 6, 7, 5, 6, 6, 7, 12, 0, 0, 3, 2, 5, 3, 0, 4, 6, 4, 6, 6, 13

Offset: 1

Andrey Zabolotskiy, Feb 10 2020

			2/7 = 1/(3+1/2), so T(2, 7) = 3 + 2 = 5.
The table begins:
0 2 3 4 5 6 7 8 9 ...
0 0 3 2 4 3 5 4 6 ...
0 2 0 4 4 2 5 5 3 ...
0 0 3 0 5 3 5 2 6 ...
0 2 3 4 0 6 5 5 6 ...
0 0 0 2 5 0 7 4 3 ...
...

Andrey Zabolotskiy, Table of n, a(n) for n = 1..4950 (antidiagonals 1..99)

Cf. A332343, A058027, A058299, A071316.

t[n_,k_] := Total@ ContinuedFraction@ FractionalPart[n/k];
Flatten[Table[t[nk+k-1,k], {nk,10}, {k,nk}]]

def cofr(p, q):
    return [] if q == 0 else [p // q] + cofr(q, p % q)
def t(n, k):
    return sum(cofr(n, k)[1:])
tr = []
for nk in range(1, 20):
    for k in range(1, nk+1):
        tr.append(t(nk+1-k, k))
print(tr)

A111047 Product of continued fraction terms of H(n) = Sum_{k=1..n} 1/k.

Original entry on oeis.org

1, 2, 5, 24, 48, 32, 100, 140, 840, 1872, 54000, 12960, 51840, 533871, 322371, 31104, 709632, 1921500, 4147200, 3701376, 124416, 262080, 2488320, 21811680, 403107840, 146966400, 2538086400, 1074954240, 14370048000, 415704960000

Offset: 1

Leroy Quet, Oct 06 2005

nonn

The last term of each continued fraction is considered to be >= 2, for n >= 2.

			1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 = 49/20 = 2 + 1/(2 + 1/(4 + 1/2)), so the 6th term of the sequence is 2*2*4*2 = 32.

Cf. A100398, A058027.

for(n=1,30,v=contfrac(sum(k=1,n,1/k));print1(prod(j=1,length(v),v[j]),","))

More terms from Klaus Brockhaus, Oct 08 2005

cp's OEIS Frontend

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

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A100398 Array where n-th row (of A055573(n) terms) is the continued fraction terms for the n-th harmonic number, sum{ k=1 to n} 1/k.

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A110020 Final term of the simple continued fraction for H(n), where H(n) = Sum_{k=1..n} 1/k.

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A112286 a(n) = numerator of sum of reciprocals of the terms of the continued fraction for H(n) = Sum_{k=1..n} 1/k.

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A112287 a(n) = denominator of sum of reciprocals of the terms of the continued fraction for H(n) = Sum_{k=1..n} 1/k.

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A139001 Partial sums of A055573 = number of terms in continued fraction of H(n)=sum(1/k,k=1..n).

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Formula

A332342 Table T(n, k) read by antidiagonals upwards: sum of the terms of the continued fraction for the fractional part of n/k (n>=1, k>=1).

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Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

Python

A111047 Product of continued fraction terms of H(n) = Sum_{k=1..n} 1/k.

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