cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 23 results. Next

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

Views

Author

Leroy Quet, Dec 30 2004

Keywords

Comments

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

Examples

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

Links

Crossrefs

m-th harmonic number H(m) = A001008(m)/A002805(m).
Cf. A055573, A058027, A110020, A112286, A112287.

Programs

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

Extensions

Extended by Ray Chandler, Sep 17 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

Views

Author

M. F. Hasler, May 31 2008

Keywords

Comments

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.

Links

Crossrefs

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

Programs

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

Formula

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

A113123 Numerator of next-best approximation to harmonic numbers. a(n) = Numerator of (A055573(n)-1)th convergent of n-th harmonic number, Sum_{k=1..n} 1/k.

Original entry on oeis.org

0, 1, 2, 2, 16, 22, 70, 106, 1836, 2639, 14281, 21167, 167857, 87932, 169452, 923889, 3590229, 950596, 40366604, 23213361, 517630, 1391957, 160363133, 222528683, 10125035246, 4324958013, 81828906108, 71315450571, 4320297286472
Offset: 1

Views

Author

Leroy Quet, Oct 14 2005

Keywords

Comments

A100398 gives terms of continued fractions of harmonic numbers.
For n >= 2, a(n) = the denominator of the ratio equal to the continued fraction made by reversing the order of the terms of the continued fraction of the n-th harmonic number. (The numerator of this ratio is the numerator of the n-th harmonic number, A001008(n).) - Leroy Quet, Dec 24 2006

Examples

			H(6) = 49/20 = 2 +1/(2 +1/(4 +1/2)), so a(6) = numerator of 2 +1/(2 +1/4) = 22/9.

Crossrefs

Cf. A100398, A055573, A113124.

Extensions

More terms from Joshua Zucker, May 08 2006

A113124 Denominator of next-best approximation to harmonic numbers. a(n) = Denominator of (A055573(n)-1)th convergent of n-th harmonic number, Sum_{k=1..n} 1/k.

Original entry on oeis.org

1, 1, 1, 1, 7, 9, 27, 39, 649, 901, 4729, 6821, 52783, 27043, 51067, 273281, 1043807, 271979, 11378119, 6452207, 141997, 377141, 42943389, 58933037, 2653340203, 1122077597, 21027833867, 18159496967, 1090528730477, 236529224117
Offset: 1

Views

Author

Leroy Quet, Oct 14 2005

Keywords

Comments

A100398 gives terms of continued fractions of harmonic numbers.

Examples

			H(6) = 49/20 = 2 +1/(2 +1/(4 +1/2)), so a(6) = denominator of 2 +1/(2 +1/4) = 22/9.

Crossrefs

Cf. A100398, A055573, A113123.

Extensions

More terms from Joshua Zucker, May 08 2006

A058027 Sum of terms of continued fraction for n-th harmonic number, 1 + 1/2 + 1/3 + ... + 1/n.

Original entry on oeis.org

1, 3, 7, 14, 15, 10, 16, 19, 26, 35, 72, 41, 38, 79, 83, 42, 59, 143, 68, 61, 70, 51, 50, 78, 74, 82, 130, 113, 111, 315, 235, 1190, 211, 407, 112, 122, 142, 246, 693, 133, 138, 162, 1904, 243, 170, 539, 363, 210, 197, 518, 275, 502, 527, 316, 1729, 224, 228, 909
Offset: 1

Views

Author

Leroy Quet, Nov 15 2000

Keywords

Comments

Is anything known about the asymptotics of this sequence?
Should be asymptotic to D*n^(3/2) with D=0.4.... - Benoit Cloitre, Dec 23 2003

Examples

			1 + 1/2 +1/3 = 11/6 = 1 + 1/(1 + 1/5). So sum of terms of continued fraction is 1 + 1 + 5 = 7.

Links

Crossrefs

m-th harmonic number H(m) = A001008(m)/A002805(m).
Cf. A055573, A100398, A110020, A112286, A112287.

Programs

  • Mathematica
    Table[Plus @@ ContinuedFraction[HarmonicNumber[n]], {n, 60}] (* Ray Chandler, Sep 17 2005 *)
  • PARI
    a(n) = vecsum(contfrac(sum(k=1, n, 1/k))); \\ Michel Marcus, Mar 23 2017

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

Views

Author

Leroy Quet, Sep 03 2005

Keywords

Examples

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

Links

Crossrefs

m-th harmonic number H(m) = A001008(m)/A002805(m).
Cf. A055573, A058027, A100398, A112286, A112287.

Programs

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

Extensions

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

Views

Author

Leroy Quet, Sep 01 2005

Keywords

Examples

			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.

Links

Crossrefs

m-th harmonic number H(m) = A001008(m)/A002805(m).
Cf. A055573, A058027, A100398, A110020, A112287.

Programs

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

Extensions

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

Views

Author

Leroy Quet, Sep 01 2005

Keywords

Examples

			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.

Links

Crossrefs

m-th harmonic number H(m) = A001008(m)/A002805(m).
Cf. A055573, A058027, A100398, A110020, A112286.

Programs

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

Extensions

Extended by Hans Havermann and Ray Chandler, Sep 06 2005

A069880 Number of terms in the simple continued fraction for Sum_{k=1..n} 1/k!.

Original entry on oeis.org

1, 2, 3, 5, 6, 9, 9, 13, 14, 18, 19, 20, 24, 24, 23, 29, 33, 36, 31, 38, 41, 42, 46, 50, 53, 58, 56, 57, 70, 73, 77, 69, 76, 76, 78, 77, 80, 85, 89, 101, 101, 105, 106, 104, 106, 112, 115, 124, 113, 126, 124, 124, 130, 144, 144, 148, 140, 149, 141, 151, 157, 158, 172
Offset: 1

Views

Author

Benoit Cloitre, May 04 2002

Keywords

Examples

			For n=4, Sum_{k=1..n} 1/k! = 1/1! + 1/2! + 1/3! + 1/4! = 1/1 + 1/2 + 1/6 + 1/24 = 41/24 = 1 + 1/(1 + 1/(2 + 1/(2 + 1/3))) = CF[1;1,2,2,1], so a(4) = 5.

Links

Crossrefs

Cf. A055573, A061354, A061355, A070267.

Programs

  • Mathematica
    lcf[f_] := Length[ContinuedFraction[f]]; lcf /@ Accumulate[Table[1/k!, {k, 1, 100}]] (* Amiram Eldar, Apr 30 2022 *)

Formula

Does lim_{n->infinity} a(n)/(n * log(log(n))) = C = 2.XXX...?

A069887 Number of terms in the simple continued fraction expansion for (1+1/n)^n.

Original entry on oeis.org

1, 2, 5, 7, 7, 10, 14, 16, 24, 16, 20, 29, 39, 40, 42, 39, 46, 42, 44, 57, 59, 55, 66, 55, 57, 70, 68, 81, 86, 81, 91, 109, 106, 108, 119, 117, 123, 118, 124, 118, 120, 133, 142, 147, 164, 155, 159, 164, 167, 163, 177, 176, 168, 171, 198, 198, 201, 201, 205, 206, 227
Offset: 1

Views

Author

Benoit Cloitre, May 04 2002

Keywords

Comments

Limit_{n -> infinity} (1+1/n)^n = e.
For any natural number N, limit_{n->infinity} (log(N)^(1/n) + 1/n)^n = e*log(N). - Alexander R. Povolotsky, Dec 06 2007

Examples

			The simple continued fraction for (1+1/10)^10 is [2, 1, 1, 2, 5, 1, 128, 1, 2, 12, 5, 3, 46, 1, 11, 7] which contains 16 elements, hence a(10) = 16.

Links

Crossrefs

Cf. A001113, A055573, A069655.

Programs

  • Mathematica
    Table[Length[ContinuedFraction[(1+1/n)^n]],{n,70}] (* Harvey P. Dale, Jun 12 2013 *)

Formula

Asymptotically it seems that a(n) ~ C*n*log(n) where C = 0.84... is close to the constant described in A055573.
Showing 1-10 of 23 results. Next

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