A175441 -id:A175441 - cp's OEIS frontend

A001008 a(n) = numerator of harmonic number H(n) = Sum_{i=1..n} 1/i.

1, 3, 11, 25, 137, 49, 363, 761, 7129, 7381, 83711, 86021, 1145993, 1171733, 1195757, 2436559, 42142223, 14274301, 275295799, 55835135, 18858053, 19093197, 444316699, 1347822955, 34052522467, 34395742267, 312536252003, 315404588903, 9227046511387

Offset: 1

N. J. A. Sloane

H(n)/2 is the maximal distance that a stack of n cards can project beyond the edge of a table without toppling.
By Wolstenholme's theorem, p^2 divides a(p-1) for all primes p > 3.
From Alexander Adamchuk, Dec 11 2006: (Start)
p divides a(p^2-1) for all primes p > 3.
p divides a((p-1)/2) for primes p in A001220.
p divides a((p+1)/2) or a((p-3)/2) for primes p in A125854.
a(n) is prime for n in A056903. Corresponding primes are given by A067657. (End)
a(n+1) is the numerator of the polynomial A[1, n](1) where the polynomial A[genus 1, level n](m) is defined to be Sum_{d = 1..n - 1} m^(n - d)/d. (See the Mathematica procedure generating A[1, n](m) below.) - Artur Jasinski, Oct 16 2008
Better solutions to the card stacking problem have been found by M. Paterson and U. Zwick (see link). - Hugo Pfoertner, Jan 01 2012
a(n) = A213999(n, n-1). - Reinhard Zumkeller, Jul 03 2012
a(n) coincides with A175441(n) if and only if n is not from the sequence A256102. The quotient a(n) / A175441(n) for n in A256102 is given as corresponding entry of A256103. - Wolfdieter Lang, Apr 23 2015
For a very short proof that the Harmonic series diverges, see the Goldmakher link. - N. J. A. Sloane, Nov 09 2015
All terms are odd while corresponding denominators (A002805) are all even for n > 1 (proof in Pólya and Szegő). - Bernard Schott, Dec 24 2021

			H(n) = [ 1, 3/2, 11/6, 25/12, 137/60, 49/20, 363/140, 761/280, 7129/2520, ... ].
Coincidences with A175441: the first 19 entries coincide because 20 is the first entry of A256102. Indeed, a(20)/A175441(20) = 55835135 / 11167027 = 5 = A256103(1). - _Wolfdieter Lang_, Apr 23 2015

John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 258-261.
H. W. Gould, Combinatorial Identities, Morgantown Printing and Binding Co., 1972, # 1.45, page 6, #3.122, page 36.
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 259.
G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, page 347.
D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, Vol. 1, p. 615.
G. Pólya and G. Szegő, Problems and Theorems in Analysis, volume II, Springer, reprint of the 1976 edition, 1998, problem 251, p. 154.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Kenny Lau, Table of n, a(n) for n = 1..2295 (first 200 terms provided by T. D. Noe)
David H. Bailey, Jonathan M. Borwein, and Roland Girgensohn, Experimental evaluation of Euler sums, Exper. Math. 3(1) (1994), 17-30; they evaluate the constants Sum_{k>=1} H_k^m/(k+1)^n.
Hongwei Chen, Evaluations of Some Variant Euler Sums, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.3.
Hongwei Chen, Interesting Series Associated with Central Binomial Coefficients, Catalan Numbers and Harmonic Numbers, J. Int. Seq. 19 (2016), #16.1.5.
R. M. Dickau, Harmonic numbers and the book-stacking problem.
Leo Goldmakher, A short(er) proof of the divergence of the harmonic series.
Antal Iványi, Leader election in synchronous networks, Acta Univ. Sapientiae, Mathematica, 5, 2 (2013), 54-82.
Fredrik Johansson, How (not) to compute harmonic numbers. Feb 21 2009.
Masanobu Kaneko, The Akiyama-Tanigawa algorithm for Bernoulli numbers, J. Integer Sequences, 3 (2000), #00.2.9.
Romeo Mestrovic, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv preprint arXiv:1111.3057 [math.NT], 2011.
Jerry Metzger and Thomas Richards, A Prisoner Problem Variation, Journal of Integer Sequences, Vol. 18 (2015), Article 15.2.7.
Hisanori Mishima, Factorizations of Wolstenholme numbers, n=1..100, n=101..200, n=201..300.
Mike Paterson et al., Maximum Overhang.
Maxie D. Schmidt, Generalized j-Factorial Functions, Polynomials, and Applications, J. Int. Seq. 13 (2010), 10.6.7, Section 4.3.
Peter Shiu, The denominators of harmonic numbers, arXiv:1607.02863 [math.NT], 2016.
N. J. A. Sloane, Illustration of initial terms.
Jonathan Sondow and Eric W. Weisstein, MathWorld: Harmonic Number.
Roberto Tauraso, Some Congruences for Central Binomial Sums Involving Fibonacci and Lucas Numbers, JIS 19 (2016) # 16.5.4.
Eric Weisstein's World of Mathematics, Book Stacking Problem, Wolstenholme's Theorem, Harmonic Mean, Digamma Function.
Wikipedia, Harmonic number.

Cf. A002805 (denominators), A007406, A007408, A007410, A075135, A001220, A125854, A121999, A014566, A056903, A067657, A177427, A177690.
Cf. A145609-A145640. - Artur Jasinski, Oct 16 2008
Cf. A003506. - Paul Curtz, Nov 30 2013
The following fractions are all related to each other: Sum 1/n: A001008/A002805, Sum 1/prime(n): A024451/A002110 and A106830/A034386, Sum 1/nonprime(n): A282511/A282512, Sum 1/composite(n): A250133/A296358.
Cf. A195505.

List([1..30],n->NumeratorRat(Sum([1..n],i->1/i))); # Muniru A Asiru, Dec 20 2018

import Data.Ratio ((%), numerator)
a001008 = numerator . sum . map (1 %) . enumFromTo 1
a001008_list = map numerator $ scanl1 (+) $ map (1 %) [1..]
-- Reinhard Zumkeller, Jul 03 2012

[Numerator(HarmonicNumber(n)): n in [1..30]]; // Bruno Berselli, Feb 17 2016

A001008 := proc(n)
    add(1/k,k=1..n) ;
    numer(%) ;
end proc:
seq( A001008(n),n=1..40) ; # Zerinvary Lajos, Mar 28 2007; R. J. Mathar, Dec 02 2016

Table[Numerator[HarmonicNumber[n]], {n, 30}]
(* Procedure generating A[1,n](m) (see Comments section) *) m =1; aa = {}; Do[k = 0; Do[k = k + m^(r - d)/d, {d, 1, r - 1}]; AppendTo[aa, k], {r, 1, 20}]; aa (* Artur Jasinski, Oct 16 2008 *)
Numerator[Accumulate[1/Range[25]]] (* Alonso del Arte, Nov 21 2018 *)
Numerator[Table[((n - 1)/2)*HypergeometricPFQ[{1, 1, 2 - n}, {2, 3}, 1] + 1, {n, 1, 29}]] (* Artur Jasinski, Jan 08 2021 *)

A001008(n) = numerator(sum(i=1,n,1/i)) \\ Michael B. Porter, Dec 08 2009

H1008=List(1); A001008(n)={for(k=#H1008,n-1,listput(H1008,H1008[k]+1/(k+1))); numerator(H1008[n])} \\ about 100x faster for n=1..1500. - M. F. Hasler, Jul 03 2019

from sympy import Integer
[sum(1/Integer(i) for i in range(1, n + 1)).numerator() for n in range(1, 31)]  # Indranil Ghosh, Mar 23 2017

def harmonic(a, b):  # See the F. Johansson link.
    if b - a == 1:
        return 1, a
    m = (a+b)//2
    p, q = harmonic(a,m)
    r, s = harmonic(m,b)
    return p*s+q*r, q*s
def A001008(n): H = harmonic(1,n+1); return numerator(H[0]/H[1])
[A001008(n) for n in (1..29)] # Peter Luschny, Sep 01 2012

H(n) ~ log n + gamma + O(1/n). [See Hardy and Wright, Th. 422.]
log n + gamma - 1/n < H(n) < log n + gamma + 1/n [follows easily from Hardy and Wright, Th. 422]. - David Applegate and N. J. A. Sloane, Oct 14 2008
G.f. for H(n): log(1-x)/(x-1). - Benoit Cloitre, Jun 15 2003
H(n) = sqrt(Sum_{i = 1..n} Sum_{j = 1..n} 1/(i*j)). - Alexander Adamchuk, Oct 24 2004
a(n) is the numerator of Gamma/n + Psi(1 + n)/n = Gamma + Psi(n), where Psi is the digamma function. - Artur Jasinski, Nov 02 2008
H(n) = 3/2 + 2*Sum_{k = 0..n-3} binomial(k+2, 2)/((n-2-k)*(n-1)*n), n > 1. - Gary Detlefs, Aug 02 2011
H(n) = (-1)^(n-1)*(n+1)*n*Sum_{k = 0..n-1} k!*Stirling2(n-1, k) * Stirling1(n+k+1,n+1)/(n+k+1)!. - Vladimir Kruchinin, Feb 05 2013
H(n) = n*Sum_{k = 0..n-1} (-1)^k*binomial(n-1,k)/(k+1)^2. (Wenchang Chu) - Gary Detlefs, Apr 13 2013
H(n) = (1/2)*Sum_{k = 1..n} (-1)^(k-1)*binomial(n,k)*binomial(n+k, k)/k. (H. W. Gould) - Gary Detlefs, Apr 13 2013
E.g.f. for H(n) = a(n)/A002805(n): (gamma + log(x) - Ei(-x)) * exp(x), where gamma is the Euler-Mascheroni constant, and Ei(x) is the exponential integral. - Vladimir Reshetnikov, Apr 24 2013
H(n) = residue((psi(-s)+gamma)^2/2, {s, n}) where psi is the digamma function and gamma is the Euler-Mascheroni constant. - Jean-François Alcover, Feb 19 2014
H(n) = Sum_{m >= 1} n/(m^2 + n*m) = gamma + digamma(1+n), numerators and denominators. (see Mathworld link on Digamma). - Richard R. Forberg, Jan 18 2015
H(n) = (1/2) Sum_{j >= 1} Sum_{k = 1..n} ((1 - 2*k + 2*n)/((-1 + k + j*n)*(k + j*n))) + log(n) + 1/(2*n). - Dimitri Papadopoulos, Jan 13 2016
H(n) = (n!)^2*Sum_{k = 1..n} 1/(k*(n-k)!*(n+k)!). - Vladimir Kruchinin, Mar 31 2016
a(n) = Stirling1(n+1, 2) / gcd(Stirling1(n+1, 2), n!) = A000254(n) / gcd(A000254(n), n!). - Max Alekseyev, Mar 01 2018
From Peter Bala, Jan 31 2019: (Start)
H(n) = 1 + (1 + 1/2)*(n-1)/(n+1) + (1/2 + 1/3)*(n-1)*(n-2)/((n+1)*(n+2)) + (1/3 + 1/4)*(n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) + ... .
H(n)/n  = 1 + (1/2^2 - 1)*(n-1)/(n+1) + (1/3^2 - 1/2^2)*(n-1)*(n-2)/((n+1)*(n+2)) + (1/4^2 - 1/3^2)*(n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) + ... .
For odd n >= 3, (1/2)*H((n-1)/2) = (n-1)/(n+1) + (1/2)*(n-1)*(n-3)/((n+1)*(n+3)) + 1/3*(n-1)*(n-3)*(n-5)/((n+1)*(n+3)*(n+5)) + ... . Cf. A195505. See the Bala link in A036970. (End)
H(n) = ((n-1)/2) * hypergeom([1,1,2-n], [2,3], 1) + 1. - Artur Jasinski, Jan 08 2021
Conjecture: for nonzero m, H(n) = (1/m)*Sum_{k = 1..n} ((-1)^(k+1)/k) * binomial(m*k,k)*binomial(n+(m-1)*k,n-k). The case m = 1 is well-known; the case m = 2 is given above by Detlefs (dated Apr 13 2013). - Peter Bala, Mar 04 2022
a(n) = the (reduced) numerator of the continued fraction 1/(1 - 1^2/(3 - 2^2/(5 - 3^2/(7 - ... - (n-1)^2/(2*n-1))))). - Peter Bala, Feb 18 2024
H(n) = Sum_{k=1..n} (-1)^(k-1)*binomial(n,k)/k (H. W. Gould). - Gary Detlefs, May 28 2024

Edited by Max Alekseyev, Oct 21 2011

Changed title, deleting the incorrect name "Wolstenholme numbers" which conflicted with the definition of the latter in both Weisstein's World of Mathematics and in Wikipedia, as well as with OEIS A007406. - Stanislav Sykora, Mar 25 2016

A002805 Denominators of harmonic numbers H(n) = Sum_{i=1..n} 1/i.

Original entry on oeis.org

1, 2, 6, 12, 60, 20, 140, 280, 2520, 2520, 27720, 27720, 360360, 360360, 360360, 720720, 12252240, 4084080, 77597520, 15519504, 5173168, 5173168, 118982864, 356948592, 8923714800, 8923714800, 80313433200, 80313433200, 2329089562800, 2329089562800, 72201776446800

Offset: 1

N. J. A. Sloane

H(n)/2 is the maximal distance that a stack of n cards can project beyond the edge of a table without toppling.
If n is not in {1, 2, 6} then a(n) has at least one prime factor other than 2 or 5. E.g., a(5) = 60 has a prime factor 3 and a(7) = 140 has a prime factor 7. This implies that every H(n) = A001008(n)/A002805(n), n not from {1, 2, 6}, has an infinite decimal representation. For a proof see the J. Havil reference. - Wolfdieter Lang, Jun 29 2007
a(n) = A213999(n,n-1). - Reinhard Zumkeller, Jul 03 2012
From Wolfdieter Lang, Apr 16 2015: (Start)
a(n)/A001008(n) = 1/H(n) is the solution of the following version of the classical cistern and pipes problem. A cistern is connected to n different pipes of water. For the k-th pipe it takes k time units (say, days) to fill the empty cistern, for k = 1, 2, ..., n. How long does it take for the n pipes together to fill the empty cistern? 1/H(n) gives the answer as a fraction of the time unit.
In general, if the k-th pipe needs d(k) days to fill the empty cistern then all pipes together need 1/Sum_{k=1..n} 1/d(k) = HM(d(1), ..., d(n))/n days, where HM denotes the harmonic mean HM. For the described problem, HM(1, 2, ..., n)/n = A102928(n)/(n*A175441(n)) = 1/H(n).
For a classical cistern and pipes problem see, e.g., the Hunger-Vogel reference (in Greek and German) given in A256101, problem 27, p. 29, where n = 3, and d(1), d(2) and d(3) are 6, 4 and 3 days. On p. 97 of this reference one finds remarks on the history of such problems (called in German 'Brunnenaufgabe'). (End)
From Wolfdieter Lang, Apr 17 2015: (Start)
An example of the above mentioned cistern and pipes problems appears in Chiu Chang Suan Shu (nine books on arithmetic) in book VI, problem 26. The numbers are there 1/2, 1, 5/2, 3 and 5 (days) and the result is 15/75 (day). See the reference (in German) on p. 68.
A historical account on such cistern problems is found in the Johannes Tropfke reference, given in A256101, section 4.2.1.2 Zisternenprobleme (Leistungsprobleme), pp. 578-579.
In Fibonacci's Liber Abaci such problems appear on p. 281 and p. 284 of L. E. Sigler's translation. (End)
All terms > 1 are even while corresponding numerators (A001008) are all odd (proof in Pólya and Szegő). - Bernard Schott, Dec 24 2021

			H(n) = [ 1, 3/2, 11/6, 25/12, 137/60, 49/20, 363/140, 761/280, 7129/2520, ... ] = A001008/A002805.

Chiu Chang Suan Shu, Neun Bücher arithmetischer Technik, translated and commented by Kurt Vogel, Ostwalds Klassiker der exakten Wissenschaften, Band 4, Friedr. Vieweg & Sohn, Braunschweig, 1968, p. 68.
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 258-261.
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 259.
J. Havil, Gamma, (in German), Springer, 2007, p. 35-6; Gamma: Exploring Euler's Constant, Princeton Univ. Press, 2003.
D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, Vol. 1, p. 615.
G. Pólya and G. Szegő, Problems and Theorems in Analysis, volume II, Springer, reprint of the 1976 edition, 1998, problem 251, p. 154.
L. E. Sigler, Fibonacci's Liber Abaci, Springer, 2003, pp. 281, 284.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Kenny Lau, Table of n, a(n) for n = 1..2308 [First 200 terms computed by T. D. Noe]
R. M. Dickau, Harmonic numbers and the book stacking problem.
Haight, Frank A., and Robert B. Jones., "A probabilistic treatment of qualitative data with special reference to word association tests." Journal of Mathematical Psychology 11.3 (1974): 237-244. [Annotated scanned copy]
Frank Haight and N. J. A. Sloane, Correspondence, 1975.
Antal Iványi, Leader election in synchronous networks, Acta Univ. Sapientiae, Mathematica, 5, 2 (2013) 54-82.
Fredrik Johansson, How (not) to compute harmonic numbers, Feb 21 2009.
Peter Shiu, The denominators of harmonic numbers, arXiv:1607.02863 [math.NT], 2016.
N. J. A. Sloane, Illustration of initial terms.
Jonathan Sondow and Eric W. Weisstein, MathWorld: Harmonic Number.
Eric Weisstein's World of Mathematics, Book Stacking Problem.
Bing-Ling Wu, Yong-Gao Chen, On the denominators of harmonic numbers, arXiv:1711.00184 [math.NT], 2017.

Cf. A001008 (numerators), A075135, A025529, A203810, A203811, A203812.
Partial sums: A027612/A027611 = 1, 5/2, 13/3, 77/12, 87/10, 223/20,...
The following fractions are all related to each other: Sum 1/n: A001008/A002805, Sum 1/prime(n): A024451/A002110 and A106830/A034386, Sum 1/nonprime(n): A282511/A282512, Sum 1/composite(n): A250133/A296358, Sum 1/n^2: A007406/A007407, Sum 1/n^3: A007408/A007409.

List([1..30],n->DenominatorRat(Sum([1..n],i->1/i))); # Muniru A Asiru, Dec 20 2018

import Data.Ratio ((%), denominator)
a002805 = denominator . sum . map (1 %) . enumFromTo 1
a002805_list = map denominator $ scanl1 (+) $ map (1 %) [1..]
-- Reinhard Zumkeller, Jul 03 2012

[Denominator(HarmonicNumber(n)): n in [1..40]]; // Vincenzo Librandi, Apr 16 2015

seq(denom(sum((2*k-1)/k, k=1..n), n=1..30); # Gary Detlefs, Jul 18 2011
f:=n->denom(add(1/k, k=1..n)); # N. J. A. Sloane, Nov 15 2013

Denominator[ Drop[ FoldList[ #1 + 1/#2 &, 0, Range[ 30 ] ], 1 ] ] (* Harvey P. Dale, Feb 09 2000 *)
Table[Denominator[HarmonicNumber[n]], {n, 1, 40}] (* Stefan Steinerberger, Apr 20 2006 *)
Denominator[Accumulate[1/Range[25]]] (* Alonso del Arte, Nov 21 2018 *)

a(n)=denominator(sum(k=2,n,1/k)) \\ Charles R Greathouse IV, Feb 11 2011

from fractions import Fraction
def a(n): return sum(Fraction(1, i) for i in range(1, n+1)).denominator
print([a(n) for n in range(1, 30)]) # Michael S. Branicky, Dec 24 2021

def harmonic(a, b): # See the F. Johansson link.
    if b - a == 1 : return 1, a
    m = (a+b)//2
    p, q = harmonic(a,m)
    r, s = harmonic(m,b)
    return p*s+q*r, q*s
def A002805(n) : H = harmonic(1,n+1); return denominator(H[0]/H[1])
[A002805(n) for n in (1..29)] # Peter Luschny, Sep 01 2012

a(n) = Denominator(Sum_{k=1..n} (2*k-1)/k). - Gary Detlefs, Jul 18 2011
a(n) = n! / gcd(Stirling1(n+1, 2), n!) = n! / gcd(A000254(n),n!). - Max Alekseyev, Mar 01 2018
a(n) = the (reduced) denominator of the continued fraction 1/(1 - 1^2/(3 - 2^2/(5 - 3^2/(7 - ... - (n-1)^2/(2*n-1))))). - Peter Bala, Feb 18 2024

Definition edited by Daniel Forgues, May 19 2010

A102928 Numerator of the harmonic mean of the first n positive integers.

Original entry on oeis.org

1, 4, 18, 48, 300, 120, 980, 2240, 22680, 25200, 304920, 332640, 4684680, 5045040, 5405400, 11531520, 208288080, 73513440, 1474352880, 62078016, 108636528, 113809696, 2736605872, 8566766208, 223092870000, 232016584800

Offset: 1

Eric W. Weisstein, Jan 19 2005

See A175441 - denominators of the harmonic means of the first n positive integers. - Jaroslav Krizek, May 16 2010
a(n) is also the denominator of H(n-1)/n + 1/n^2 = -Integral_{x=0..1} x^n*log(1-x) with H(n) = A001008(n)/A002805(n) the harmonic number of order n. - Groux Roland, Jan 08 2011 [corrected by Gary Detlefs, Oct 06 2011]
Equivalently, a(n) is the reduced denominator of the arithmetic mean of the reciprocals of the first n positive integers (corresponding reduced numerator is A175441(n)). - Rick L. Shepherd, Jun 15 2014
n divides a(n) iff n is not from the sequence A256102. - Wolfdieter Lang, Apr 23 2015

			1, 4/3, 18/11, 48/25, 300/137, 120/49, 980/363, 2240/761, ...
Division property: The first n not dividing a(n) is 20 because 20 = A256102(1). Indeed, a(20) = 62078016. - _Wolfdieter Lang_, Apr 23 2015

Stefano Spezia, Table of n, a(n) for n = 1..2000
Eric Weisstein's World of Mathematics, Harmonic Mean

Cf. A001008, A175441, A256102.

Table[Numerator[n/HarmonicNumber[n]], {n, 26}]

a(n) = numerator(n/sum(k=1, n, 1/k)); \\ Michel Marcus, Jul 29 2022

a(n) = denominator(EulerGamma/n + PolyGamma(0, 1 + n)/n). - Artur Jasinski, Nov 02 2008
a(n) = numerator(n/H(n)), where H(n) is the n-th harmonic number. - Gary Detlefs, Sep 10 2011
a(n) = denominator((1/n)*Sum_{k=1..n} k + 1/k). - Stefano Spezia, Jul 27 2022
a(n) = denominator(Sum_{k>0} 1/(k*(k+n))). - Mohammed Yaseen, Jun 23 2024

Definition edited by N. J. A. Sloane, Jan 24 2024

A256102 Numbers m such that gcd(A001008(m), m) > 1, in increasing order.

Original entry on oeis.org

20, 42, 77, 110, 156, 272, 342, 506, 812, 930, 1247, 1332, 1640, 1806, 2162, 2756, 3422, 3660, 4422, 4970, 5256, 6162, 6806, 7832, 9312, 9328, 10100, 10506, 11342, 11772, 12656, 16002, 17030, 18632, 19182, 22052, 22650, 24492, 26406, 27722, 29756, 31862, 32580, 36290, 37056, 38612, 39402

Offset: 1

Wolfdieter Lang, Apr 16 2015

For the corresponding values of GCD(A001008(a(n)), a(n)) see A256103(n).
A001008(a(n))/A175441(a(n)) = A256103(n), n >= 1.
This means that for all values n not in the present sequence the numerator of the harmonic sum (HS) of the first n positive integers coincides with the denominator of the harmonic mean (HM) of the first n positive integers. That is, n divides the HM(n) numerator A102928(n) for n not in the present sequence.
Of course, HS(n)*HM(n) = n, n >= 1, where HS(n) = A001008(n)/A002805(n) and HM(n) = A102928(n)/A175441(n).
All terms are composite. Sequences contains all numbers of the form p*(p - 1), where p is a prime >= 5. This is because p^2 divides numerator(Sum_{i=1..p-1} 1/(k*p + i)), and p divides numerator(Sum_{i=1..p-1} 1/(i*p)), so p divides numerator(Sum_{i=1..p*(p-1)} 1/i). - Jianing Song, Dec 24 2018

			n = 1: gcd(A001008(20), 20) = gcd(55835135, 20) = 5 = A256103(1) > 1.
A001008(20)/A175441(20) = 55835135/11167027 = 5 = A256103(1).
Because 19 is not in this sequence 1 = gcd(A001008(19), 19) = gcd(275295799, 19).

Amiram Eldar, Table of n, a(n) for n = 1..500

Cf. A256103, A001008, A002805, A102928, A175441.

Select[Range[10^4], !CoprimeQ[#, Numerator @ HarmonicNumber[#]] &] (* Amiram Eldar, Feb 24 2020 *)

a(n) is the n-th smallest element of the set M:= {m positive inter | gcd(A001008(m), m) > 1}, n >= 1.

A232193 Numerators of the expected value of the length of a random cycle in a random n-permutation.

Original entry on oeis.org

1, 3, 23, 55, 1901, 4277, 198721, 16083, 14097247, 4325321, 2132509567, 4527766399, 13064406523627, 905730205, 13325653738373, 362555126427073, 14845854129333883, 57424625956493833, 333374427829017307697, 922050973293317, 236387355420350878139797

Offset: 1

Geoffrey Critzer, Nov 20 2013

In this experiment we randomly select (uniform distribution) an n-permutation and then randomly select one of the cycles from that permutation. Cf. A102928/A175441 which gives the expected cycle length when we simply randomly select a cycle.

			Expectations for n=1,... are 1/1, 3/2, 23/12, 55/24, 1901/720, 4277/1440, 198721/60480, 16083/4480, ... = A232193/A232248
For n=3 there are 6 permutations.  We have probability 1/6 of selecting (1)(2)(3) and the cycle size is 1.  We have probability 3/6 of selecting a permutation with cycle type (1)(23) and (on average) the cycle length is 3/2.  We have probability 2/6 of selecting a permutation of the form (123) and the cycle size is 3.  1/6*1 + 3/6*3/2 + 2/6*3 = 23/12.

Alois P. Heinz, Table of n, a(n) for n = 1..250

Denominators are A232248.
Cf. A028417(n)/n! the expected value of the length of the shortest cycle in a random n-permutation.
Cf. A028418(n)/n! the expected value of the length of the longest cycle in a random n-permutation.

with(combinat):
b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0,
      expand(add(multinomial(n, n-i*j, i$j)/j!*(i-1)!^j
      *b(n-i*j, i-1) *x^j, j=0..n/i))))
    end:
a:= n->numer((p->add(coeff(p, x, i)/i, i=1..n))(b(n$2))/(n-1)!):
seq(a(n), n=1..30);  # Alois P. Heinz, Nov 21 2013
# second Maple program:
a:= n-> numer(add(abs(combinat[stirling1](n, i))/i, i=1..n)/(n-1)!):
seq(a(n), n=1..30);  # Alois P. Heinz, Nov 23 2013

Table[Numerator[Total[Map[Total[#]!/Product[#[[i]],{i,1,Length[#]}]/Apply[Times,Table[Count[#,k]!,{k,1,Max[#]}]]/(Total[#]-1)!/Length[#]&,Partitions[n]]]],{n,1,25}]

a(n) = Numerator( 1/(n-1)! * Sum_{i=1..n} A132393(n,i)/i ). - Alois P. Heinz, Nov 23 2013

a(n) = numerator(Sum_{k=0..n} A002657(k)/A091137(k)) (conjectured). - Michel Marcus, Jul 19 2019

A256103 a(n) = gcd(A001008(m(n)), m(n)), with m(n) = A256102(n), n >= 1.

Original entry on oeis.org

5, 7, 11, 11, 13, 17, 19, 23, 29, 31, 43, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 11, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223

Offset: 1

Wolfdieter Lang, Apr 16 2015

See A256102. The entry a(n) gives the quotient of the numerator of the harmonic sum of the first A256102(n) positive integers and the denominator of the harmonic mean of the same numbers. For each positive integer values m not from A256102 this quotient is 1.

			n = 1: gcd(A001008(20), 20) = gcd(55835135, 20) = 5. A001008(20)/A175441(20) = 55835135/11167027 = 5.
Because 19 is not from A256102 one has A001008(19) = A175441(19) = 275295799.

Cf. A256102, A002805, A001008, A175441.

a(n) = gcd(A001008(m(n)), m(n)), with m(n) = A256102(n), n >= 1.

a(n) = A001008(m(n))/A175441(m(n)), with m(n) = A256103(n), n >= 1.

A368372 a(n) = numerator of AM(n)-HM(n), where AM(n) and HM(n) are the arithmetic and harmonic means of the first n positive integers.

Original entry on oeis.org

0, 1, 4, 29, 111, 103, 472, 2369, 12965, 30791, 197346, 452993, 3337271, 7485915, 4160656, 18358463, 170991927, 124184839, 1278605110, 110351535, 98802055, 211524139, 2595194516, 16562041459, 219589922071, 464651871609, 2207044831642, 4649180818987, 70862100349605, 148699793966557

Offset: 1

N. J. A. Sloane, Jan 24 2024

			0, 1/6, 4/11, 29/50, 111/137, 103/98, 472/363, 2369/1522, 12965/7129, 30791/14762, 197346/83711, 452993/172042, 3337271/1145993, 7485915/2343466, 4160656/1195757, 18358463/4873118, ...

Paolo Xausa, Table of n, a(n) for n = 1..2000

Cf. A102928/A175441, A368366, A368373.

AM:=proc(n) local i; (add(i,i=1..n)/n); end;
HM:=proc(n) local i; (add(1/i,i=1..n)/n)^(-1); end;
s1:=[seq(AM(n)-HM(n),n=1..50)];

A368372[n_] := Numerator[(n+1)/2 - n/HarmonicNumber[n]];
Array[A368372, 35] (* Paolo Xausa, Jan 29 2024 *)

a368372(n) = numerator((n+1)/2 - n/harmonic(n)) \\ Hugo Pfoertner, Jan 25 2024

from fractions import Fraction
from itertools import count, islice
def agen(): # generator of terms
    A = H = 0
    for n in count(1):
        A += n
        H += Fraction(1, n)
        yield ((A*Fraction(1, n) - n/H)).numerator
print(list(islice(agen(), 30))) # Michael S. Branicky, Jan 24 2024

from fractions import Fraction
from sympy import harmonic
def A368372(n): return (Fraction(n+1,2)-Fraction(n,harmonic(n))).numerator # Chai Wah Wu, Jan 25 2024

A368373 a(n) = denominator of AM(n)-HM(n), where AM(n) and HM(n) are the arithmetic and harmonic means of the first n positive integers.

Original entry on oeis.org

1, 6, 11, 50, 137, 98, 363, 1522, 7129, 14762, 83711, 172042, 1145993, 2343466, 1195757, 4873118, 42142223, 28548602, 275295799, 22334054, 18858053, 38186394, 444316699, 2695645910, 34052522467, 68791484534, 312536252003, 630809177806, 9227046511387, 18609365660294, 290774257297357

Offset: 1

N. J. A. Sloane, Jan 24 2024

			0, 1/6, 4/11, 29/50, 111/137, 103/98, 472/363, 2369/1522, 12965/7129, 30791/14762, 197346/83711, 452993/172042, 3337271/1145993, 7485915/2343466, 4160656/1195757, 18358463/4873118, ...

Paolo Xausa, Table of n, a(n) for n = 1..2000

Cf. A102928/A175441, A368366, A368372.

AM:=proc(n) local i; (add(i,i=1..n)/n); end;
HM:=proc(n) local i; (add(1/i,i=1..n)/n)^(-1); end;
s1:=[seq(AM(n)-HM(n),n=1..50)];

A368373[n_] := Denominator[(n+1)/2 - n/HarmonicNumber[n]];
Array[A368373, 35] (* Paolo Xausa, Jan 29 2024 *)

a368373(n) = denominator((n+1)/2 - n/harmonic(n)) \\ Hugo Pfoertner, Jan 25 2024

from fractions import Fraction
from itertools import count, islice
def agen(): # generator of terms
    A = H = 0
    for n in count(1):
        A += n
        H += Fraction(1, n)
        yield ((A*Fraction(1, n) - n/H)).denominator
print(list(islice(agen(), 31))) # Michael S. Branicky, Jan 24 2024

from fractions import Fraction
from sympy import harmonic
def A368373(n): return (Fraction(n+1,2)-Fraction(n,harmonic(n))).denominator # Chai Wah Wu, Jan 25 2024

cp's OEIS Frontend

A001008 a(n) = numerator of harmonic number H(n) = Sum_{i=1..n} 1/i.

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A002805 Denominators of harmonic numbers H(n) = Sum_{i=1..n} 1/i.

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A102928 Numerator of the harmonic mean of the first n positive integers.

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A256102 Numbers m such that gcd(A001008(m), m) > 1, in increasing order.

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A232193 Numerators of the expected value of the length of a random cycle in a random n-permutation.

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A256103 a(n) = gcd(A001008(m(n)), m(n)), with m(n) = A256102(n), n >= 1.

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